Outer joins are equijoins and?
Manual says Yes, but I think their only equijoins, insofar as the matched data and include other data as well. so technically should not fall outside the category of the equijoins?
Hello
Outer joins can be are equijoins. In other words, some outer joins are of equijoins, others are not.
Tags: Database
Similar Questions
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outer join on the aggregate query
This is probably a relatively simple matter, as long as I explain it well enough:
I have two tables:
categorycodes and properties
categorycodes is a lookup table.
both tables have a catcode field which is a char (1) that contains matching data (only the numbers 1 to 6)
CREATE
TABLE CATEGORYCODES
(
CATCODE CHAR (1 BYTE) NOT NULL,
DESCRIPTION VARCHAR2 (25 BYTE) NOT NULL,
CONSTRAINT CATEGORYCODES_PK PRIMARY KEY (CATCODE) ALLOW
)
catCode
1
2
3
4
5
6
The properties table has approximately 600 000 records. The properties table also has a field named parcelno which is a tank (9). It contains a string of figures and numbers only.
What I would like is:
catCode, count (*)
1 580
2 300
3 3000
4 235
5 0
6 80
I limited the results of the query to make sure it was a game that would not all catcodes in it. I have trouble to get the one with zero to display. I know that this has to do with how I do the join, but I don't know what.
It is a sample of what I've tried:
Select i.
Of
(select catcode, count (*)
property p
where substr (parcelno, 1, 3) = ' 871 "
Catcode group) i
outer join right categorycodes cc
We i.catcode = cc.catcode;
I'm not worried about the situations where catcode is null in the properties. Parcelno cannot be null.
Hello
Looks like your query should work; except that you won't COUNT (*); That would make each issue at least 1. COUNT (*) means that count the total number of lines, no matter what is on them, so he'll see the line with just the catcode of the lookup table that matches nothing and which count as 1. You want to count the number of rows in the table of properties, so expect a column of the properties that cannot be NULL.
Here is a slightly different way
SELECT c.catcode
EARL of (p.catcode) AS cnt
OF categorycodes c
P ON p.catcode = c.catcode LEFT OUTER JOIN properties
AND SUBSTR (p.parcelno
1
3
) = ' 871 "
;
If the condition about 871' ' part of the join condition, then you don't need a subquery.
.
I hope that answers your question.
If not, post a small example data (CREATE TABLE and only relevant columns, INSERT statements) for all of the tables involved and also publish outcomes from these data.
Point where the above statement is erroneous results, and explain, using specific examples, how you get the right result of data provided in these places.Always say what version of Oracle you are using (for example, 11.2.0.2.0).
See the FAQ forum: https://forums.oracle.com/message/9362002
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Hello
I want a report that shows data for all combinations of dimensions, although in fact no data exists for them.
http://download.Oracle.com/docs/CD/E11882_01/server.112/e10578/tdpdw_sql.htm#TDPDW0072
Here's my data model, inventory with a status as 'sold out' or 'available' and the designer of the elementOracle Database 11g Enterprise Edition Release 11.2.0.2.0 - 64bit Production
and some examples of dataDROP TABLE inventory / DROP TABLE inv_status / DROP TABLE designer / CREATE TABLE inv_status( id NUMBER PRIMARY KEY ,description VARCHAR2(12) ) / CREATE TABLE designer( id NUMBER PRIMARY KEY ,name VARCHAR2(12) ) / CREATE TABLE inventory( id NUMBER PRIMARY KEY ,fk_ist NUMBER CONSTRAINT fk_inv_status REFERENCES inv_status (id) ,fk_des NUMBER CONSTRAINT fk_designer REFERENCES designer (id) ,name VARCHAR2(20) ,description VARCHAR2(12) ) /
I need a simple outer join to find each status as well as the objects (if available)INSERT INTO inv_status (id,description) VALUES (25,'sold out'); INSERT INTO inv_status (id,description) VALUES (26,'available'); INSERT INTO inv_status (id,description) VALUES (27,'comming soon'); INSERT INTO inv_status (id,description) VALUES (28,'not in use'); INSERT INTO designer (id,name) VALUES (111,'Marco'); INSERT INTO designer (id,name) VALUES (112,'Tommy'); INSERT INTO designer (id,name) VALUES (113,'Daisy'); INSERT INTO designer (id,name) VALUES (114,'Laura'); INSERT INTO inventory (id,fk_ist,fk_des,name,description) VALUES (1,25,112,'moon boot','k'); INSERT INTO inventory (id,fk_ist,fk_des,name,description) VALUES (2,25,113,'high heel','r'); INSERT INTO inventory (id,fk_ist,fk_des,name,description) VALUES (3,26,114,'sandal','f'); INSERT INTO inventory (id,fk_ist,fk_des,name,description) VALUES (4,26,113,'flip-flop','u'); INSERT INTO inventory (id,fk_ist,fk_des,name,description) VALUES (5,27,112,'horseshoe','j'); INSERT INTO inventory (id,fk_ist,fk_des,name,description) VALUES (6,27,114,'magic pair of boots','o'); INSERT INTO inventory (id,fk_ist,fk_des,name,description) VALUES (7,27,113,'runner','r'); INSERT INTO inventory (id,fk_ist,fk_des,name,description) VALUES (8,27,112,'loafer',NULL); INSERT INTO inventory (id,fk_ist,fk_des,name,description) VALUES (9,27,113,'climbing spur',NULL); COMMIT;
The same I get with a left join partitionSELECT ist.id ist ,inv.id ,inv.name ,ist.description FROM inv_status ist LEFT JOIN inventory inv ON (ist.id = inv.fk_ist) ORDER BY ist.id, inv.id; /* IST ID NAME DESCRIPTION --- --- -------------------- ------------ 25 1 moon boot sold out 25 2 high heel sold out 26 3 sandal available 26 4 flip-flop available 27 5 horseshoe comming soon 27 6 magic pair of boots comming soon 27 7 runner comming soon 27 8 loafer comming soon 27 9 climbing spur comming soon 28 not in use 10 rows selected */
But what is the right partition join? Why do I have multiple lines for each State where no items exist? Or in other words: someone can tell me what I asked of the DB and it makes sense to ask him: -?SELECT ist.id ist ,inv.id ,inv.name ,ist.description FROM inv_status ist PARTITION BY (ist.id) LEFT JOIN inventory inv ON (ist.id = inv.fk_ist) ORDER BY ist.id, inv.id; /* IST ID NAME DESCRIPTION --- --- -------------------- ------------ 25 1 moon boot sold out 25 2 high heel sold out 26 3 sandal available 26 4 flip-flop available 27 5 horseshoe comming soon 27 6 magic pair of boots comming soon 27 7 runner comming soon 27 8 loafer comming soon 27 9 climbing spur comming soon 28 not in use 10 rows selected */
For any combination of x as well as objects Designer item (if available) again a simple Cartesian product is easy to do.SELECT ist.id ist ,inv.id ,inv.name ,ist.description FROM inventory inv PARTITION BY (inv.fk_ist) RIGHT JOIN inv_status ist ON (ist.id = inv.fk_ist) ORDER BY ist.id, inv.id; /* IST ID NAME DESCRIPTION --- --- -------------------- ------------ 25 1 moon boot sold out 25 2 high heel sold out 25 sold out 25 sold out 26 3 sandal available 26 4 flip-flop available 26 available 26 available 27 5 horseshoe comming soon 27 6 magic pair of boots comming soon 27 7 runner comming soon 27 8 loafer comming soon 27 9 climbing spur comming soon 27 comming soon 27 comming soon 28 not in use 28 not in use 28 not in use 18 rows selected */
But I just can't get with the join of LEFT or RIGHT partitionSELECT ist.id ist ,des.id des ,inv.id ,inv.name ,ist.description ,des.name designer FROM inv_status ist CROSS JOIN designer des LEFT JOIN inventory inv ON (ist.id = inv.fk_ist AND des.id = inv.fk_des ) ORDER BY ist.id,des.id,inv.id; /* IST DES ID NAME DESCRIPTION DESIGNER --- --- --- -------------------- ------------ ------------ 25 111 sold out Marco 25 112 1 moon boot sold out Tommy 25 113 2 high heel sold out Daisy 25 114 sold out Laura 26 111 available Marco 26 112 available Tommy 26 113 4 flip-flop available Daisy 26 114 3 sandal available Laura 27 111 comming soon Marco 27 112 5 horseshoe comming soon Tommy 27 112 8 loafer comming soon Tommy 27 113 7 runner comming soon Daisy 27 113 9 climbing spur comming soon Daisy 27 114 6 magic pair of boots comming soon Laura 28 111 not in use Marco 28 112 not in use Tommy 28 113 not in use Daisy 28 114 not in use Laura 18 rows selected */SELECT inv_ist.ist ,des.id des ,inv_ist.id ,inv_ist.name ,inv_ist.description ,des.name designer FROM designer des PARTITION BY (des.id) LEFT JOIN ( SELECT ist.id ist ,inv.fk_des ,inv.id ,inv.name ,ist.description FROM inv_status ist PARTITION BY (ist.id) LEFT JOIN inventory inv ON (ist.id = inv.fk_ist) ) inv_ist ON (des.id = inv_ist.fk_des OR inv_ist.fk_des IS NULL) ORDER BY inv_ist.ist,des.id,inv_ist.id; /* IST DES ID NAME DESCRIPTION DESIGNER --- --- --- -------------------- ------------ ------------ 25 112 1 moon boot sold out Tommy 25 113 2 high heel sold out Daisy 26 113 4 flip-flop available Daisy 26 114 3 sandal available Laura 27 112 5 horseshoe comming soon Tommy 27 112 8 loafer comming soon Tommy 27 113 7 runner comming soon Daisy 27 113 9 climbing spur comming soon Daisy 27 114 6 magic pair of boots comming soon Laura 28 111 not in use Marco 28 112 not in use Tommy 28 113 not in use Daisy 28 114 not in use Laura 13 rows selected */
Does anyone know a simple description, something like 'Partition of outer join for DummiesSELECT inv_ist.ist ,des.id des ,inv_ist.id ,inv_ist.name ,inv_ist.description ,des.name designer FROM ( SELECT ist.id ist ,inv.fk_des ,inv.id ,inv.name ,ist.description FROM inventory inv PARTITION BY (inv.fk_ist) RIGHT JOIN inv_status ist ON (ist.id = inv.fk_ist) ) inv_ist PARTITION BY (inv_ist.fk_des) RIGHT JOIN designer des ON (des.id = inv_ist.fk_des) ORDER BY inv_ist.ist,des.id,inv_ist.id; /* IST DES ID NAME DESCRIPTION DESIGNER --- --- --- -------------------- ------------ ------------ 25 112 1 moon boot sold out Tommy 25 113 2 high heel sold out Daisy 26 113 4 flip-flop available Daisy 26 114 3 sandal available Laura 27 112 5 horseshoe comming soon Tommy 27 112 8 loafer comming soon Tommy 27 113 7 runner comming soon Daisy 27 113 9 climbing spur comming soon Daisy 27 114 6 magic pair of boots comming soon Laura 111 Marco 111 Marco 111 Marco 111 Marco 112 Tommy 112 Tommy 112 Tommy 113 Daisy 113 Daisy 113 Daisy 114 Laura 114 Laura 114 Laura 22 rows selected */
Concerning
Marcus
Edited by: Marwim the 28.12.2010 09:11
TypoHey, Marcus,
Marwim wrote:
And that's exactly my problem: what kind of problems can I solve with outer joins partitioned?Programming would be much easier (and therefore less interesting and less well-paid) if it has nice, short answers to these questions.
According to my experience, cross joins are the best when you dimension tables and partitioned outer joins are good when you don't.
Cionsider the following partitioned outer join, which shows all the possible combinations of the job and the Department:SELECT d.dname , e.deptno , e.job , e.ename FROM scott.dept d LEFT OUTER JOIN scott.emp e PARTITION BY (e.job) ON d.deptno = e.deptno ORDER BY d.deptno , e.job ;Output:
DNAME DEPTNO JOB ENAME -------------- ---------- --------- --------- ACCOUNTING ANALYST ACCOUNTING 10 CLERK MILLER ACCOUNTING 10 MANAGER CLARK ACCOUNTING 10 PRESIDENT KING ACCOUNTING SALESMAN ... OPERATIONS ANALYST OPERATIONS CLERK OPERATIONS MANAGER OPERATIONS PRESIDENT OPERATIONS SALESMANIt's a great use of a partitioned outer join, because there is no dimension table for employment (that is, there is no table with one line per job, which may be the target of a key foregin reference). You can use a join Cross (more an external koin) to get the same results, but it would be a subquery such as
WITH all_jobs AS ( SELECT DISTINCT job FROM scott.emp ) ...to create a table dimesnion to the fly.
There may be exceptions. For example, if you don't have a table, but there are some jobs that actually occur in the emp table, and you want the output to include only the jobs that exist in the emp table, then the easiest thing is just to ignore the jobs table and do the partitioned outer join illustrated above.
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Hello
I have create a join foreign key between the table dimension and in the physical layer. Is it possible to create an outer join on the columns, the option (driving and type) in properties is gray and is by default 'indoor '. How you can activate this option?
Thank youHello
You create outer joins in the business layer, but the OBIEE outer joins are not like SQL outer join. They can lead to strange errors.
Its better to do these things in ETL.
Thank you
Sandeep -
Show all dates between date range (time Dimension is left outer join)
All,
I did some research on this issue, but in all positions on date variables, date prompts and date filtering I have not seen one exactly to my question (perhaps that they are and I don't have my head around it properly yet).
My requirement of report is to allow a user to select a start date and an end of day. The report is expected to show the activity of these two days - AND display 0/null on days where there is no activity. This second part is where I am getting hung up.
The paintings in question are:
Timedim
EventFact
CustomerDim
My MDB is configured as follows:
Left outer join of Timedim EventFact
Inner join CustomerDim EventFact
If I run a report by selecting the DAYS of Timedim and an EventFact measure1 with range day 01/01/2010-31/12/2010... A record for each day and it looks perfect because of the left outer join between Timedim and CustomerDim.
But... If I add a CustomerDim field, Select TimeDim.DAY, CustomerDim.CUSTNAME, EventFact.MEASURE1, OBIEE returns only records for the days that have record EventFact.
This is due to the fact that the Timedim is always external joined in EventFact, but adding in fact CustomerDim OBIEE set up an inner join between tables that will only return data where there are data EventFact.
There is a way around it in this simple case, and that is to define the relationship between CustomerDim and EventFact as an outer join as well. This will give the desired effect (but an outer join between the two tables is not the real relationship) and I have add an extra dimension and add additional sources of logic to a single dimension in MDB it becomes complicated and messy.
Also, ive ruined with the definition of the conduct in the relationship table, etc... but he gave not the desired effect.
Has anyone ever met the need for force display all dates within a range specified with a fact table that does not have an entry for each date?
Thanks in advance.
K
Published by: user_K on April 27, 2010 11:32Hi there, the easiest way is to the LTS himself. Double-click your LTS, go to the tab with the column mappings. Make sure you have checked "show no mapped" column.
You should see your new dummy column in the list, in the central part of mapping (IE not the right) just enter 0, or launch the expression editor and enter 0 in there.
simple! -
[8i] need help with full outer join combined with a cross join...
I can't understand how to combine a full outer join with a different type of join... is it possible?
Here are some create table and insert for examples of database:
And, the results that I want to get:CREATE TABLE my_tab1 ( record_id NUMBER NOT NULL , workstation VARCHAR2(4) , my_value NUMBER CONSTRAINT my_tab1_pk PRIMARY KEY (record_id) ); INSERT INTO my_tab1 VALUES(1,'ABCD',10); INSERT INTO my_tab1 VALUES(2,'ABCD',15); INSERT INTO my_tab1 VALUES(3,'ABCD',5); INSERT INTO my_tab1 VALUES(4,'A123',5); INSERT INTO my_tab1 VALUES(5,'A123',10); INSERT INTO my_tab1 VALUES(6,'A123',20); INSERT INTO my_tab1 VALUES(7,'????',5); CREATE TABLE my_tab2 ( workstation VARCHAR2(4) , wkstn_name VARCHAR2(20) CONSTRAINT my_tab2_pk PRIMARY KEY (workstation) ); INSERT INTO my_tab2 VALUES('ABCD','WKSTN 1'); INSERT INTO my_tab2 VALUES('A123','WKSTN 2'); INSERT INTO my_tab2 VALUES('B456','WKSTN 3'); CREATE TABLE my_tab3 ( my_nbr1 NUMBER , my_nbr2 NUMBER ); INSERT INTO my_tab3 VALUES(1,2); INSERT INTO my_tab3 VALUES(2,3); INSERT INTO my_tab3 VALUES(3,4);
I tried a number of different things, google my problem and no luck yet...workstation sum(my_value) wkstn_name my_nbr1 my_nbr2 --------------------------------------------------------------- ABCD 30 WKSTN 1 1 2 ABCD 30 WKSTN 1 2 3 ABCD 30 WKSTN 1 3 4 A123 35 WKSTN 2 1 2 A123 35 WKSTN 2 2 3 A123 35 WKSTN 2 3 4 B456 0 WKSTN 3 1 2 B456 0 WKSTN 3 2 3 B456 0 WKSTN 3 3 4 ???? 5 NULL 1 2 ???? 5 NULL 2 3 ???? 5 NULL 3 4
So, what I want, it's a full outer join of t1 and t2 on workstation and a cross join of one with the t3. I wonder if I can't find examples of it online because it is not possible...SELECT t1.workstation , SUM(t1.my_value) , t2.wkstn_name , t3.my_nbr1 , t3.my_nbr2 FROM my_tab1 t1 , my_tab2 t2 , my_tab3 t3 ...
Note: I'm stuck dealing with Oracle 8i
Thank you!!Hello
The query I posted yesterday is a little more complex that it should be.
My_tab2.workstation is unique, there is no reason to make a separate subquery as mt1. We can join my_tab1 to my_tab2 and get the SUM in a subquery.SELECT foj.workstation , foj.sum_my_value , foj.wkstn_name , mt3.my_nbr1 , mt3.my_nbr2 FROM ( -- Begin in-line view foj for full outer join SELECT mt1.workstation , SUM (mt1.my_value) AS sum_my_value , mt2.wkstn_name FROM my_tab1 mt1 , my_tab2 mt2 WHERE mt1.workstation = mt2.workstation (+) GROUP BY mt1.workstation , mt2.wkstn_name -- UNION ALL -- SELECT workstation , 0 AS sum_my_value , wkstn_name FROM my_tab2 WHERE workstation NOT IN ( -- Begin NOT IN sub-query SELECT workstation FROM my_tab1 WHERE workstation IS NOT NULL ) -- End NOT IN sub-query ) foj -- End in-line view foj for full outer join , my_tab3 mt3 ORDER BY foj.wkstn_name , foj.workstation , mt3.my_nbr1 , mt3.my_nbr2 ;Thanks for posting the CREATE TABLE and INSERT statements, and very clear expected results!
user11033437 wrote:
... So, what I want, it's a full outer join of t1 and t2 on workstation and a cross join of one with the t3.She, exactly!
The trickiest part is when and how get SUM (my_value). You could address the question of exactly what my_tab3 must be attached to a cross that's exactly what should look like the result set of the full outer join between my_tab1 and my_tab2 to. To do this, take your desired results, remove columns that do not come from the outer join complete and delete duplicate rows. You will get:workstation sum(my_value) wkstn_name ----------- ------------- ---------- ABCD 30 WKSTN 1 A123 35 WKSTN 2 B456 0 WKSTN 3 ???? 5 NULLSo the heart of the problem is how to get these results of my_tab1 and my_tab2, which is done in the subquery FOJ above.
I tried to use auto-documenté in my code names. I hope you can understand.
I could spend hours explaining the different parts of this query more in detail, but I don't know that I would lose some of that time, explain things that you already understand. If you want an explanation of the specific element (s), let me know. -
Hello
Pls help me my request. I tried the following, but it does not give the expected results.
Tab1
EmplId, RepDt, Code, Hrs
1/100,1/2009,199,8
1/100,1/2009,200,4
1/100,1/2009,255,3
200,1/1/2009,200,4
100.5/1/2009,199,8
Tab2
EmplId, RepDt, Code, Hrs
1/100,1/2009,200,6
200,1/1/2009,200,3
I need output like this - 4 rows - need of all the rows from TAB1 when EMPLID, REPDT match TAB2, field values: need at a time when Code is also, when emlid, correspondence of the date, the Code missing so need A hrs, display ZERO as Tab2 hours.
1/100,1/2009,199,8,0 - did not exist is not in tab2, so hrs Tab2 is ZERO
1/100,1/2009,200,4,6 - Emplid, RepDt, Code match existence in Tab2, Tab1 Hrs 8, 6 Hrs Tab2
1/100,1/2009,255,3,0 - did not exist is not in Tab2, so Tab2 hrs is equal to ZERO
200,1/1/2009,200,4,3 - Emplid, Repdt, Code Match existed in Tab2, 4 Hrs of Tab1, Tab2 3 Hrs
We are in 10g, Oracle.
I tried the following
T1. EmplId = T2.emplid AND T1.repdt = T2.repdt AND T1.code, T2.code = (+) - returns the unique corresponding lines, 2 rows.
Pls help.
Thanks in advance.
Published by: NL 23 February 2009 09:20Hello
See the Boneist first message in this thread:
select t1.emplid, t1.repdt, t1.code, t1.hrs, nvl(t2.hrs, 0) from tab1 t1, tab2 t2 where t1.emplid = t2.emplid (+) and t1.repdt = t2.repdt (+) and t1.code = t2.code (+) and (t1.emplid, t1.repdt) in (select emplid, repdt from tab2) order by repdt, emplid, code;As Boneist said, this fact corresponds to option (b)
"(b) do just an outer join, with a condition EXISTS (or IN) in the WHERE clause to find corresponding repdts.
The(+)an outer join are signs.
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Outer join or a function of the level line
Hi all
I have a question which involves five tables T1, R1, R2, R3, and R4. T1 is an operating table, while R1, R2, R3, and R4 are tables of references (tables parent, with the foreign keys defined in T1). Table T1 contains always referenced data R2 and R1. BUT table T1 can sometimes contain NULL for R3 and R4.
Now my question is simple;
Should I use an OUTER Join for R3 and R4 in the query? as
< code >
Select T1.col1, R1.col2, R2.col2, R3.col2, R4.col2
T1, R1, R2, R3, R4
where T1.col2 = R1.col1
and T1.col3 = R2.col1
and T1.col4 = R3.col1 (+)
and T1.col5 = R4.col1 (+)
< code >
OR
Can I use level functions online for R3 and R4, as
< code >
Select T1.col1, R1.col2, R2.col2,
(Select R3.col2 in R3 where R3.col1 = T1.col4),
(Select R4.col2 in R4 where R4.col1 = T1.col5)
T1, R1, R2
where T1.col2 = R1.col1
and T1.col3 = R2.col1
< code >
Which approach is better and why?
Records in T1 = 2 000 000
R1 = 1000 records
Records in R2 = 300
Records in R3 = 1800
Records in R4 = 200
Please note that all foreign keys are indexed, there are primary keys for all the tables in R
Thank you
QQ.According to the 'official' documentation the outer join should be the by far best choice, because I still think the Oracle documentation says that the subquery will be executed for each row of the main query.
But the Oracle execution engine is a lot smarter (I think from Oracle 8i) and offers a "Value for the filter" feature supposedly which also applies to the "scalar subqueries" you are talking about to (but only if the subquery is considered to be deterministic, i.e. returns always the same value for the same input value of the main request).
Basically, this Treaty optimization the subquery as a function of an input value (the link to the main request) and an output value (the result of the query) and maintains a hash table in memory of pairs of input/output value. The size of the changes table in memory from one version to the other, I think in 9i, it is set by default to 256 entries but in 10g, it is limited in size and therefore the number of entries depends on the size of the values to store.
The effectiveness of this optimization of memory-search in the table depends on certain factors such as the number of pairs of distinct value, the order of the incoming values and collisions in the table which can be occur according to hash values.
Jonathan Lewis has an in-depth coverage of these mechanism in its "cost-based Oracle: Fundamentals" book.
To make a long story short, since your tables R3 and R4 are relatively low, you might want to give a chance, because these optimizations could work very well in your particular case, but unfortunately the real result is simply unpredictable due to the above constraints, including the order of the incoming values.
Kind regards
RandolfOracle related blog stuff:
http://Oracle-Randolf.blogspot.com/SQLTools ++ for Oracle (Open source Oracle GUI for Windows):
http://www.sqltools-plusplus.org:7676 /.
http://sourceforge.NET/projects/SQLT-pp/ -
OUTER JOIN query returns the results of JOIN IN-HOUSE 11.2.0.1.0
I'm data transfer in 11.2.01.0 (Windows XP 32-bit) and I wanted to compare the sizes of table with the same Table name in two different patterns, ML and SILENT, with a FULL OUTER JOIN (to account for all the tables and NULL values in a diagram).
The scheme of ML has 176 tables: schema TUT a 133 tables. The use of a standard INNER JOIN gives 131 paintings.
I get precisely the results with a FULL OUTER JOIN I get with an INTERNAL JOIN (not the same NULL values so I know they exist).
This happens in SQL-Plus, SQL_Developer and using Oracle Wire pilot of Data_Direct.
Here is the code:
Login: SYS as SYSDBA or SYSTEM (same results for either)
SELECT M.TABLE_NAME, M.NUM_ROWS, T.TABLE_NAME, T.NUM_ROWS
OF SYS. ALL_TABLES M FULL OUTER JOIN SYS. ALL_TABLES T ON M.TABLE_NAME = T.TABLE_NAME
WHERE
M.OWNER = 'ML' AND
T.OWNER = 'TUT';
Produce the same results with LEFT OUTER joins and RIGHT OUTER joins in ASI and Oracle (+) syntax.
Any thoughts?Hello
If you read what I posted, forget it. MScallion (below) gave the correct answerr.
If conditions such as
owner = 'ML'in the WHERE clause, and then they will reject the rows of the result set formed by the join condition.
The inner join returns only 131 lines where the two 'paintings' have the same table_names.
The outer joins return multiple lines (133, 176 or 178) before the place WHERE the provision is applied , but the WHERE clause eliminates all lines except the 131 found by the inner join.Published by: Frank Kulash, July 10, 2010 14:23
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outer join when there are several tables are involved
Could not put up the question correctly to the last channel, my problem is with the join when there are several tables are involved, this is just one example of the task that I have to carry.
Tab1 aura model id retailer_id information for all the weeks (from the first Monday) of the month of JUNE with cost and Helen
Tab1
model_id
retailer_id
sell_date
cost
Helene
1
12
June 3, 13
100
40
1
12
June 10, 13
200
20
1
12
17 June 13
300
20
1
12
24 June 13
400
20
2
12
June 3, 13
300
10
2
12
June 10, 13
200
20
2
12
17 June 13
300
20
2
12
24 June 13
400
20
Tab2:
each retailer belongs to a dealer, under the table has the same information
retailer_id
Dealer_id
12
100
13
100
14
101
15
101
16
101
Tab 3
There is a third layer where each dealership is having a garage band
Dealer_id
Dealer_group
100
1001
101
1001
102
2001
103
2001
104
3001
105
3001
Tab4:
Of this table for each model and dealer discount information for the month of June (every week)
model_id
Dealer_group
discount_date
discount
1
1001
June 3, 13
10
1
1001
June 10, 13
20
1
1001
17 June 13
10
1
1001
24 June 13
30
2
1001
June 3, 13
10
2
1001
June 10, 13
20
2
1001
17 June 13
10
2
1001
24 June 13
30
3
2001
June 3, 13
10
3
2001
June 10, 13
20
3
2001
17 June 13
10
3
2001
24 June 13
30
Master_info:
It's the main table which is the master table for model /retailer information
Model_id
retailer_id
1
12
2
12
3
12
4
12
1
13
2
13
Output
model_id
retailer_id
sell_date
cost
Helene
Final (cost-helene-discount)
1
12
June 3, 13
100
40
50
1
12
June 10, 13
200
20
160
1
12
17 June 13
300
20
270
1
12
24 June 13
400
20
350
2
12
June 3, 13
300
10
280
2
12
June 10, 13
200
20
160
2
12
17 June 13
300
20
270
2
12
24 June 13
400
20
350
3
12
June 3, 13
0
0
0
3
12
June 10, 13
0
0
0
3
12
17 June 13
0
0
0
3
12
24 June 13
0
0
0
4
12
June 3, 13
0
0
0
4
12
June 10, 13
0
0
0
4
12
17 June 13
0
0
0
4
12
24 June 13
0
0
0
1
13
June 3, 13
0
0
0
1
13
June 10, 13
0
0
0
1
13
17 June 13
0
0
0
1
13
24 June 13
0
0
0
2
13
June 3, 13
0
0
0
2
13
June 10, 13
0
0
0
2
13
17 June 13
0
0
0
1
13
24 June 13
0
0
0
For highted above records (model_id / retailer_id combination) there is no record in tab1 but they have entered in master_info then the recordings should come for all model_id/retailer_id with all the 0 values
Hello
Thanks for posting the sample data.
It is unclear what dates you want to include in the output. The following query shows how you can generate every Monday in a given range. If you only want to include the dates that are actually present in tabl1 and/or tab4, you can simplify this a bit.
WITH date_range AS
(
SELECT TRUNC (TO_DATE (' 3 June 2013', 'DD-Mon-YYYY'))
, 'IW '.
) AS first_monday
, TRUNC (TO_DATE (24 June 2013 ', 'DD-Mon-YYYY') + 6)
, 'IW '.
) AS last_monday
OF the double
)
all_mondays AS
(
First_monday SELECT + (7 * (LEVEL - 1)) AS sell_date
OF date_range
CONNECT BY LEVEL<= 1="" +="" (="" (last_monday="" -="">
/ 7
)
)
SELECT mi.model_id
mi.retailer_id
am.sell_date
, Cost of NVL (t1.cost, 0) AS
, NVL (t1.rebat, 0) IN the refund
, NVL (t1.cost, 0)
-(NVL (t1.rebat, 0))
+ NVL (t4.discount, 0)
) AS final
E master_info
CROSS JOIN all_mondays am
LEFT OUTER JOIN tab1 t1 ON t1.model_id = mi.model_id
AND t1.retailer_id = mi.retailer_id
AND t1.sell_date = am.sell_date
LEFT OUTER JOIN tab2 t2 ON t2.retailer_id = mi.retailer_id
LEFT OUTER JOIN tab 3 t3 ON t3.dealer_id = t2.dealer_id
LEFT OUTER JOIN tab4 t4 ON t4.model_id = t1.model_id
AND t4.dealer_group = t3.dealer_group
AND t4.discount_date = t1.sell_date
ORDER BY mi.retailer_id
mi.model_id
am.sell_date
;
The results are not exactly what said you you wanted. I suspect it's because of typos in that you posted.
-
BAD RESULTS WITH OUTER JOINS AND TABLES WITH A CHECK CONSTRAINT
HII All,
Could any such a me when we encounter this bug? Please help me with a simple example so that I can search for them in my PB.
Bug:-8447623
Bug / / Desc: BAD RESULTS WITH OUTER JOINS AND TABLES WITH a CHECK CONSTRAINT
I ran the outer joins with check queries constraint 11G 11.1.0.7.0 and 10 g 2, but the result is the same. Need to know the scenario where I will face this bug of your experts and people who have already experienced this bug.
Version: -.
SQL> select * from v$version; BANNER -------------------------------------------------------------------------------- Oracle Database 11g Enterprise Edition Release 11.1.0.7.0 - 64bit Production PL/SQL Release 11.1.0.7.0 - Production CORE 11.1.0.7.0 Production TNS for Solaris: Version 11.1.0.7.0 - Production NLSRTL Version 11.1.0.7.0 - ProductionWhy do you not use the description of the bug test case in Metalink (we obviously can't post it here because it would violate the copyright of Metalink)? Your test case is not a candidate for the elimination of the join, so he did not have the bug.
Have you really read the description of the bug in Metalink rather than just looking at the title of the bug? The bug itself is quite clear that a query plan that involves the elimination of the join is a necessary condition. The title of bug nothing will never tell the whole story.
If you try to work through a few tens of thousands of bugs in 11.1.0.7, of which many are not published, trying to determine whether your application would be affected by the bug? Wouldn't be order of magnitude easier to upgrade the application to 11.1.0.7 in a test environment and test the application to see what, if anything, breaks? Understand that the vast majority of the problems that people experience during an upgrade are not the result of bugs - they are the result of changes in behaviour documented as changes in query plans. And among those who encounter bugs, a relatively large fraction of the new variety. Even if you have completed the Herculean task of verifying each bug on your code base, which would not significantly easier upgrade. In addition, at the time wherever you actually performed this analysis, Oracle reportedly released 3 or 4 new versions.
And at this stage would be unwise to consider an upgrade to 11.2?
Justin
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outer join and LignesMax problem left
I'm having a problem with the method and an sql join. My left table includes some documents that I want in the list, max 25 per page. These records have some notes are related in another table that is outer joined. When I specify the method in my cfoutput tag it includes my external joined table rows. So I could only go 5 records in my table left and 20 of my attachment table. What I want is 25 records in my table on the left and however the number of records in the table on the other that could be associated with these 25 records. Is this possible?I make two requests and avoid the outer join in this case:
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Problem with Outer join and filter
Hello
I join two tables in the source using a left outer join. Outside of the join, I have a filter specified with condition TabA.C1 > TabB.C2.
Now, when ODI generates the query it puts the left outer join on the filter condition as well. So he puts filter as
where
(1 = 1)
And ((TabA.C1 = TabB.C1 (+)) AND)
(TabA.C2 = TabB.C2 (+))
And TabA.C10 > TabB.C14 (+)
How to avoid this problem. I tried this performance on stage as well, always generated query remains the same.
I use the incremental update of the IKM Oracle. My source and target are both on the same PB.
~ ChikkHi Chikk,
If you analyze the data, you'll see it's OK to have the "(+)" to the filter...
Anyway, if you want to drop it, leave it as inner join and put the "(+)" manually to the join object.
This help you?
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Left outer join query and check the status
Hello
I have two tables as tables below
table_1
MI_ACC_IDENTIFIER CHARGE_START_DATE CHARGE_END_DATE PBA_INT_AMT 2000000000 01 SEP-05 00.00.00 29 SEP-05 00.00.00 0.26 2000000000 30 SEP-05 00.00.00 31 OCTOBER 05 00.00.00 1.92 2000000000 1 NOVEMBER 05 00.00.00 NOVEMBER 30 05 00.00.00 0.34 2000000000 1 AUGUST 06 00.00.00 31 AUGUST 06 00.00.00 0.47 2000000000 31 MARCH 06 00.00.00 27 APRIL 06 00.00.00 0.34 2000000000 30 DECEMBER 05 00.00.00 31 JANUARY 06 00.00.00 1.92 2000000000 1 MARCH 05 00.00.00 31 MARCH 05 00.00.00 0.26 Table_2
MI_ACC_IDENTIFIER CHARGE_START_DATE CHARGE_END_DATE TOT_INT_AMT_OVER_25P 2000000000 30 SEP-05 00.00.00 31 OCTOBER 05 00.00.00 0 2000000000 1 NOVEMBER 05 00.00.00 NOVEMBER 30 05 00.00.00 0.81756 2000000000 1 DECEMBER 05 00.00.00 29 DECEMBER 05 00.00.00 0.64724 2000000000 30 DECEMBER 05 00.00.00 31 JANUARY 06 00.00.00 5.51555 Power required:
MI_ACC_IDENTIFIER CHARGE_START_DATE CHARGE_END_DATE NVL(B.PBA_INT_AMT,0) TOT_INT_AMT_OVER_25P 2000000000 1 NOVEMBER 05 00.00.00 NOVEMBER 30 05 00.00.00 0.34 0.81756 2000000000 30 DECEMBER 05 00.00.00 31 JANUARY 06 00.00.00 1.92 5.51555 2000000000 1 DECEMBER 05 00.00.00 29 DECEMBER 05 00.00.00 0 0.64724 I have to check if TOT_INT_AMT_OVER_25P > B.PBA_INT_AMT and also required to display if there is no matching record in table_1 and exist in table_2 then display as well
Queries for the table:
CREATE TABLE 'TABLE_1 '.
(
ACTIVATE THE "MI_ACC_IDENTIFIER" NUMBER (10,0) NOT NULL,
DATE OF THE "CHARGE_START_DATE."
DATE OF THE "CHARGE_END_DATE."
NUMBER (15.2) "PBA_INT_AMT".
);
Insert into TABLE_1 (MI_ACC_IDENTIFIER, CHARGE_START_DATE, CHARGE_END_DATE, PBA_INT_AMT) values (2000000000, to_date ('01 - SEP - 05 00.00.00','DD-MON-RR HH24.MI.)) SS'), to_date (29-SEP-05 00.00.00','DD-MON-RR HH24.MI.) SS'), 0.26);
Insert into TABLE_1 (MI_ACC_IDENTIFIER, CHARGE_START_DATE, CHARGE_END_DATE, PBA_INT_AMT) values (2000000000, to_date (30-OCT-05 00.00.00','DD-MON-RR HH24.MI.)) SS'), to_date (31 October 05 00.00.00','DD-MON-RR HH24.MI.) SS'), 1.92);
Insert into TABLE_1 (MI_ACC_IDENTIFIER, CHARGE_START_DATE, CHARGE_END_DATE, PBA_INT_AMT) values (2000000000, to_date (1 November 05 00.00.00','DD-MON-RR HH24.MI.)) To_date SS'), (30 November 05 00.00.00','DD-MON-RR HH24.MI.) SS'), 0.34).
Insert into TABLE_1 (MI_ACC_IDENTIFIER, CHARGE_START_DATE, CHARGE_END_DATE, PBA_INT_AMT) values (2000000000, to_date (1 August 06 00.00.00','DD-MON-RR HH24.MI.)) SS'), to_date (31 August 06 00.00.00','DD-MON-RR HH24.MI.) SS'), 0.47);
Insert into TABLE_1 (MI_ACC_IDENTIFIER, CHARGE_START_DATE, CHARGE_END_DATE, PBA_INT_AMT) values (2000000000, to_date (31 March 06 00.00.00','DD-MON-RR HH24.MI.)) To_date SS'), (27 April 06 00.00.00','DD-MON-RR HH24.MI.) SS'), 0.34).
Insert into TABLE_1 (MI_ACC_IDENTIFIER, CHARGE_START_DATE, CHARGE_END_DATE, PBA_INT_AMT) values (2000000000, to_date (30 December 05 00.00.00','DD-MON-RR HH24.MI.)) To_date SS'), (31 January 06 00.00.00','DD-MON-RR HH24.MI.) SS'), 1.92);
Insert into TABLE_1 (MI_ACC_IDENTIFIER, CHARGE_START_DATE, CHARGE_END_DATE, PBA_INT_AMT) values (2000000000, to_date (1 March 05 00.00.00','DD-MON-RR HH24.MI.)) SS'), to_date (31 March 05 00.00.00','DD-MON-RR HH24.MI.) SS'), 0.26);
CREATE TABLE 'TABLE_2.
(
ACTIVATE THE "MI_ACC_IDENTIFIER" NUMBER (10,0) NOT NULL,
DATE OF THE "CHARGE_START_DATE."
DATE OF THE "CHARGE_END_DATE."
"TOT_INT_AMT_OVER_25P" NUMBER (15.5)
);
Insert in TABLE_2 (MI_ACC_IDENTIFIER, CHARGE_START_DATE, CHARGE_END_DATE, TOT_INT_AMT_OVER_25P) values (2000000000, to_date (30-OCT-05 00.00.00','DD-MON-RR HH24.MI.)) SS'), to_date (31 October 05 00.00.00','DD-MON-RR HH24.MI.) SS'), 0);
Insert in TABLE_2 (MI_ACC_IDENTIFIER, CHARGE_START_DATE, CHARGE_END_DATE, TOT_INT_AMT_OVER_25P) values (2000000000, to_date (1 November 05 00.00.00','DD-MON-RR HH24.MI.)) To_date SS'), (30 November 05 00.00.00','DD-MON-RR HH24.MI.) SS'), 0.81756);
Insert in TABLE_2 (MI_ACC_IDENTIFIER, CHARGE_START_DATE, CHARGE_END_DATE, TOT_INT_AMT_OVER_25P) values (2000000000, to_date (1 December 05 00.00.00','DD-MON-RR HH24.MI.)) SS'), to_date (29 December 05 00.00.00','DD-MON-RR HH24.MI.) SS'), 0.64724);
Insert in TABLE_2 (MI_ACC_IDENTIFIER, CHARGE_START_DATE, CHARGE_END_DATE, TOT_INT_AMT_OVER_25P) values (2000000000, to_date (30 December 05 00.00.00','DD-MON-RR HH24.MI.)) To_date SS'), (31 January 06 00.00.00','DD-MON-RR HH24.MI.) SS'), 5.51555);
Query, I used:
SELECT A.MI_ACC_IDENTIFIER, A.CHARGE_START_DATE, A.CHARGE_END_DATE, NVL (B.PBA_INT_AMT, 0), a.TOT_INT_AMT_OVER_25P OF TABLE_2 A, B FROM TABLE_1 WHERE A.MI_ACC_IDENTIFIER = B.MI_ACC_IDENTIFIER (+) AND A.CHARGE_START_DATE = B.CHARGE_START_DATE (+) AND A.CHARGE_END_DATE = B.CHARGE_END_DATE (+) and A.TOT_INT_AMT_OVER_25P > B.PBA_INT_AMT (+); I have been using the syntax of ANSI join for some time. They are readable and code looks more elegant. A reason do not want to use it?
I don't care
SQL> select t2.mi_acc_identifier 2 , t2.charge_start_date 3 , t2.charge_end_date 4 , nvl(t1.pba_int_amt, 0) pba_int_amt 5 , t2.tot_int_amt_over_25p 6 from table_1 t1 7 , table_2 t2 8 where t1.mi_acc_identifier (+)= t2.mi_acc_identifier 9 and t1.charge_start_date (+)= t2.charge_start_date 10 and t1.charge_end_date (+)= t2.charge_end_date 11 and (t1.pba_int_amt < t2.tot_int_amt_over_25p or t1.mi_acc_identifier is null); MI_ACC_IDENTIFIER CHARGE_ST CHARGE_EN PBA_INT_AMT TOT_INT_AMT_OVER_25P ----------------- --------- --------- ----------- -------------------- 2000000000 01-NOV-05 30-NOV-05 .34 .81756 2000000000 30-DEC-05 31-JAN-06 1.92 5.51555 2000000000 01-DEC-05 29-DEC-05 0 .64724 SQL> -
Hi gurus,
Left outer join:
-------------------
Select * from A LEFT OUTER JOIN B on A.col = B.col;
by definition, a left outer join brings the rows that match the join condition and lines not corresponding to table A.
My question here is "corresponding to B and no matching rows in A" is that nothing, but... SELECT * FROM A;
can someone pls clarity...
Thank you.
Imagine that you had:
TableA
COLUMN1 COLUMN2
'A' 1
'B' 2
'C' 3
TABLEB
COLUMN1 COLUMN2
'A' 'X1'
'A' 'X2'
'B' 'Y'
'D' 'Z'
SELECT * FROM TABLEA;
A 1
B 2
C 3
Now, if you want to join (first column is either in table A or B, (deuxieme from tableA, third table B)
SELECT * FROM TABLEA A JOIN TABLEB B ON (A.COLUMN1 = B.COLUMN1)
A 1 X 1
A 1 X 2
B 2 Y
SELECT * FROM TABLE LEFT B EXTERNAL ON (A.COLUMN1 = B.COLUMN1) JOIN TABLE
A 1 X 1
A 1 X 2
B 2 Y
C 3 {null}
SELECT * FROM TABLE A TABLE RIGHT OUTER JOIN B (A.COLUMN1 = B.COLUMN1)
A 1 X 1
A 1 X 2
B 2 Y
D {null} Z
SELECT * FROM TABLE A TABLE FULL OUTER JOIN B (A.COLUMN1 = B.COLUMN1)
A 1 X 1
A 1 X 2
B 2 Y
C 3 {null}
D {null} Z
HTH
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