当其中一列和另一列大于

Question

Consider the following dataset:

+---------------------+
| ID | NAME  | VALUE  |
+---------------------+
| 1  | a     | 0.2    |
| 1  | b     | 8      |  
| 1  | c     | 3.5    |
| 1  | d     | 2.2    |
| 2  | b     | 4      |
| 2  | c     | 0.5    |
| 2  | d     | 6      |
| 3  | a     | 2      |
| 3  | b     | 4      |
| 3  | c     | 3.6    |
| 3  | d     | 0.2    |
+---------------------+

I'm tying to develop a sql select statement that returns the top or distinct ID where NAME 'a' and 'b' both exist and both of the corresponding VALUE's are >= '1'. Thus, the desired output would be:

+---------------------+
| ID | NAME  | VALUE  |
+---------------------+
| 3  | a     | 2      |
+----+-------+--------+

Appreciate any assistance anyone can provide.

Answer 1

You can try to use MIN window function and some condition to make it.

SELECT * FROM (
    SELECT *,
    MIN(CASE WHEN NAME = 'a' THEN [value] end) OVER(PARTITION BY ID) aVal,
    MIN(CASE WHEN NAME = 'b' THEN [value] end) OVER(PARTITION BY ID) bVal
    FROM T
) t1
WHERE aVal >1 and bVal >1 and aVal = [Value]

sqlfiddle

Answer 2

This seems like a group by and having query:

select id
from t
where name in ('a', 'b')
having count(*) = 2 and
       min(value) >= 1;

No subqueries or join s are necessary.

The where clause filters the data to only look at the "a" and "b" records. The count(*) = 2 checks that both exist. If you can have duplicates, then use count(distinct name) = 2 .

Then, you want the minimum value to be 1, so that is the final condition.

I am not sure why your desired results have the "a" row, but if you really want it, you can change the select to:

select id, 'a' as name,
       max(case when name = 'a' then value end) as value

Answer 3

you can use in and sub-query

select top 1 * from t
where t.id in
(
select id from t
where name in ('a','b')
group by id 
having sum(case when value>1 then 1 else 0)>=2
)
order by id