SQL | 如何 GROUP BY 并根据两个条件获得不同的 COUNT？

Question

Given a table with two attributes "ID" and "bug" where "bug" is valued either 0 or 1, meaning
person with that "ID" either made a commit with a bug or with no bug. The same "ID" may re-appear in the table. The task is to print the distinct ID's with the number of buggy commits and non-buggy commits. The schema and an example table are as follows

CREATE TABLE commits (
ID INT NOT NULL,
bug INT);

The table is

 _ID__|bugs_
 | 121|1|
 | 121|1|
 | 121|0|
 | 111|1|
 | 111|0|
 | 111|1|
 | 131|0|
 | 131|0|
 | 121|1|
 | 111|0|
 | 121|1|
 | 111|0|
 | 121|1|
 | 131|0|
 --------

The solution should be(not exactly, but like this)

111|3|4
121|6|2
131|1|5

The way I went about it was to create two tables, one where "bug" equals 1 and the other where it is 0. Then used GROUP BY and count on these to get the tables with buggy and non buggy commits. Then I INNER JOINED them using "ID". Finally, accessed the ID and the count of "buggy" & "notbuggy" columns with COUNT(). This was my Query, which worked

select distinct buggy.id,buggy.buggycount,notbuggy.notbuggycount from 
(select id,count(bug) as buggycount 
from (select id,bug from commits where bug = 1) a group by id) buggy 
inner join 
(select id,count(bug) as notbuggycount
from (select id,bug from commits where bug = 0) a group by id) notbuggy
where buggy.id = notbuggy.id
order by buggy.id;

Is there a better and shorter way to do this? Ideally, without any INNER JOIN. Thank You.

Answer 1

Use conditional aggregation:

SELECT id,
       SUM(bug = 0) notbuggycount,
       SUM(bug = 1) buggycount
FROM commits
GROUP BY id;

or:

SELECT id,
       SUM(1 - bug) notbuggycount,
       SUM(bug) buggycount
FROM commits
GROUP BY id;

See the demo .