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The issue is caused by the following setup:

Duplicate entries for BSDK Bestseller application in FND_PRODUCT_INSTALLATIONS table.

+++++++++++++++++++++++++++++++++++++++++++++++++++++

When the following queries are executed, two entries are found for the same application in FND_PRODUCT_INSTALLATIONS table with same application name and application id but different oracle id.

1 . SELECT APPLICATION_ID,COUNT(*)
FROM FND_PRODUCT_INSTALLATIONS
GROUP BY APPLICATION_ID
HAVING COUNT(*) > 1;
———————————————————-
Upon execution the result will be as follows:

APPLICATION_ID COUNT(*)
————– ———-
20045  3

———————————————————-

2. SELECT FPI.APPLICATION_ID, FPI.ORACLE_ID, FAL.APPLICATION_NAME
FROM FND_PRODUCT_INSTALLATIONS FPI, FND_APPLICATION_TL FAL
WHERE FPI.APPLICATION_ID= ” ID from above Query”
AND FPI.APPLICATION_ID = FAL.APPLICATION_ID;
———————————————————-
Upon execution the result will be as follows:-

APPLICATION_ID  ORACLE_ID APPLICATION_NAME
————– ———- ————————————————–
20045        300 XXX Custom Application
20045      20034 XXX Custom Application

Note: The data over here is customer specific, it may be different for case to case basis.
——————————————————–

Solution

SELECT LAST_UPDATE_DATE, APPLICATION_ID, STATUS, ORACLE_ID FROM
FND_PRODUCT_INSTALLATIONS WHERE APPLICATION_ID = <–this value is customer specific–>;

NOTE: There will be two duplicate entries in the FND_PRODUCT_INSTALLATIONS table.

SQL> /

LAST_UPDATE_DATE   APPLICATION_ID S  ORACLE_ID
—————— ————– – ———-
12-Aug-14                   20045 I        300
12-Aug-14                   20045 I      20034

3. In the FND_PRODUCT_INSTALLATIONS table, check for the STATUS = ‘L’ for each application which from the above query when executed.

4. Remove the application which has values for STATUS = ‘L’ and application_id same as the other one.

(STATUS =>, I means installed, L means custom, N means not installed, and S means installed as shared product in FND_PRODUCT_INSTALLATIONS table).

5. Take backup of FND_PRODUCT_INSTALLATIONS table (either with export or with create table)

6. SQL> delete from FND_PRODUCT_INSTALLATIONS where APPLICATION_ID= 20023 and ORACLE_ID=20045;

7. SQL> SELECT FPI.APPLICATION_ID, FPI.ORACLE_ID, FAL.APPLICATION_NAME FROM FND_PRODUCT_INSTALLATIONS FPI, FND_APPLICATION_TL FAL WHERE FPI.APPLICATION_ID=20023 AND FPI.APPLICATION_ID = FAL.APPLICATION_ID;

APPLICATION_ID  ORACLE_ID APPLICATION_NAME
————– ———- ————————————————–

SQL> SELECT APPLICATION_ID,COUNT(*)
FROM FND_PRODUCT_INSTALLATIONS
GROUP BY APPLICATION_ID
HAVING COUNT(*) > 1;  2    3    4

no rows selected

8. SQL> commit