Reputation: 59
ID | Material | Material_Num | User | Assign User |
---|---|---|---|---|
1 | stick | 1111 | Billy Bob | "select list of users" |
2 | stone | 1112 | Jane Doe | "select list of users" |
3 | rock | 1113 | John Deer | "select list of users" |
4 | slab | 1114 | "select list of users" | |
5 | brick | 1115 | "select list of users" |
There will be a save button which will update the user column with the value in the assign user column. If the select list can be consolidated under the user column and keep the updated values that would be even better.
These are the steps I have taken so far:
SELECT DISTINCT
MD.ID,
MD.MATERIAL,
MD.MATERIAL_NUMBER,
FIRST_NAME || ' ' || LAST_NAME USER,
USER AS DISPLAY_VALUE
FROM MATERIAL_DATA MD
INNER JOIN MATERIAL_OWNER_DATA MOD
ON MD.USER = MOD.OWNER
SELECT
T1.DISPLAY_VALUE as DISPLAY_VALUE
FROM (
( SELECT
TM.FIRST_NAME || ' ' || TM.LAST_NAME as DISPLAY_VALUE,
upper(TM.NUMBER) as RETURN_VALUE,
1 as ORD
FROM
TEAM_MEMBERS TM
WHERE
TM.NUMBER IS NOT NULL
AND TM.BOOL = 'Y'
)
UNION ALL
( SELECT
'Not Required' as DISPLAY_VALUE,
'Not Required' as RETURN_VALUE,
0 as ORD
FROM DUAL
)
ORDER BY 3, 1
) T1
So far as I have searched I have not seen a step-by-step way to add the select list to the grid. It's possible I've oversimplified my process but this is the best way I could describe it. If answering this is too cumbersome a link to a good source would also be appreciated.
My company won't allow the use of css or javascript so this has to be done strictly with sql and what's already available in apex_oracle.
Upvotes: 0
Views: 2321
Reputation: 142705
The 2nd query you posted should be
SELECT t1.display_value, t1.return_value FROM (...) t1
(at least, looks like to me).
Upvotes: 0