SQL to check for all values in column - sql

Question

I have the following table in Oracle DB.
ID VALUE
-----------
1 1
1 2
1 3
2 1
2 2
3 1
3 2
3 3
4 1
How can I select ID's which have all 3 values (1,2,3)

Answer 1

The simplest option is generally something like this
SQL> ed
Wrote file afiedt.buf
1 with x as (
2 select 1 id, 1 val from dual union all
3 select 1 id, 2 val from dual union all
4 select 1 id, 3 val from dual union all
5 select 2 id, 1 val from dual union all
6 select 2 id, 2 val from dual union all
7 select 3 id, 1 val from dual union all
8 select 3 id, 2 val from dual union all
9 select 3 id, 3 val from dual union all
10 select 4 id, 1 val from dual
11 )
12 select id
13 from x
14 where val in (1,2,3)
15 group by id
16* having count(distinct val) = 3
SQL> /
ID
----------
1
3
The WHERE clause identifies the values you're interested in. The HAVING clause tells you how many of those values need to exist. If you wanted all the rows that had at least 2 of the 3 values, for example, you'd change the HAVING clause to look for a COUNT of 2.
If a particular val is guaranteed to occur at most once per id, you can eliminate the distinct in the HAVING clause.

Answer 2

Try this:
SELECT ID
FROM TABLENAME T
WHERE EXISTS (SELECT *
FROM TABLENAME T1
WHERE T1.ID = T.ID AND T1.VALUE = '1')
AND EXISTS (SELECT *
FROM TABLENAME T2
WHERE T1.ID = T.ID AND T2.VALUE = '2')
AND EXISTS (SELECT *
FROM TABLENAME T3
WHERE T1.ID = T.ID AND T2.VALUE = '3')
or
SELECT ID
FROM TABLENAME T
WHERE (SELECT COUNT( * )
FROM (SELECT VALUE
FROM TABLENAME T1
WHERE T1.ID = T.ID
GROUP BY VALUE)) = 3;
where 3 is number of values which can be calculated by a
SELECT COUNT( * )
FROM TABLENAME T1
GROUP BY VALUE
so this will be general purpose:
SELECT ID
FROM TABLENAME T
WHERE (SELECT COUNT( * )
FROM (SELECT VALUE
FROM TABLENAME T1
WHERE T1.ID = T.ID
GROUP BY VALUE)) = (SELECT COUNT( * )
FROM TABLENAME T2
GROUP BY VALUE)

Answer 3

Here's an option... each expression in the HAVING clause is counting the number of values that are found equal to 1, 2, or 3. If any of these counts is less than 1, then the ID will not be returned.
http://sqlfiddle.com/#!4/00fdc/8
SELECT ID
FROM myTable
GROUP BY ID
HAVING
SUM(DECODE(VALUE, 1, 1, 0)) > 0 AND
SUM(DECODE(VALUE, 2, 1, 0)) > 0 AND
SUM(DECODE(VALUE, 3, 1, 0)) > 0
EDIT - To require value 1, and either 2 or 3:
SELECT ID
FROM myTable
GROUP BY ID
HAVING
SUM(DECODE(VALUE, 1, 1, 0)) > 0 AND
(
SUM(DECODE(VALUE, 2, 1, 0)) > 0 OR
SUM(DECODE(VALUE, 3, 1, 0)) > 0
)

Answer 4

select id from (select id,sum(case when value=1 then 1 else 0 end) as 'v1',
sum(case when value=2 then 1 else 0 end) as 'v2',
sum(case when value=3 then 1 else 0 end) as 'v3'
from orac group by id) as final
where v1>0 and v2>0 and v3>0

Answer 5

With this option you will get more than the IDs, up to your application to select the column you want:
SELECT ID,
sum(CASE WHEN VALUE = 1 THEN 1 ELSE 0 END) AS ONE,
sum(CASE WHEN VALUE = 2 THEN 1 ELSE 0 END) AS TWO,
sum(CASE WHEN VALUE = 3 THEN 1 ELSE 0 END) AS THREE
FROM MYTABLE
GROUP BY ID
HAVING ONE >= 1 AND TWO >= 1 AND THREE >= 1;
alternatively if your case is specific (only values 1, 2, 3 are possible, and no duplicate values are allowed), then you could try the following one:
SELECT ID,
count(VALUE) AS VALUECOUNT
FROM MYTABLE
GROUP BY ID
HAVING VALUECOUNT = 3;
I would take care before going that way, as you might get side effects if later you want to add additional values. But it's still worth proposing if your current case fits the restrictions given above.
And, of course, if you don't like the idea of fetching these intermediate counts, enclose the queries I gave within another select
SELECT ID FROM (
...
)