Questions joining MFN

I am in the United Kingdom and have been sent a survey to join. However, when I get to the step to enter my IMEI, it won't let me join, because apparently my IMEI is in an "invalid format".

I checked several times, it is OK, and the necessary length. I have attached an image so this question. For security reasons, I've hidden in the middle of my IMEI.

What should I do? I really want to test the new software!

His 15 characters, because he says it should be. It is copied exactly from * #06 #, I'm sure it's correct!

EDIT: Misclicked the button 'best answer '. I have not yet found a solution!

Tags: Motorola Phones

Similar Questions

When the bike g will get the update?

Help!

Rule #1 any attempt of impregnation is to not say anything about the trial of impregnation.

If please stop asking people to post as if you are in one you can not message to the public.

Join MFN and sit and relax. Wait an invitation If chosen.

So do not message that you have received an invitation.

I'm closing this thread since its not allowed to be discussed.

Text formatting?

I have problems with the taking of a word document, the text in the text box in indesign and it change the character formatting to what I want. Here is what I do (I asked this before, but I never went forward with it and I can't find my original post.) I take a text document lets say its arial font with some bold inside. I want it to change automatically to Helvetica and keep the "BOLD" bold. At the moment I can do either or. I do not see how to keep normal Normal text and "BOLD" in the same paragraph "BOLD". EX below. How to put this in indesign and retain the formatting but replace the font Helvetica?
Learn Adobe InDesign, read answers to common questions, join our online community, or get instant help from Adobe support. Learn Adobe InDesign, read answers to common questions, join our online community, or get instant assistance from Adobe support. Learn Adobe InDesign, read answers to common questions, join our online community, or get instant assistance from Adobe support. Learn Adobe InDesign, read answers to common questions, join our online community, or for immediate help from Adobe support. Learn Adobe InDesign, read years communitywers to questions, join our com online, or get immediate help from Adobe support.

Thank you
Alain bombaert

If you want to know everything about using Word and InDesign together, take a look at this comprehensive online video tutorial:

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SQL question on natural joins

I am a newbie to SQL and I studied the book of Fundamentals SQL Server Oracle OCA 12 c, but a section on natural joins to confuse me.
Basically, the author says:
SELECT r.region_name, d.department_name, l.city, c.country_name
DEPARTMENTS d
NATURAL JOIN places l, c countries, regions r;
The natural join between DEPARTMENTS and LOCATIONS creates a provisional result, consisting of 27 lines since they are implicitly attached on the column of location_id. This set is then Cartesian-joined to the table of the COUNTRY as a join condition is not implicitly or explicitly specified. 27 interim lines are attached to 25 lines in the COUNTRY table, which gives a new temp results set with 675 (27 × 25) rows and three columns: DEPARTMENT_NAME, CITY and COUNTRY_NAME. This set is then attached to the REGION table. Once more, a Cartesian join occurs because column REGION_ID is absent from any join condition. The final result set contains rows and four columns (675 × 4) 2700.
I can understand because you evaluate joins from the left. But then he wrote:
The JOIN... USE and JOIN... ON the syntaxes are better suited to join multiple tables. The following query joins four tables using the natural join syntax:
SELECT country_id region_id, c.country_name, l.city, d.department_name
DEPARTMENTS d
NATURAL JOIN places l
NATURAL JOIN country c
NATURAL JOIN region r;
This query generates correctly 27 lines in the final results set since the required join columns are listed in the SELECT clause. The following query illustrates how the JOIN... CLAUSE is used to extract the same 27 lines. A join condition can reference columns only in its scope. In the following example, the join of DEPARTMENTS slots can not reference columns in the COUNTRIES or REGIONS of tables, but the join between the COUNTRIES and REGIONS may refer to any column of four tables involved in the query.
This second method of the part of the natural writing joined confuses me. I don't understand the logic behind the 2nd series of States of Natural Join. For me, it seems that the first series and the 2nd set look alike, but they are apparently not.
Can someone tell me the differences?
Thank you!!

Hello

The more important thing to learn more about NATURAL JOIN is: never use it. If you add new columns, joins can get different results. I've never heard of someone uisng NATURAL JOIN apart from a manual or a question like yours.

There are a lot of things in Oracle that take the time to learn and are useful. All the time you spend learning things is better spent on them.

Join question / unlink you domain Windows 8

I recently performed a clean installation of Windows 8 without problem. I tried to join to a domain. Everything seems to work as expected and I even got the standard windows of information indicating that I have was greeted at the field and I had to restart. After the reboot, I am unable to log on to the domain and I get a message indicating that the trust relationship is broken. Unfortunately, I can't use my local admin account either... the system is telling me no matter what I try to do requires high privileges which I am unable to access the connection with the Act that the workstation has not correctly joined to the domain. In addition, for the domain controller, I can not find a computer account for this system. Is there an easy way to separate this pc "is not entirely-joints" the domain and return it to a State of "stand alone" so I can try to join again? Thanx.

Hello

For the Domain Join questions, I recommend that ask you this question on the following forums.

Windows 8 IT Pro category:

http://social.technet.Microsoft.com/forums/en-us/category/w8itpro

Concerning

Question about the order of evaluation of the clause WHERE CLAUSE when the Oracle OF the syntax used to join tables

Hello
Oracle version: 11.1.0.7.0 - 64 bit
I read the documentation online at joins. The page is avialable here: joins at
My question is about the join order of evaluation of the conditions in clause and the conditions of those
are not the join conditions and are placed in the WHERE clause.
Consider the following pseudocode
SELECT
T1. Col1,
T2.Col1
Of
Table1 t1 LEFT OUTER JOIN table2 t2
WE
(condition_expression1)
WHERE
(condition_expression2)
Is it correct to say that if there is no column on the status of join (condition_expression1) in condition_expression2, then condition_expression2 is executed before condition_expression1? In other words, oracle always trying to filter based on the WHERE clause individually each table as much as possible before joining them based on the conditions on the article?
Thanks in advance,

Hello

dariyoosh wrote:

Hello

Oracle version: 11.1.0.7.0 - 64 bit

I read the documentation online at joins. The page is avialable here: joins at

My question is about the join order of evaluation of the conditions in clause and the conditions of those

are not the join conditions and are placed in the WHERE clause.

Consider the following pseudocode

SELECT

T1. Col1,

T2.Col1

Of

Table1 t1 LEFT OUTER JOIN table2 t2

WE

(condition_expression1)

WHERE

(condition_expression2)

Is it correct to say that if there is no column on the status of join (condition_expression1) in condition_expression2, then condition_expression2 is executed before condition_expression1? In other words, oracle always trying to filter based on the WHERE clause individually each table as much as possible before joining them based on the conditions on the article? ...

The reverse is actually closer to the truth, but we can't really make general statements like that.

SQL is not a language of the proceedings. Looking at the code SQL, we could say that the code does, but we cannot say much about how that code it. In other words, SQL is a language that describes the results you get, not the way to get them.

The optimizer will do everything what he thinks is faster if it does not change the results. If any order in which they are applied (in outer joins or CONNECT BY queries, for example), then think of the join is done first, and the value of the WHERE clause is applied to the result of the join.

Here is a query looks very much like you posted:

SELECT d.deptno

e.ename, e.sal

OF scott.dept d

LEFT OUTER JOIN scott.emp e ON e.deptno = d.deptno

WHERE e.sal > = 3000

ORDER BY d.deptno

;

Output:

DEPTNO ENAME SAL

---------- ---------- ----------

10 KING 5000

20 FORD 3000

20 3000 SCOTT

The scott.dept table contains deptnos 30 and 40; Why are they not in the result set? The query behaves as if the outer join is made first (production 15 rows), then the WHERE clause has been applied. All lines with deptno = 30 had sals down han 3000 and all single line with deptno = 40 was NULL in the sal column, then these lines are excluded (as well as other lines of deptnos 10 and 20), and only 3 lines above are left.

First, I'm joining the forum with a question CS4 and CS5 Suite for Mac. Can someone tell me why it is NOT a place to submit a question as it is here?

First, I'm joining the forum with a question CS4 and CS5 Suite for Mac. Can someone tell me why it is NOT a place to submit a question as it is here?

Creative Suites Mac forum seems to be closed, so moved that Creative Suites Windows the Forum of Creative Suites "base".

Then... Post your question and someone may be able to help... is your question about the installation of the old software on a new Mac?

IF El Capitan Mac read below

CS6 and previous programs have not been tested and will not be updated to run on Mac El Capitan

-which means you are trying to use CS6 and earlier at YOUR risk of having problems

-You can get CS6 and previous programs to install and run, or you can not (some do, some don't)

-IF not, Details of the message from the error messages and a person may be able to help (just not Adobe)

This information is a MUST to install old programs on Mac El Capitan

-You can't get the same error message, but here are some links that CAN help with old programs

-Java https://helpx.adobe.com/dreamweaver/kb/dreamweaver-java-se-6-runtime.html can help

Install CS5 on Mac 10.11 https://forums.adobe.com/thread/2003455 can help (also for others than CS5)

-also a TEMPORARY security change https://forums.adobe.com/thread/2039319

-http://mac-how-to.wonderhowto.com/how-to/open-third-party-apps-from-unidentified-developer s-mac-os-x-0158095 /

-the guardian https://support.apple.com/en-au/HT202491

I have been using LR for at least 4 years now and have continually updated. Current version LR6. I now want to join CC and thus access the PS and all the great stuff. Question: Can I get a refund or the recognition of the fact that I bought LR? Yo

I have been using LR for at least 4 years now and have continually updated. Current version LR6. I now want to join CC and thus access the PS and all the great stuff. Question: Can I get a refund or the recognition of the fact that I bought LR? Your answer please. PS - no doubt a question you get more often. !@

Hi Philippe,.

Please see: pricing and membership creative cloud plans | Adobe Creative Cloud

Otherwise, you can join Adobe sales for rate applications.

Kind regards

Sheena

Application of self-join interview question?

Hi guys!
I asked this question in an interview for the job as a developer of database.
I received a select based on symbol table and asked to drow the printout and then explain works step by step this statement.
Here are the instructions create & insert
/**********************************************/
create table tbl (number x, y varchar2 (50));
INSERT INTO tbl (x, y) VALUES ('A', 0);
INSERT INTO tbl (x, y) VALUES (1, 'B');
COMMIT;
/***********************************************/
Below the query (select statement)
SELECT tbl1.x
OF tbl tbl1 tbl2 tbl
WHERE tbl1.x = 0 OR tbl2.x = 1;
Please explain me the output and work step by step this statement...
Thank you...

Hello usman_noshahi,

If you omit the where the query clause will result as below

SELECT tbl1.x, tbl2.x

OF tbl tbl1 tbl2 tbl

/

X          X

----- ----------

0          0

0          1

1          0

1          1

Request granted will result as below based on where condition on each line on top 4 rows... Or Condition will be returs real if one condition is met with the data set.

SQL > SELECT tbl1.x, tbl2.x

2 tbl tbl1 tbl2 tbl

3. WHERE tbl1.x = 0 tbl2.x = 1 GOLD

4.

X          X

---------- ----------

0          0

0          1

1          1

SQL >

You can get the information in a different way as below...

SQL > explain plan set statement_id = "ORACLEFORUMS" for

2. SELECT tbl1.x

3 tbl tbl1 tbl2 tbl

4 WHERE tbl1.x = 0 OR tbl2.x = 1;

He explained.

SQL > select LPad (' ', 2 *(Level-1)). Level | '.' || NVL (position 0). ' ' ||

Operation 2. ' ' || Options of | ' ' || Object_name | ' ' || Object_type

3   || ' ' || Decode (id, 0, Statement_Id |) "Cost = ' | Position) | cost

4   || ' ' || Object_Node "Query Plan.

plan_table 5

6 start with id = 0 and statement_id = 'ORACLEFORUMS. '

7. connect by prior id = parent_id

8 and statement_id = 'ORAFAQ.

9   /

Query plan

--------------------------------------------------

1.7 STATEMENT SELECT ORACLEFORUMS cost = 77

2.1 NESTED LOOPS 7

3.1 ACCESS FULL TBL TABLE 3 TABLE

3.2 TABLE ACCESS FULL TBL TABLE 2

SQL > SELECT *.

TABLE 2 (DBMS_XPLAN. DISPLAY ('PLAN_TABLE', 'ORACLEFORUMS', 'BASIC'));

PLAN_TABLE_OUTPUT

------------------

Hash value of plan: 1718268327

-----------------------------------

| ID | Operation | Name |

-----------------------------------

|   0 | SELECT STATEMENT |      |

|   1. NESTED LOOPS |      |

|   2.   TABLE ACCESS FULL | TBL |

|   5:   TABLE ACCESS FULL | TBL |

-----------------------------------

10 selected lines.

SQL >

SQL > select LPAD (' ', 2 *(LEVEL-1)). operation 'OPERATION', 'options ',.

2 DECODE (TO_CHAR (id), '0',' COST = ' |) NVL (to_char (position), n / 'a').

3 object_name) 'OBJECTNAME', id | » -'|| NVL (parent_id, 0) | » -'||

4 NVL (position 0), 'ORDER', SUBSTR (optimizer, 1, 6) 'OPT '.

plan_table 5

6. start with id = 0

7 and statement_id = 'ORACLEFORUMS. '

8 connect by prior id = parent_id

9 and statement_id = 'ORACLEFORUMS. '

10.

OPTIO OBJECTNAME OPT OPERATION ORDER

------------------------- ----- ----------------------------------------------- ---------- ---------------

COST OF THE SELECT = 7 0-0-7 ALL_RO

NESTED LOOPS                                                                  1-0-1

TABLE ACCESS FULL TBL 2-1-1

TABLE 3-1-2 TBL FULL ACCESS

SQL >

ID ID-PARENT as follows...

00

10

21

31

Here level 1 has 2 childs (2,3)

Thus the sequential order of enforcement will be 2,3,1.

For more reference I recommend you read below, of course it's Oracle V8 but it gives more information.

Good reading!

Optimization of joins

JOIN the question... Join two tables without omitting lines

I came across a problem that should be an easy fix (I hope), but I'm having a hard time to come up with a solution.

Basically I have two tables, one with the actual amounts with the budget. I have to write a sql statement select that joins these tables together and includes all of their lines. I was able to join the tables by using JOIN, LEFT JOIN and RIGHT JOIN, but it always fails the lines I need.

Below, I have examples of my tables (AMOUNT_TABLE and BUDGET_TABLE). For simplicity, I built the examples to show the same values in the first four columns, with the 5th and 6th columns (SUB_ACCOUNT, AMOUNT, BUDGET) as the only values that are different. My actual tables are not quite that simple, but I don't think it was relavent to this issue.

AMOUNT_TABLE

FISCAL_YEAR	PERIOD	ACCT_UNIT	ACCOUNT	SUB_ACCOUNT	AMOUNT
2013

111111555555000010020131111115555551000100201311111155555520001002013111111555555300010020131111115555554000100

BUDGET_TABLE

FISCAL_YEAR	PERIOD	ACCT_UNIT	ACCOUNT	SUB_ACCOUNT	BUDGET
2013

1111115555553000200201311111155555540002002013111111555555500020020131111115555556000200

Here's what I hope. Note that SUB_ACCOUNTs 0000, 1000, and 2000 show without budget amounts since there is not a corresponding line in the BUDGET_TABLE. And same for SUB_ACCOUNTs 5000 and 6000, they show budgets with no amount because there is not a corresponding line in the AMOUNT_TABLE.

(exit)

FISCAL_YEAR	PERIOD	ACCT_UNIT	ACCOUNT	SUB_ACCOUNT	AMOUNT	BUDGET
2013	1	11111	555555	0000	100	0
2013	1	11111	555555	1000	100	0
2013	1	11111	555555	2000	100	0
2013	1	11111	555555	3000	100	200
2013	1	11111	555555	4000	100	200
2013	1	11111	555555	5000	0	200
2013	1	11111	555555	6000	0	200

If all goes well, my question is clear. Any help on this would be greatly appreciated. Thanks in advance.

Use the join ANSI - FULL OUTER JOIN syntax:

with amount_table like)

Select fiscal_year 2013, 1 period, 11111 acct_unit, 555555, 0000 sub_account, amount 100 of all the double union

Select 2013,1,11111,555555,1000,100 from all the double union

Select 2013,1,11111,555555,2000,100 from all the double union

Select 2013,1,11111,555555,3000,100 from all the double union

Select double 2013,1,11111,555555,4000,100

budget_table like)

Select 1 time, 11111 acct_unit, account 555555, sub_account 3000, 2013 fiscal_year, budget of 200 Union double all the

Select 2013,1,11111,555555,4000,200 from all the double union

Select 2013,1,11111,555555,5000,200 from all the double union

Select double 2013,1,11111,555555,6000,200

)

Select nvl (a.fiscal_year, b.fiscal_year) fiscal_year,

period of NVL (a.period, b.period),

NVL (a.acct_unit, b.acct_unit) acct_unit.

account of NVL (a.Account, b.Account),

NVL (a.sub_account, b.sub_account) sub_account.

Amount NVL(a.amount,0),

NVL(b.budget,0) budget

of amount_table one

full join

budget_table b

on)

a.Fiscal_Year = b.fiscal_year

and

a.period = b.period

and

a.acct_unit = b.acct_unit

and

a.Account = b.account

and

a.sub_account = b.sub_account

)

FISCAL_YEAR PERIOD ACCT_UNIT ACCOUNT SUB_ACCOUNT AMOUNT BUDGET
----------- ---------- ---------- ---------- ----------- ---------- ----------
2013 1 11111 555555 0 100 0
2013 1 11111 555555 1000 100 0
2013 1 11111 555555 2000 100 0
2013 1 11111 555555 3000 100 200
2013 1 11111 555555 4000 100 200
2013 1 11111 555555 6000 0 200
2013 1 11111 555555 5000 0 200

7 selected lines.

SQL >

SY.

Simple join question
Hi guys, I'm confused.

I was always told that one table can have ONLY 1 primary key. is this true or false? Or a table can cause several primary keys?

Second question is, if I join two tables that are formed as a result of columns.

Table 1:
Account_number
Account_create_month
Customer_address

Table 2:
Account_number
Account_create_month
Customer_phone

Should I join on account_number AND account_create_month? If yes why? and why not? Sorry if this is a stupid question. I'm just very confused. Thank you
Hello

Z KHAN wrote:
Hi guys, I'm confused.

I was always told that one table can have ONLY 1 primary key. is this true or false? Or a table can cause several primary keys?

A table can have 1 primary key. There may be any number of columns (1, 2, 3 or more) in the primary key.
A table can contain any number of unique keys.

Second question is, if I join two tables that are formed as a result of columns.

Table 1:
Account_number
Account_create_month
Customer_address

Table 2:
Account_number
Account_create_month
Customer_phone

Should I join on account_number AND account_create_month? If yes why? and why not? Sorry if this is a stupid question. I'm just very confused. Thank you

No, you do not have to connect using the two. If you use the two in a join given or not depends on what the data means, and what results you want.

two questions. question 1: join the clouds will give me access to QuarkXpress.

two questions. question 1: join the clouds will give me access to QuarkXpress.

QuarkXpress is not an Adobe product. The QuarkXpress product page is here http://www.quark.com/Products/QuarkXPress/.

To see which products are included with Adobe Creative Software cloud go to this page https://creative.adobe.com/apps. There are learn more links for each of the products and services included.

Gites join Question!
Hi guys,.

I need a report that displays the comm total of 'sale' and the total of the "rent" comm
Based on the following table.

Transaction Type Total Comm
62 - 1000 SALE
61 9 APRIL 11 - RENT 500
63 - SALE 1000
41 16 APR 11 - RENT 500

--
Best regards
Hello

You can do this with a self-join, but it is much more effective to do it like this:
```
SELECT       SUM (CASE WHEN type = 'SALE' THEN comm END)     AS total_sale
,       SUM (CASE WHEN type = 'RENT' THEN comm END)     AS total_rent
FROM       table_x
WHERE     type    IN ('RENT', 'SALE')
-- GROUP BY  ...          -- If wanted
;
```
I hope that answers your question.
If not, post a small example data (CREATE TABLE and only relevant columns, INSERT statements), and the results you want from this data.
Explain how you get these results from these data.
Always tell what version of Oracle you are using.

questions about join quter
Hi all
I am now learning on the outer join.
so far, I've learned the syntax of the "right outer join" / «left outer join»»»
now, I try to understand the other syntax with the (+) sign sign.which I tried to find explanations to this sign
without success

lets talk about the left outer join:
```
select a.id,b.name from 
employees a left outer join names b 
on a.id=b.id 
```
issues related to the:
1. in the syntax with the sign of the operator (+), I need to like this left join BOF:
```
select a.id,b.name from 
employees a, names b 
where a.id = b.id (+)
```
My question is why?
What this sign means? If the left join supposed to return all lines
the left table independently to corresponding records in the right table isn't this supposed to be
```
where a.id(+) = b.id
```
Please explain?
2. If we continue with the outer join operator (+),
How am I supposed to do no outer join? which means: can I leave the
side of the location of the table the same and just change the location of the operator (+) side? as it is, we mean:
If it is not join:
```
select a.id,b.name from 
employees a, names b 
where a.id = b.id (+)

does the right join  will look like this :
select a.id,b.name frin 
names b , employess a 
where b.id = a.id(+) 

or  can i  do also like this ? meaning just moving the operator (+) sign ? 
select a.id,b.name from 
employees a, names b 
where a.id (+) = b.id 
is the last syntax o.k and legal ?
```
3 are the columns of the tables assume appear in the query and in where clause in the same order?
lets say:
```
select  *a*.id, *b*.name from 
employees *a*, names *b* 
where *a*.id = *b*.id(+) 
or is it o.k to do like this too : 
select *b*.name,*a*.id from //the column name appers on the left while the table employess b appear on the right 
employees *a*, names *b* 
where *a*.id = *b*.id(+) // in the condition relation are they suppose to appear in the same side as their tables ? 
```
4. What is the significance of this kind of syntax:
```
select b.name,a.id from 
employees a, names b 
where a.id = b.id(+) and 
a.street = b.street (+) 
```
5. What is the meaning of the sign < or > in the outer join:
example:
```
select .name,a.id from 
employees a, names b 
where a.id > b.id(+) 
```
6.
y at - it or the outer join condition too?
7. in what way is best to write the outer join?
with the sign of the operator (+) or with the "... outer join" sentence?
Thanks in advance
Naama

Under the direction of: naama on December 23, 2010 05:56

Under the direction of: naama on December 23, 2010 06:11
Hi, Naama,

Sorry, I don't understand the question.
Naama wrote:
Hi Frank
I have drawn on your explanation, I got a few concludes we will see if I can:
I can say that if I have an outer join complete with the State "and" which refers to one of the table, as in this command:
```
select t0.comp as comp_T0 ,
t0.type_i           AS type_i_t0     ,          -- For debugging only
t2.comp as comp_T2,
t2.type_i           AS type_i_t2     ,          -- For debugging only
t0.amt type_0,
t2.amt type_2
from table_summary t0
*full join* table_summary t2
on t0.comp= t2.comp --linked column
and t2.type_i = 2 --addition to the condition that relates to just one of the table (the "extra" condition) 
```
Please makes it difficult for people to help you. The format of your code. In particular, he dash so that it is easy to find the main proposals, including FROM and WHERE.

so I can say that the table with the 'extra' condition becomes the table of conduct (according to my specific example: T2 is the driving table).

Do you mean that t2 is the 'engine', but is not the case of the t0?
I don't understand what you mean by "table of conduct." Earlier you seemed to use 'conduct table' to refer to the table of in
```
FROM            d
LEFT OUTER JOIN       t     ON ...
```
or
```
FROM            t
RIGHT OUTER JOIN  d     ON ...
```
in other words, table in these asymmetrical relations whose lines will appear in the result set, even if that line has no match in t.
FULL OUTER JOINs are symmetric; two tables are used the same way. In a FULL OUTER JOIN, rows in each table must be included in the game even if they have no match in the other table. If one of the tables is 'flying', how can you say that the other table is not "led" as well?
and the query is corrlated as if I was doing:
```
select t0.comp as comp_T0 ,
t0.type_i           AS type_i_t0     ,          -- For debugging only
t2.comp as comp_T2,
t2.type_i           AS type_i_t2     ,          -- For debugging only
t0.amt type_0,
t2.amt type_2
from table_summary t0
*right join* table_summary t2
on t0.comp= t2.comp --linked column
and t2.type_i = 2 --addition to the condition that relates to just one of the table ,
```
Instead,.
because through the example of all, I saw that there were none concern the t0 is an outer join means that the conduct at the table, only in the table back.
Am I wrong?
Ask yourself if the FULL OUTER JOIN you posted at the beginning of your last message is equivalent to the RIGHT OUTER JOIN you posted more far in this message?
No, they are not equivalent. With the sample data that John Spencer posted on 30 December, the two queries above happens to get the same results, but, generally, they will not get the same results. For example, if you add this line in the sample data:
```
SELECT  9 AS comp,  0 AS type_i,  860 AS amt     FROM dual     UNION ALL
```
then the set of results of the FULL OUTER JOIN will include these two production lines:
```
`  COMP_T0  TYPE_I_T0    COMP_T2  TYPE_I_T2     TYPE_0     TYPE_2
---------- ---------- ---------- ---------- ---------- ----------
         9          0                              860
                               9          0                   860
```
but the RIGHT OUTER JOIN that you posted will get one of these lines:
```
`  COMP_T0  TYPE_I_T0    COMP_T2  TYPE_I_T2     TYPE_0     TYPE_2
---------- ---------- ---------- ---------- ---------- ----------
                               9          0                   860
```

Question on join
Hello world

I have a question about the join.

I have 3 tables... USER FROM POINT 3 OF the AGENDA) 1 2))

Suppose if I want to know the total QUANTITY of ITEM for a userid of the user = 10, I need to join all the 3 tables.

I have no information, 3 tables do not join to the rest of the users.

I tried several ways... like nested box instructions... .but no luck.
```
select case when u.userid=10 then i. itemquntity
else null end as ITEMQUANTITY 
from user u
join order o
on o.id=u.orderid -- and u.userid=10
join item i
on i.id=o.itemid

NOTE: My query is in view that can be called by all the other users. Only USERID=10 needs ITEMQUANTITY rest of the users i want to avoid that joins (NULL will be fine)
```
How to write a simple select query?

version 11.1.0.7

any ideas...

Thank you
Mike
Hi, Mike,.

Mike wrote:
Hello world

I have a question about the join.

I have 3 tables... USER FROM POINT 3 OF the AGENDA) 1 2))

The USER is not a name very good table. It is the name of a built-in function (in fact, you can use it in this problem) and will be confusing. Why not call the table USER_TABLE or MY_USERS?

... How to write a simple select query?

You want a single query? Two separate petitions be much simpler and more effective?

If you want everyone to say the same thing (for example ('SELECT...) X"), but they have x means a different view for the user = 10 ID it does for everyone, then create a special view just for userid = 10, either in the scheme of this user, or create a synonym of have in this schema of the user by referring to the special view.

If you create a query (or view: I'll just say view later), then you will have to join three tables and hide some of the information form everyone except userid = 10. Maybe it's not difficult. You might be able to outer join the other tables to user_table and 'AND u.userid = 10' part of the join condition. All the columns of these tables will be NULL for all users except userid = 10.
It will be ineffective. You will have the expense of do outer joins whenever someone uses the view, even if a single user will see ever all data from joined tables.

I hope that answers your question.
If not, post a small example of data (CREATE TABLE and only relevant columns, INSERT statements) for all tables and also post two sets the outcomes from these data:
(1) the userid = 10 results should see
(2) the results, everyone should see

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