Improve company productivity with a Business Account.Sign Up

x
?
Solved

Oracle Query Help

Posted on 2011-03-17
12
Medium Priority
?
285 Views
Last Modified: 2012-05-11
Dear Experts

I need your help

I have query as follows

/* Formatted on 18/03/2011 11:42:48 AM (QP5 v5.136.908.31019) */
SELECT CITY.CITYDESCRIPTION,
       COMPANY.COMPANYACCNO,
       COMPANY.COMPANYNAME,
       COMPANY.ADDRESS,
       COMPANY.PINCODE,
       COMPANY.PHONENO1,
       MEMBER.MEMBERACCNO,
       MEMBER.MEMBERNAME,
       TITLE.TITLEDESCRIPTION
  FROM CITY
       JOIN COMPANY
          ON (CITY.CITY_ID = COMPANY.CITY_ID)
       JOIN MEMBER
          ON (COMPANY.COMPANY_ID = MEMBER.COMPANY_ID)
       JOIN TITLE
          ON (MEMBER.SPOUSETITLE_ID = TITLE.TITLE_ID)
 WHERE (MEMBER.FLAG = 0)

In attached excel sheet, I have given actual output and required output.
I need to alter this query to get required output
Urgent.xls
0
Comment
Question by:GRChandrashekar
  • 6
  • 4
  • 2
12 Comments
 
LVL 41

Expert Comment

by:Sharath
ID: 35163202
How many max. no. of MEMBERNAMEs you can have for a COMPANYACCNO?
0
 

Author Comment

by:GRChandrashekar
ID: 35163205
THREE
0
 
LVL 41

Expert Comment

by:Sharath
ID: 35163236
Can you check this?
SELECT CITYDESCRIPTION, COMPANYACCNO, COMPANYNAME, ADDRESS, PINCODE, PHONENO1, MEMBERACCNO, 
         MAX(CASE 
               WHEN rn = 1 THEN MEMBERNAME 
             END) MEMBERNAME, 
         MAX(CASE 
               WHEN rn = 2 THEN MEMBERNAME 
             END) MEMBERNAME2, 
         MAX(CASE 
               WHEN rn = 3 THEN MEMBERNAME 
             END) MEMBERNAME3, 
         TITLEDESCRIPTION 
    FROM (SELECT CITY.CITYDESCRIPTION, COMPANY.COMPANYACCNO, COMPANY.COMPANYNAME, COMPANY.ADDRESS, COMPANY.PINCODE, 
                 COMPANY.PHONENO1, MEMBER.MEMBERACCNO, MEMBER.MEMBERNAME, TITLE.TITLEDESCRIPTION, 
                 ROW_NUMBER() 
                   OVER(PARTITION BY COMPANY.COMPANYACCNO ORDER BY MEMBER.MEMBERNAME) rn 
            FROM CITY 
                 JOIN COMPANY 
                   ON (CITY.CITY_ID = COMPANY.CITY_ID) 
                 JOIN MEMBER 
                   ON (COMPANY.COMPANY_ID = MEMBER.COMPANY_ID) 
                 JOIN TITLE 
                   ON (MEMBER.SPOUSETITLE_ID = TITLE.TITLE_ID) 
           WHERE (MEMBER.FLAG = 0)) t1 
GROUP BY CITYDESCRIPTION, COMPANYACCNO, COMPANYNAME, ADDRESS, PINCODE, PHONENO1, MEMBERACCNO, TITLEDESCRIPTION

Open in new window

0
Free Tool: Site Down Detector

Helpful to verify reports of your own downtime, or to double check a downed website you are trying to access.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

 

Author Comment

by:GRChandrashekar
ID: 35163246
FOR COMPANYACCNO

THERE WILL BE
3 MEMBERACCNO
3 TITLE
3 MEMBERNAMES
0
 
LVL 41

Expert Comment

by:Sharath
ID: 35163255
Run my above query and see if you are able to get MemberName in one record. The same logic can be implemented for other columns also.
0
 

Author Comment

by:GRChandrashekar
ID: 35163262
Yes it works request help to include
3 MEMBERACCNO
3 TITLE
3 MEMBERNAMES
0
 
LVL 41

Accepted Solution

by:
Sharath earned 1800 total points
ID: 35163398
check this.
SELECT CITYDESCRIPTION, 
         COMPANYACCNO, 
         COMPANYNAME, 
         ADDRESS, 
         PINCODE, 
         PHONENO1, 
         MAX(CASE 
               WHEN rn = 1 THEN MEMBERACCNO 
             END) MEMBERACCNO, 
         MAX(CASE 
               WHEN rn = 2 THEN MEMBERACCNO 
             END) MEMBERACCNO2, 
         MAX(CASE 
               WHEN rn = 3 THEN MEMBERACCNO 
             END) MEMBERACCNO3, 
         MAX(CASE 
               WHEN rn = 1 THEN MEMBERNAME 
             END) MEMBERNAME, 
         MAX(CASE 
               WHEN rn = 2 THEN MEMBERNAME 
             END) MEMBERNAME2, 
         MAX(CASE 
               WHEN rn = 3 THEN MEMBERNAME 
             END) MEMBERNAME3, 
         MAX(CASE 
               WHEN rn = 1 THEN TITLEDESCRIPTION 
             END) TITLEDESCRIPTION, 
         MAX(CASE 
               WHEN rn = 2 THEN TITLEDESCRIPTION 
             END) TITLEDESCRIPTION2, 
         MAX(CASE 
               WHEN rn = 3 THEN TITLEDESCRIPTION 
             END) TITLEDESCRIPTION3 
    FROM (SELECT CITY.CITYDESCRIPTION, 
                 COMPANY.COMPANYACCNO, 
                 COMPANY.COMPANYNAME, 
                 COMPANY.ADDRESS, 
                 COMPANY.PINCODE, 
                 COMPANY.PHONENO1, 
                 MEMBER.MEMBERACCNO, 
                 MEMBER.MEMBERNAME, 
                 TITLE.TITLEDESCRIPTION, 
                 ROW_NUMBER() 
                   OVER(PARTITION BY COMPANY.COMPANYACCNO ORDER BY MEMBER.MEMBERNAME) rn 
            FROM CITY 
                 JOIN COMPANY 
                   ON (CITY.CITY_ID = COMPANY.CITY_ID) 
                 JOIN MEMBER 
                   ON (COMPANY.COMPANY_ID = MEMBER.COMPANY_ID) 
                 JOIN TITLE 
                   ON (MEMBER.SPOUSETITLE_ID = TITLE.TITLE_ID) 
           WHERE (MEMBER.FLAG = 0)) t1 
GROUP BY CITYDESCRIPTION, 
         COMPANYACCNO, 
         COMPANYNAME, 
         ADDRESS, 
         PINCODE, 
         PHONENO1

Open in new window

0
 

Author Comment

by:GRChandrashekar
ID: 35163686
One last help. I am tyring to split the address into 4 parts but not sure where to split so it gives error

SELECT CITYDESCRIPTION,
         COMPANYACCNO,
         COMPANYNAME,
         
   TRIM (SUBSTR (COMPANY.ADDRESS,
                             1,
                             INSTR (COMPANY.ADDRESS, '$', 1, 1) - 1
                            )
                    ) AS ADDRESS1,
               TRIM (SUBSTR (COMPANY.ADDRESS,
                             INSTR (COMPANY.ADDRESS, '$', 1, 1) + 1,
                               (INSTR (COMPANY.ADDRESS, '$', 1, 2) - 1
                               )
                             - (INSTR (COMPANY.ADDRESS, '$', 1, 1))
                            )
                    ) AS ADDRESS2,
               TRIM (SUBSTR (COMPANY.ADDRESS,
                             INSTR (COMPANY.ADDRESS, '$', 1, 2) + 1,
                               (INSTR (COMPANY.ADDRESS, '$', 1, 3) - 1
                               )
                             - (INSTR (COMPANY.ADDRESS, '$', 1, 2))
                            )
                    ) AS ADDRESS3,
               TRIM (SUBSTR (COMPANY.ADDRESS,
                             INSTR (COMPANY.ADDRESS, '$', 1, 3) + 1
                            )
                    ) AS ADDRESS4,
         PINCODE,
         PHONENO1,
         MAX(CASE
               WHEN rn = 1 THEN MEMBERACCNO
             END) MEMBERACCNO,
         MAX(CASE
               WHEN rn = 2 THEN MEMBERACCNO
             END) MEMBERACCNO2,
         MAX(CASE
               WHEN rn = 3 THEN MEMBERACCNO
             END) MEMBERACCNO3,
         MAX(CASE
               WHEN rn = 1 THEN MEMBERNAME
             END) MEMBERNAME,
         MAX(CASE
               WHEN rn = 2 THEN MEMBERNAME
             END) MEMBERNAME2,
         MAX(CASE
               WHEN rn = 3 THEN MEMBERNAME
             END) MEMBERNAME3,
         MAX(CASE
               WHEN rn = 1 THEN TITLEDESCRIPTION
             END) TITLEDESCRIPTION,
         MAX(CASE
               WHEN rn = 2 THEN TITLEDESCRIPTION
             END) TITLEDESCRIPTION2,
         MAX(CASE
               WHEN rn = 3 THEN TITLEDESCRIPTION
             END) TITLEDESCRIPTION3
    FROM (SELECT CITY.CITYDESCRIPTION,
                 COMPANY.COMPANYACCNO,
                 COMPANY.COMPANYNAME,
                 COMPANY.ADDRESS,
                 COMPANY.PINCODE,
                 COMPANY.PHONENO1,
                 MEMBER.MEMBERACCNO,
                 MEMBER.MEMBERNAME,
                 TITLE.TITLEDESCRIPTION,
                 ROW_NUMBER()
                   OVER(PARTITION BY COMPANY.COMPANYACCNO ORDER BY MEMBER.MEMBERNAME) rn
            FROM CITY
                 JOIN COMPANY
                   ON (CITY.CITY_ID = COMPANY.CITY_ID)
                 JOIN MEMBER
                   ON (COMPANY.COMPANY_ID = MEMBER.COMPANY_ID)
                 JOIN TITLE
                   ON (MEMBER.SPOUSETITLE_ID = TITLE.TITLE_ID)
           WHERE (MEMBER.FLAG = 0)) t1
GROUP BY CITYDESCRIPTION,
         COMPANYACCNO,
         COMPANYNAME,
         PINCODE,
         PHONENO1
0
 
LVL 32

Assisted Solution

by:awking00
awking00 earned 200 total points
ID: 35165525
Use regular expressions to split the address. I think it's a little cleaner. See attached.
query.txt
0
 

Author Comment

by:GRChandrashekar
ID: 35165791
Can you please help me re-write my query
0
 
LVL 32

Expert Comment

by:awking00
ID: 35165841
You should just be able to replace this -
TRIM (SUBSTR (COMPANY.ADDRESS,
                             1,
                             INSTR (COMPANY.ADDRESS, '$', 1, 1) - 1
                            )
                    ) AS ADDRESS1,
               TRIM (SUBSTR (COMPANY.ADDRESS,
                             INSTR (COMPANY.ADDRESS, '$', 1, 1) + 1,
                               (INSTR (COMPANY.ADDRESS, '$', 1, 2) - 1
                               )
                             - (INSTR (COMPANY.ADDRESS, '$', 1, 1))
                            )
                    ) AS ADDRESS2,
               TRIM (SUBSTR (COMPANY.ADDRESS,
                             INSTR (COMPANY.ADDRESS, '$', 1, 2) + 1,
                               (INSTR (COMPANY.ADDRESS, '$', 1, 3) - 1
                               )
                             - (INSTR (COMPANY.ADDRESS, '$', 1, 2))
                            )
                    ) AS ADDRESS3,
               TRIM (SUBSTR (COMPANY.ADDRESS,
                             INSTR (COMPANY.ADDRESS, '$', 1, 3) + 1
                            )
                    ) AS ADDRESS4,
with this -
regexp_substr(address,'^[^$]+') as address1,
trim('$' from regexp_substr(address,'\$[^$]+')) as address2,
trim('$' from regexp_substr(address,'\$[^$]+',1,2)) as address3,
trim('$' from regexp_substr(address,'\$[^$]+',1,3)) as address4,
0
 

Author Comment

by:GRChandrashekar
ID: 35170104
@Sharath_123:

Request help: Please alter the query to split address. Please help
0

Featured Post

Get your problem seen by more experts

Be seen. Boost your question’s priority for more expert views and faster solutions

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

Join & Write a Comment

How to Unravel a Tricky Query Introduction If you browse through the Oracle zones or any of the other database-related zones you'll come across some complicated solutions and sometimes you'll just have to wonder how anyone came up with them.  …
Shell script to create broker configuration file using current broker Configuration, solely for purpose of backup on Linux. Script may need to be modified depending on OS-installation. Please deploy and verify the script in a test environment.
This video shows how to recover a database from a user managed backup
This video explains what a user managed backup is and shows how to take one, providing a couple of simple example scripts.

588 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question