How to do order by min() of some column? - sql

Question

I am pretty new to SQL. I want a query which should do order by on min of some column. Below is the query i want.
SELECT *
FROM (
SELECT p.PROJECT_ID,
p.PROJECT_NAME,
p.PROJECT_TYPE
FROM PROJECT p
LEFT OUTER JOIN code c
ON p.PROJECT_ID= c.PROJECT_ID
WHERE p.PROJECT_NAME IN ('test')
ORDER BY min(c.LABEL) ASC
)
WHERE rownum <= 25;
Why i need it this way is. I have one table PROJECT.
PROJECT_ID PROJECT_NAME PROJECT_TYPE
1 a test1
2 b test2
i have another table code which has project_id as foreign key.
ID PROJECT_ID LABEL
1 1 a
2 1 b
3 1 c
4 2 d
now when i will join it on project_id and make order by on code.label it will give me 4 records three with project id 1 and 1 with project id 2. But my requirement is to sort the project based on the codes label. so logically i want two records . One for project id 1 with min vale of label of all the possible combinations of project id 1 i.e with label a and other with project id 2. So that's why i want to sort it based on min of code label. I cannot use group by as it will degrade the performance.

Answer 1

For use a MIN( )
you need a group by eg:
SELECT *
FROM (
SELECT p.PROJECT_ID,
p.PROJECT_NAME,
p.PROJECT_TYPE
FROM PROJECT p
LEFT OUTER JOIN code c
ON p.codeId=c.ID
WHERE p.PROJECT_NAME IN ('test')
GROUP BY .PROJECT_ID,
p.PROJECT_NAME,
p.PROJECT_TYPE
ORDER BY min(c.LABEL) ASC
)
WHERE rownum <= 25;
and in some db you must select the column you need for order by eg:
SELECT *
FROM (
SELECT p.PROJECT_ID,
p.PROJECT_NAME,
p.PROJECT_TYPE,
min(c.LABEL)
FROM PROJECT p
LEFT OUTER JOIN code c
ON p.codeId=c.ID
WHERE p.PROJECT_NAME IN ('test')
GROUP BY .PROJECT_ID,
p.PROJECT_NAME,
p.PROJECT_TYPE
ORDER BY min(c.LABEL) ASC
)
WHERE rownum <= 25;