选择所有具有相同ID的行

Question

I have this table:

ID | Part
1  | A
1  | B
1  | C
2  | B
2  | C
3  | A
3  | D
3  | E
4  | B
4  | D

and want a query that will grab all ID's that have an A, and return a list of all other parts with that ID.

eg: Want Parts related to B:

Part | Count
 A   | 1
 C   | 2
 D   | 1

What I have currently:

SELECT * FROM tble WHERE ID IN (SELECT DISTINCT ID FROM tble t WHERE Part = ?)
GROUP BY Part ORDER BY COUNT(Part) DESC

This works, but is quite slow and I'm looking to improve it, but having difficulty

Answer 1

Simplify this.. Once you have the logic down, then add back in the SELECT * FROM..

SELECT Part, COUNT(Part) As Count_of_Part
GROUP BY Part ORDER BY COUNT(Part) DESC

Answer 2

Do a join from the table back to itself on ID, and then count the distinct values that pop up:

SELECT b.part, COUNT(DISTINCT b.id)
FROM 
    table as a 
    INNER JOIN table as b ON
        a.id = b.id AND
        a.part <> b.part
WHERE
    a.part = 'B'
GROUP BY b.part

Answer 3

Your query is not unreasonable, although the distinct is unnecessary and I would use exists rather than in . And, the outer select needs to be fixed for the aggregation

SELECT t.part, COUNT(*)
FROM tble t
WHERE EXISTS (SELECT 1 FROM tble t2 WHERE t2.ID = t.ID AND t2.Part = ?)
GROUP BY t.Part
ORDER BY COUNT(*) DESC;

Then, to optimize this query, you want an index:

create index idx_tble_id_part on tble(id, part);

Answer 4

This can be simply done by joining back to the table:

SELECT t1.part
,count(*)
FROM tble t1
INNER JOIN tble t ON t.id = t1.id
                  AND t.part = 'B'
                  AND t1.part <> t.part
GROUP BY t1.part

SQL Fiddle Demo

Answer 5

You should be able to do this by grouping the data.

Try something like this:

SELECT part, COUNT(id) AS TotalCountId
FROM TABLE_NAME
GROUP BY ID