mysql - COUNT 正在计算 LEFT JOIN 上不存在的内容。为什么？

Question

我遇到了问题。

我有一张名为 usersbycourse 的表，它显示了以下信息:

+------------+-----------------+--------+-----------+-------+-----------------+-----------------+
| instanceid | shortname       | userid | firstname | logid | lastaccessdelta | modulesfinished |
+------------+-----------------+--------+-----------+-------+-----------------+-----------------+
|          2 | PJU             |     74 | Robin     |   766 |         1662246 |               0 |
|          3 | Fundgest-GRHN1A |     75 | Batman    |   867 |         1576725 |               0 |
|          3 | Fundgest-GRHN1A |     77 | Abigobeu  |  1004 |          610480 |               0 |
+------------+-----------------+--------+-----------+-------+-----------------+-----------------+

和这个 SQL:

SELECT
mdl_course.id,
mdl_course.shortname,
COUNT(CASE WHEN usersbycourse.modulesfinished = 1 THEN NULL ELSE 1 END) AS studentcount
FROM mdl_course LEFT JOIN usersbycourse ON mdl_course.id = usersbycourse.instanceid
GROUP BY mdl_course.id;

SQL 的结果是:

+----+-----------------+--------------+
| id | shortname       | studentcount |
+----+-----------------+--------------+
|  1 | Unity I         |            1 |
|  2 | PJU             |            1 |
|  3 | Fundgest-GRHN1A |            2 |
|  4 | asdzxc2         |            1 |
+----+-----------------+--------------+

但是为什么？在 SQL 内部没有 Unity I，也没有 asdzxc2。我如何产生这样的结果:

+----+-----------------+--------------+
| id | shortname       | studentcount |
+----+-----------------+--------------+
|  1 | Unity I         |            0 |
|  2 | PJU             |            1 |
|  3 | Fundgest-GRHN1A |            2 |
|  4 | asdzxc2         |            0 |
+----+-----------------+--------------+

?

编辑:

我只想计算 modulesfinished = 0 的行

Answer 1

你要找的是SUM而不是COUNT，即

SELECT
mdl_course.id,
mdl_course.shortname,
SUM(CASE WHEN usersbycourse.modulesfinished = 0 THEN 1 ELSE 0 END) AS studentcount
FROM mdl_course LEFT JOIN usersbycourse ON mdl_course.id = usersbycourse.instanceid
GROUP BY mdl_course.id;