connect before between two tables

Similar Questions

• I got a prob while creating a link/link between two tables in a jdeveloper11g

  Hi guys...

  I got a prob while creating a link/link between two tables in a jdeveloper11g, help me.

  I use two table employee, Department.

  Unfortunately there is no primary key and foreign key in these tables.now i updated on those of table structure.

  I want to create a form like it should be a link between the Department data, i.e. data and employee while the Department is changed then only present corresponding employees in this block...

  Table structure as follows

  Employee:

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

  employe_id,

  name,

  Department,

  GR,

  Date of birth

  Department of justice

  Department:

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

  department_id,

  department_name;

  Note: in the employee department table is the same as in the department table department_id...

  So, you want to build master / detail between Department & sub_departments?

  In the same way, create a new, yet another link between departments & sub_departments, so that the connection is:

  Department.Department_Id = sub_departments.department_id

• I need to implement Drag N Drop between two tables that saves the two records

  I need to implement Drag N Drop between two tables that saves the two records on a third page, using drag 'n drop.

  Check out this video http://baigsorcl.blogspot.com/2011/01/drag-and-drop-collection-in-oracle-adf.html

• two relationships between two tables

  Hi all
  
  I created a cube in my db (using OWB) of two relationships to my time dimension. A relationship shows the start time of a process, that the other represents the date of termination or end.
  
  example of the cube:
  
  Fact1
  Fact2
  Dim_Time1
  Dim_Time2
  
  where Dim_time1 and Dim_Time2 link to the same dimension.
  
  Unfortunately I have trouble this manually import in obiee administrator. I imported the fact table and the dimension of the physical layer table. Can I move to the business layer and create the two joins between two tables. But the overall consistency check up a mistake, I shouldn't have two relationships between two tables and remove a foreign key.
  
  What I am doing wrong?
  
  Kind regards
  
  Knut

  Knut,

  In your example, you should have a time dimension in the phyiscal layer and alias of this 2 x, for Dim_time1 and for Dim_time2, in this way aliases are separate tables as the BI server point of view and you define releations separate on the same table, you can define specific to each alias dim relevant time joins.

  In this way, you will have two time decreases intensity in MDB layer also, has 'Dim-Start Time' and a ' Sun - term time "or similar.

  Hope this helps,
  Alastair

• Use a where clause clause if there is a relationship between two tables

  Is it possible to use a where clause clause if there is a relationship between two tables.
  I have a relationship between two tables:
  And a.customer = b.customer (+)
  
  If b.customer exists in a.customer I would add this in where clause
  And b.info = 'X '.
  But how?

  It's the outer joins by chance?

  If so, why not just:

  and  b.customer(+) = a.customer
  and  b.info(+) = 'X'
  

  BTW, I put the two "b" on the left column to make it easier to read. Some people mix that makes me crazy, but makes no difference to the query.

• How to remove the extra space between two Table HTML

  Hello
  
  I wrote a code for printing costs. But there is more space between two Table Html, how I can remove it.
  
  Please run that Code
  
  Start
  
  HTP.p (')
  < html >
  (< body > ');
  
  HTP.p (')
  < TABLE align = "left" width = "500" cellspacing = "0" cellpadding = "0" border = "1" >
  < tr > < td align = 'right' white-space: nowrap; > Ph-2201751 < table >
  < td > < table >
  < /tr >
  < tr > < td align = "center" white-space: nowrap; > < B > < table > < /B > SATYAM MODERN PUBLIC SCHOOL
  < td > < table >
  < /tr >
  < b >
  < td align = "center" white-space: nowrap; > < i > < H3 > (AFFILIATED to THE CBSC, NEW DELHI CODE No. 53544) < / H3 > < /I > < table >
  < td > < table >
  < /tr >
  < tr > < td align = "center" white-space: nowrap; > new colony Braham, - 131001 (h) < table >
  < td > < table >
  < /tr >
  < tr > < td align = "center" white-space: nowrap; > RECEPTION COSTS < table >
  < td > < table >
  < /tr >
  < /table >
  < TABLE width = "500" cellspacing = "0" cellpadding = "0" border = "1" >
  < b >
  < td width = "100" white-space: nowrap; > receipt no.: < table >
  < td width = "100" white-space: nowrap; > Date received: < table >
  < /tr >
  < b >
  < td width = "100" white-space: nowrap; > name: < table >
  < td width = "100" white-space: nowrap; > father name: < table >
  < /tr >
  < b >
  < td width = "100" white-space: nowrap; > class & s: < table >
  < td width = "100" white-space: nowrap; > A/C No. : < table >
  < /tr >
  < b >
  < td width = "100" white-space: nowrap; > from: < table >
  < td width = "100" white-space: nowrap; > to: < table >
  < /tr >
  
  (< /table > ');
  HTP.p (')
  < / body >
  (< / html > ');
  end;
  
  
  Thank you
  
  Ed

  Hello

  HTML is valid, that your code print?

  Wouldn't be as below?

  BR, Jari

  Published by: jarola on December 18, 2009 15:22
  I test your code
  http://Apex.Oracle.com/pls/OTN/f?p=40323:25
  I can't see any extra space between the tables

• Cannot connect call between two SX20

  Hello world

  I've implemented two Cisco SX20 units at two sites. When you attempt to place the call between two units, it is unstable. In particular, sometimes the call is connected, sometimes, it is not connected even if ping, tracert are Ok. It's very strange. I'm sure that the SX20 unit works well. If I make a conclusion that there is a problem with my network but I don't exactly know why.

  All the world is facing this problem before?

  Dzung.

  Hey yours,

  Disable NAT if you do not use.

  Also useful to configure QoS, check on your link packet loss and bandwidth.

• Comparison between two tables

  Hello

  I have two tables USERS and LISTS:

  And I want that there is a comparison between these two tables executed when user table USERS connect to the application.

  They must compare the columns between the SEXES and AGE and if some users have the same attributes in its ranks as in a line in the table LISTS (here is unique each line) so that the user gets in his line in the column 'LIST_ID' the ID of the line with which attributes between GENDER and AGE.

  Example:

  USERS TABLE: exampleuser | Boby | McGrahem | Male | 18. -------

  LISTS OF TABLE: 28 | examplelist | Male | 18

  Once the Login of the user Boby gets in its ranks the ID = 28 because the SEX and AGE corresponds to the row of table LIST:

  Result:

  USERS TABLE: exampleuser | Boby | McGrahem | Male | 18. 28

  
  CREATE TABLE  "USERS" 
     (    "USERNAME" VARCHAR2(15) NOT NULL ENABLE, 
          "FORRNAME" VARCHAR2(15) NOT NULL ENABLE,                  
          "SURNAME" VARCHAR2(15) NOT NULL ENABLE, 
          "GENDER" VARCHAR2(10) NOT NULL ENABLE,   
          "AGE" NUMBER, 
          "LIST_ID" NUMBER, 
           CONSTRAINT " USERS_PK" PRIMARY KEY ("USERNAME")
    USING INDEX  ENABLE
  
  

  CREATE TABLE  "LISTS" 
     (    " LIST_ID" NUMBER NOT NULL ENABLE, 
          " LISTNAME" VARCHAR2(50) NOT NULL ENABLE, 
          "GENDER" VARCHAR2(10) NOT NULL ENABLE, 
          "AGE" VARCHAR2(3) NOT NULL ENABLE,  
           CONSTRAINT " LISTS_PK" PRIMARY KEY ("LIST_ID")
    USING INDEX  ENABLE
     
  

  I need help for coding this, if someone be so good.

  This is my essay:

  
  

  DECLARE  
      v_listid varchar2(100);   
      BEGIN  
        
      SELECT  LIST_ID into v_listgid
       FROM LISTS
       WHERE GENDER = (SELECT GENDER
                            FROM USERS 
                          WHERE UPPER(TRIM(USERNAME)) = UPPER(TRIM(:APP_USER)))
       AND AGE = (SELECT AGE
                            FROM USERS  
                          WHERE UPPER(TRIM(USERNAME)) = UPPER(TRIM(:APP_USER)));
  
       UPDATE USERS   
      SET LIST_ID =  v_listid   WHERE UPPER(TRIM(USERNAME)) = UPPER(TRIM(:APP_USER));  
    
      END;
  

  I'd like to REALLY shy away using the USER name as the primary key, because the fact that users can create duplicate user names... YOU should NEVER base a PK on the entry of user, the better to let the database create and maintain this element of date...

  FIX:

  UPDATE to users U

  SET List_Id = (SELECT List_Id OF lists L WHERE L.Gender = U.Gender and L.SOMENUMBER = U.SOMENUMBER) AND

  Upper (U.UserName) = UPPER(:APP_USER);

  Don't really want to cut the strings, Oracle which will do since they are of type varchar2 columns...

  Thank you

  Tony Miller
  Software LuvMuffin
  Ruckersville, WILL

• Connectivity options between two Playbooks?

  What are the options of connectivity between two Playbooks?

  -What is BT pairing between possible?

  -I guess if they are on the same subnet, you should be able to speak directly on Wifi?

  You should definitely be able to talk directly between two PBs on the same wireless subnet.  Note that at least in the AIR, UDP is not supported, then you are probably limited to the protocols built on TCP, for now.  Find the other device is your main problem, but would work at the entrance of the user of an IP address, or something like QR codes.

  I don't think we will have any BT support initially, but which should eventually be possible somehow.

  We see interconnectivity NFC at some point, which would allow some ways ordained to share things.

  In some simple cases, you can use the camera and a QR code, although I do not think THAT RIM has indicated any support in the short term for a barcode API so this would probably involve some third-party libraries.

  If you wanted to get complicated, you can use the audio output on one and the MIC on the other.  Which would also be you to make a "show" mode of communication, although humans nearby could get mad.  (Probably the QR code approach allows a limited form of 'broadcast' as well).

  I think that exhausts my supply of ideas for the moment.

• The most elegant way to get the difference between two tables - not least!

  Hello
  
  
  Simplified example of what I'm trying to achieve - I have two tables ORIGINAL and REVISED.
  
  
  My aim is to compare the two, such as; -
  
  
  When there is data in the two tables I get the difference between the Budget column, and if there is no difference, so I don't want no lines.
  
  When data exists in the ORIGINAL, but not in review, I want to the inverse of the current value of the Budget column.
  
  Where the data exist in REVISED I want the REVISED value.
  
  I can see how I can do this, see below, but is there a more elegant solution?
  
  
  
  
  Data for the ORIGINAL table
  

  select '801040' entity, '2186' expense_type, 234000 budget
  from dual
  union all
  select '801040' entity, '3001' expense_type, 1000 budget
  from dual
  union all
  select '801040' entity, 'P132' expense_type, 34000 budget
  from dual
  union all
  select '801040' entity, 'P135' expense_type, 43000 budget
  from dual

  Data for the REVISED table
  

  select '801040' entity, '2186' expense_type, 235000 budget
  from dual
  union all
  select '801040' entity, 'P132' expense_type, 34000 budget
  from dual
  union all
  select '801040' entity, 'P139' expense_type, 56000 budget
  from dual

  Desired output
  
  
  
  ENTITY EXPENSE_TYPE DIFFERENCE
  ------ ------------ ----------
  801040 2186 1000
  801040 3001-1000
  801040 P135-43000
  801040 P139 56000
  
  5 selected lines.
  
  
  
  Code current to achieve this, is there a better way?
  

  select original.entity
  ,      original.expense_type
  ,       (nvl(revised.budget,0) - original.budget) as difference
  from   original
  ,      revised
  where  original.entity = revised.entity(+)
  and    original.expense_type = revised.expense_type(+)
  and   (nvl(revised.budget,0) - original.budget) != 0
  union all
  select  revised.entity
  ,       revised.expense_type
  ,       revised.budget as difference
  from   revised
  where  not exists
  (select 'x'
  from   original
  where  original.entity = revised.entity
  and    original.expense_type = revised.expense_type)
  and    revised.budget != 0

  Thanks a lot for your comments,.
  
  
  Robert.
  
  Published by: Robert Angel on January 17, 2012 03:31 to change is not equal to! = - Thanks for the heads up

  SQL> with original
    2  as
    3  (
    4    select '801040' entity, '2186' expense_type, 234000 budget
    5    from dual
    6    union all
    7    select '801040' entity, '3001' expense_type, 1000 budget
    8    from dual
    9    union all
   10    select '801040' entity, 'P132' expense_type, 34000 budget
   11    from dual
   12    union all
   13    select '801040' entity, 'P135' expense_type, 43000 budget
   14    from dual
   15  )
   16  , revised
   17  as
   18  (
   19    select '801040' entity, '2186' expense_type, 235000 budget
   20    from dual
   21    union all
   22    select '801040' entity, 'P132' expense_type, 34000 budget
   23    from dual
   24    union all
   25    select '801040' entity, 'P139' expense_type, 56000 budget
   26    from dual
   27  )
   28  select *
   29    from (
   30          select nvl(o.entity, r.entity) entity,
   31                 nvl(o.expense_type, r.expense_type) expense_type,
   32                 nvl(r.budget,0) - nvl(o.budget,0) budget
   33            from original o
   34            full join revised r
   35              on o.entity = r.entity
   36             and o.expense_type = r.expense_type
   37         )
   38   where budget <> 0
   39  /
  
  ENTITY EXPE     BUDGET
  ------ ---- ----------
  801040 2186       1000
  801040 P135     -43000
  801040 3001      -1000
  801040 P139      56000
  
  SQL>
  

• getting data between two tables

  Hi everyone, I need help with a query that seems simple, but I can't understand. consider the following data.

  with data as (

  s élire 802 cd from dual union all

  select 803 cd from dual union all

  select 804 cd from dual

  data2 as

  ( select 804 cd from dual )

  
  

  I want to join the two tables so that the result will be as follows:

  in the above data, we can see that 804 is common in both tables.  in this case, the output should be 804.

  Similarly, if 2 data include the following:

  data2 as

  ( select 804 cd from dual union all

  Select 803 cd from dual

  )

  
  

  the output is then 803 804.

  
  

  
  

  Now consider the following data
  

  with data as (

  s élire 802 cd from dual union all

  select 803 cd from dual union all

  select 804 cd from dual

  )

  take data2 table is empty

  
  

  I want to join the two tables, and in this case, the output should be 802,803,804.  data2 table is empty, then the default behavior is to get all the data in the data table. Since there is no line that corresponds to rows in the data table,

  the default behavior is to display output all data in the data table.  in this case 802,803,804
  

  can someone write a query for this?
  

  Hello

  elmasduro wrote:

  Hi Frank, thanks for the request.  These are very good questions. Here are the answers

  What happens if data2 rows, but none of them match data CD?

  It won't happen because the integrity of the data.

  ...

  I see it; a data line 2 will be in the data.  You have a FOREIGN KEY constraint, or something similar, so if cd = 804 exists in database2, then it must also exist in the data.

  What happens if the data is empty and data2 is not?

  If data is empty and data2 is not, then the data2 data should be released.

  for example

  with the data as)

  data table is empty

  data2 as

  (select cd 804 Union double all the)

  Select cd 805 double

  )

  When you join data and data2 account with the above data, then the output should be 804 805.  data2 data should be the display if the data table is empty. ...

  Doesn't that contradict what you said earlier?  Previously, it seemed to say that each row of data2 will be in the data, but now you give an example where all data2 lines are not in the data.

  If the first is true, then the solution in response #2 should work.

  If the latter is true, you can change the query like this:

  WITH got_r_num AS

  (

  SELECT NVL (d.cd, d2.cd) AS cd

  DENSE_RANK () (ORDER IN CASE

  WHEN d.cd IS NOT NULL

  AND d2.cd IS NOT NULL

  THEN "A".

  WHEN d.cd IS NOT NULL

  THEN 'B '.

  ANOTHER 'C '.

  END

  ) AS r_num

  DATA d

  FULL OUTER JOIN data2 d2 ON d.cd = d2.cd

  )

  SELECT cd

  OF got_r_num

  WHERE r_num = 1

  ORDER BY cd

  ;

• Difference between two tables (ORA-01722)

  I am trying to get the difference of two tables on two databases.
  
  Walker I run the script below, it gives me a result that is not accurate.
  

  Select name, creation_time of v$datafile@REMOTE_database.com
  2 LESS THAN
  3 select name, creation_time from v$ datafile;
  
  I get 14 ranks of foregoing.

  However, when I run this script, I get an accurate count of the difference. How can I list the exact number instead of only to count the difference? Want lists all five name missing on table 2.
  

  SQL > SELECT a.cnt - b.cnt

  > count 2 (name) select cnt (of v$datafile@REMOTE_database.com) a, b

  3 (select count (name) NTC v$ datafile);
  
  A.CNT - B. CNT
  -----------
  5

  edited by: Albert Zaza on July 19, 2010 11:19

  have you tried:

  select trim(upper(name)) from v$datafile@REMOTE_database.com
   MINUS
   select trim(upper(name)) from v$datafile;
  

• Left join between two tables using two different conditions

  I have following three tables with their data, as shown below.
  
  CREATE TABLE TIREMASTERPROCESS_TEMP
  (
  PRODUCTIONCODE TANK (12 BYTES),
  FIELDNAME CHAR (12 BYTES),
  DATACHR VARCHAR2 (60 BYTE),
  REVISIONNO TANK (3 BYTES),
  DATANUM NUMBER (9.4)
  )
  Whether the table c. and its sample data are like
  
  FIELDNAME DATACHR REVISIONNO DATANUM PRODUCTIONCODE
  AB No. Nch 1 100 0
  AB No. Nch 1 108 0
  AB No.-of-Nch 1 1 0
  Ass42 teased dishes 1 0
  Ass42 BTDrumNo1 BTD-051 1 0
  AB53 BTDrumNo1 BTD-051 104 0
  
  
  CREATE TABLE materialcode
  (
  UPPERMATERIALCODE CHAR (20),
  PROCESSNO TANK (3),
  PROCESSADOPTDATE TANK (9)
  
  )
  Whether the table b. and its sample data are like
  
  UPPERMATERIALCODE PROCESSNO PROCESSADOPTDATE
  Ass42 1 20120717
  AB53 108 20121121
  111 20111104 AS05
  104 20120928 AS30
  
  
  CREATE TABLE BUILDINGSCHEDULEMASTER
  (
  BUILDINGSIZE TANK (5 BYTES),
  GTCODE FLOAT (4 BYTES),
  TIREPERCART NUMBER (3.0).
  BUILDINGLOT NUMBER (3.0)
  )
  Whether table and its sample data are like
  
  BUILDINGSIZE GTCODE TIREPERCART BUILDINGLOT
  AB42A 4 12 ass42
  AB53A 4 88 AB53
  AS30A AS30 4 80
  BF03A 1 90 BF03
  
  
  Now, I need to get some data in these fields by using the suite of mappings.
  
  « PR_ » || Trim (a.BUILDINGSIZE) | ' / MPL2' Processid,--PRIMARY KEY
  TO_DATE (substr (NVL(b.PROCESSADOPTDATE,'19800101'), 7, 2) |) » /'|| SUBSTR (NVL(b.PROCESSADOPTDATE,'19800101'), 5, 2) | » /'|| SUBSTR (NVL(b.PROCESSADOPTDATE,'19800101'), 1, 4). (' 01:00:00 ',' dd-mm-yyyy hh24:mi:ss') starteff, to_date (SUBSTR (NVL(b.PROCESSADOPTDATE,'19800101'), 7, 2) |) » /'|| SUBSTR (NVL(b.PROCESSADOPTDATE,'19800101'), 5, 2) | » /'|| SUBSTR (NVL(b.PROCESSADOPTDATE,'19800101'), 1, 4). (' 01:00:00 ',' dd-mm-yyyy hh24:mi:ss') startDate.
  b.PROCESSNO revisioncodeid,
  a.BUILDINGLOT tirepercart ,
  CASE WHEN c.FIELDNAME = 'BTDrumNo1' then c.DATACHR end drumtype1.
  BOX WHEN (trim (C.DATACHR) = '1' and trim (C.FIELDNAME) = "No. Nch") and THEN click 'YES '.
  ANOTHER 'NO '.
  END may-BUG,
  case
  When c.FIELDNAME = 'Wavy'
  AND c.DATACHR = 'Use' THEN on 'YES '.
  When c.FIELDNAME = 'Wavy'
  AND c.DATACHR <>'Use' THEN 'NO '.
  fine wavy
  
  
  Now for the tables A and B I have the simple join as condition
  
  a.GTCODE = b.UPPERMATERIALCODE
  
  But I have to use two different left join conditions to join the table B and C to calculate the RESP fields according to the condition.
  Now, I have to use
  
  (trim (b.UPPERMATERIALCODE)). » -'|| Trim (b.PROCESSNO) = Trim (c.PRODUCTIONCODE) | » -'|| Trim (c.REVISIONNO)
  
  condition to get the data for the WAVY and DRUMTYPE1 fields.
  
  But at the same time, I have to use
  
  substr (Trim (b.UPPERMATERIALCODE), 1, 2). » -'|| Trim (b.PROCESSNO) = Trim (c.PRODUCTIONCODE) | » -'|| Trim (c.REVISIONNO)
  
  conition to derive the CHAFER field.
  
  And for this reason I am come so many duplicates and do not correct results for JUNEBUG, WAVY, DRUMTYPE1, and STARTDATE field.
  
  I need some advice as how can I do so that I get the correct results without duplicates for the PRIMARY KEY.
  
  Thank you
  
  Mahesh

  It would have been easier to answer if you gave us your example query that gives incorrect results - which both show us what you are doing and give us a starting point.
  Insert instructions for data, rather than only the data would have helped too, making it more quick to test this point.
  You do not show what you have done with your two outer join criteria, but I think you need to list TIREMASTERPROCESS_TEMP in your query twice, with alias names. Here's what I did:

  select a.BUILDINGSIZE, a.GTCODE, b.PROCESSNO ,  c1.FIELDNAME, c1.DATACHR, C2.FIELDNAME, C2.DATACHR
  , CASE WHEN c1.FIELDNAME='BTDrumNo1' then c1.DATACHR end drumtype1
  , CASE WHEN (trim(c2.DATACHR) = '1' and trim(c2.FIELDNAME) = 'No-of-Nch') THEN 'YES' ELSE 'NO' END CHAFER
  , case when c1.FIELDNAME='Wavy' AND c1.DATACHR='Use' THEN 'YES'
         when c1.FIELDNAME='Wavy' then 'NO' else null end wavy
  from BUILDINGSCHEDULEMASTER        A
  join materialcode b on a.GTCODE=b.UPPERMATERIALCODE
  left outer join TIREMASTERPROCESS_TEMP c1
     ON trim(b.UPPERMATERIALCODE)||'-'||trim(b.PROCESSNO) = trim(c1.PRODUCTIONCODE)||'-'||trim(c1.REVISIONNO)
  left outer join TIREMASTERPROCESS_TEMP c2
     ON substr(trim(b.UPPERMATERIALCODE),1,2)||'-'||trim(b.PROCESSNO) = trim(c2.PRODUCTIONCODE)||'-'||trim(c2.REVISIONNO)
  

  C1 is an alias for the TIREMASTERPROCESS_TEMP line which is necessary for drumtype1 and wavy, c2 is an alias for the TIREMASTERPROCESS_TEMP line required for JuneBug. There was some minor errors in your calculations of field I tried to correct.

  The results are:

  BUILD GTCO PRO FIELDNAME    DATACHR    FIELDNAME    DATACHR    DRUMTYPE1  CHA WAV
  ----- ---- --- ------------ ---------- ------------ ---------- ---------- --- ---
  AB53A AB53 108                         No-of-Nch    1                     YES
  AB42A AB42 1   BTDrumNo1    BTD-051    No-of-Nch    1          BTD-051    YES
  AB42A AB42 1   Wavy         NotUse     No-of-Nch    1                     YES NO
  AS30A AS30 104                                                            NO
  

  Hope this helps,
  David

• Generation of permutations of data between two tables

  Hello world
  I was wondering if someone can help me with the problem am I facing the work.
  Simply put, I'll just assume three tables (in reality there are more).
  
  I have the following tables:
  
  Employees
  Jobs
  Employee_Job_Ratings
  
  
  Thus, the Employees table contains a list of employees, the table of jobs a list of jobs and table Employee_Job_Ratings considers a number that represents how "good" the employee is on a particular task.
  
  I am trying to write a query that will determine the best way to assign people to the jobs team. That is to say. which combination gives the total utilization of skills. So, I have three jobs and three people like this:
  

  Job
  A
  B
  C
  
  Person
  1
  2
  3
  
  
  Ratings
  A/1: 4
  A/2: 2
  A/3: 4
  B/1: 3
  B/2: 1
  B/3: 2
  C/1: 2
  C/2: 3
  C/3: 4

  So - as you can see above, I could assign the people to jobs in the following permutations:
  

  *1*
  A - 1
  B - 2
  C - 3
  Total: 9
  
  *2*
  A - 1
  B - 3
  C - 2
  Total: 9
  
  *3*
  B - 1
  C - 2
  A - 3
  Total: 10
  
  *4*
  B - 1
  C - 3
  A - 2
  Total: 9
  
  *5*
  C - 1
  A - 2
  B - 3
  Total: 7
  
  *6*
  C - 1
  A - 3
  B - 2
  Total: 7

  But - I obviously want the best people on every job. It would be ridiculous for me to choose 5 on 3 because the total score of skills for the whole team's 7 vs 10.
  
  
  So what I actually want to do is work on all the possible combinations that I can have people on each heap, so I can do a SUM of their scores and actually give a 'score' for each combination. This will allow to see the best way to distribute the jobs within a team. That is to say, do what I did above, but automatically.
  
  The above is a trivial example, but in reality, we have many teams with a lot of jobs in each team. This is currently done manually but I know that this can be done automatically - all data are in our database so it would be great to be able to tell users "it's your best strategy for workloads" or at least give them an idea. People in charge of this planning can not use that told them the system, but it allows them to get at least an idea.
  
  
  I'm fine with everything above independently to obtain permutations in lines, that is to say, tables A and B with three lines each, make out the six possible combinations. A Cartesian join, unfortunately, is not doing the job.
  
  A person will only ever assigned to a single job.
  
  
  I know it's a big ask and I apologize for not providing any code that I wrote myself, but unfortunately the first step I'm stuck - once I've done then I will be hopefully good! :) It is not as simple as just put a person with a level 4 on a job because they could be one 3 on another, then everyone is a 1 and so they would be better used on this other employment (that is to say on a job level 3 instead of a 4 level).
  
  
  See you soon,.

  / dev/null wrote:
  Hello
  Thanks for the quick response. I use a 10g database. I tried using eval_number but get the following error (even when selecting just to double):

  ORA-37002: Oracle OLAP failed to initialize. Please contact Oracle OLAP technical support.
  ORA-33262: Analytic workspace EXPRESS does not exist.
  ORA-06512: at "SYS.DBMS_AW", line 203
  ORA-06512: at "SYS.DBMS_AW", line 212
  

  It is difficult to suggest what you can do differently when I don't know what you're doing.
  Post a full script (including the CREATION of a PACKAGE or CREATE FUCNTION, as well as the request) that recreates the problem.

  Here are the test data, I use:...

  Thank you; This seems good.

  If this I expect to get 18 rear ranks, group in six 'strategies '. It is possible that there are more people that jobs however, I will deal with that once I sorted it for the same numbers - I hope that the solution dimensionnera easily!

  My output in the example above would be so I hope that something like this:

  strategy  person_id   job_id  rating
  1         1           1       4
  1         2           2       1
  1         3           3       4
  2         1           1       4
  2         2           3       2
  2         3           2       3
  3         1           2       2
  3         2           1       3
  3         3           3       4
  4         1           2       2
  4         2           3       2
  4         3           1       2
  5         1           3       4
  5         2           1       3
  5         3           2       3
  6         1           3       4
  6         2           2       1
  6         3           1       2
  

  I hope that this example shows a little better?

  Now I see!
  The query I posted earlier discoveries strategy presenting the total highest. This isn't what you want.
  You want all the possible strategies, regardless of their classification. Here's a way to do it:

  WITH     got_job_num     AS
  (
       SELECT     job_id
       ,     ROW_NUMBER () OVER (ORDER BY job_id)     AS job_num
       ,     COUNT (*)     OVER ()                      AS job_cnt
       FROM     xxjobs
       WHERE     job_name  IN ('Mechanic', 'Painter', 'Welder')
  )
  ,     permutations     AS
  (
       SELECT     SYS_CONNECT_BY_PATH (p.person_id, ', ')               AS person_path
       ,     SYS_CONNECT_BY_PATH (r.rating, '+')               AS rating_path
       FROM     xxpeople     p
       JOIN     xxratings    r  ON   p.person_id  = r.person_id
       JOIN     got_job_num  j  ON   r.job_id       = j.job_id
       WHERE     LEVEL     = j.job_cnt
       START WITH         j.job_num  = 1
       CONNECT BY NOCYCLE  j.job_num  = PRIOR j.job_num + 1
  )
  SELECT       DENSE_RANK () OVER (ORDER BY  pe.person_path)     AS strategy
  ,       REGEXP_SUBSTR ( pe.person_path
                   , '[^ ,]+'
                 , 1
                 , jn.job_num
                 )          AS person_id
  ,       jn.job_id
  ,       REGEXP_SUBSTR ( pe.rating_path
                   , '[^+]+'
                 , 1
                 , jn.job_num
                 )          AS rating
  FROM          permutations     pe
  CROSS JOIN     got_job_num       jn
  WHERE       NOT REGEXP_LIKE ( pe.person_path || ','
                         , '( [^ ,]+,).*\1'
                   )
  ORDER BY  strategy
  ,            person_id
  ;
  

  Since you are not concerned with the total score, you don't have need eval_number, or something like that.

  Aside from that, rhe subqueries got_job_num and permutations are wjhat I psoted yesterday, except that they use your actual column and table names. If you do not want to use the person_name, then the xxpeople table is not necessary at all in this problem, and you can remove it from permutations.
  The main request is quite different.

  The above query will display all 6 strategies and number them 1 to 6. I guess that you really care what strategy gets what number. The above query uses = 4 strategy where you had strategy = 5 and vice versa. If it's a problem, I'm sure it can be fixed.
  This is the result I get:

  STRATEGY PERSON_ID      JOB_ID RATING
  -------- ---------- ---------- ----------
         1 1                   1 4
         1 2                   2 1
         1 3                   3 4
  
         2 1                   1 4
         2 2                   3 2
         2 3                   2 3
  
         3 1                   2 2
         3 2                   1 3
         3 3                   3 4
  
         4 1                   3 4
         4 2                   1 3
         4 3                   2 3
  
         5 1                   2 2
         5 2                   3 2
         5 3                   1 2
  
         6 1                   3 4
         6 2                   2 1
         6 3                   1 2
  

• match Word between two table

  Hello

  I want to retrieve those records where full word of the 'text' column does not match with the word in the column 'object' of another table.

  for example

  for id 101 table abc, it contains the Word as "Air Blow" which is exactly with 101 id of table of bcd. I don't want the case

  With abc as

  (

  Select 101 as id, "Air Blow" as double union text

  Select 102, 'Air' Union double

  Select 103, 'Sink' double union

  Select 104, 'Air press' Union double

  Select 105, 'Putting out fires' double

  ),

  BCD as

  (

  Select 101 as id, "Air blown near the mount" as subject, 'WE' as a double union countries

  Select 102, "Air breath seen here," 'China' of the double union

  Select 103, 'Throughput', 'canada' to double union

  Select 104, 'air pressure', 'UK' Union double

  Select 105, 'Fire engine', 'Ubdy' from dual

  )

  Select * from abc, bcd

  where abc.id = bcd.id

  and upper (abc.text) <>(bcd.subject) upper

  OUTPUT

  ID	TEXT	ID_1	Object	COUNTRY
  102	Air blown	102	Blowing air saw	China
  103	Water flow	103	Water flow	Canada
  105	Putting out fires	105	Firefighter	Ubdy

  Another corresponding original solution post results...

  SQL > ed
  A written file afiedt.buf

  1 with abc as)
  2. Select 101 as id, "Air BLOW" as double union text
  3. Select 102, 'Air' Union double
  4. Select 103, 'Sink' double union
  5. Select 104, 'Air press' Union double
  6 select 105, 'Fire extinguish' from dual
  7    ),
  8 bcd as
  9    (
  10. Select 101 as id, "Air blown near the mount" as subject, 'WE' as a double union countries
  11. Select 102, "Air breath seen here," 'China' of the double union
  12. Select 103, 'Throughput', 'canada' to double union
  13. Select 104, 'air pressure', 'UK' Union double
  14 select 105, 'Fire engine', 'Ubdy' from dual
  15)
  16. Select *.
  17 ABC join bcd on (abc.id = bcd.id)
  18 * where no regexp_like (bcd.subject, abc.text, 'i')
  SQL > /.
  TEXT ID COUNTR OBJECT ID
  ---------- --------------- ---------- ----------------------- ------
  102 air blowing here view China 102
  103 103 water flow flow of water canada
  Fire 105 105 fire Ubdy

  3 selected lines.