我有一个用户表、一个任务表和一个提醒表。我想返回每个用户的任务数和提醒数。当我只计算其中一个(提醒或任务)时，我可以让它工作，但是当我在一个查询中计算它们时，由于某种原因，它们会彼此相乘。
SQLFiddle:http://www.sqlfiddle.com/#!9/f0d6696/1/0
这是我到目前为止的查询:
SELECT
users.name,
COUNT(reminders.id),
COUNT(tasks.id)
FROM users
LEFT JOIN reminders on users.id = reminders.id
LEFT JOIN tasks on users.id = tasks.id
GROUP BY users.id
这就是我的用户表的样子:
+---------------------------------------+
| ID | Name | Email |
+---------------------------------------+
| 1 | John Smith | <a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="046e77696d706c446169656d682a676b69" rel="noreferrer noopener nofollow">[email protected]</a> |
| 2 | Mark Twain | <a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="2d40595a4c44436d4f4242465e034e4240" rel="noreferrer noopener nofollow">[email protected]</a> |
| 3 | Elon Musk | <a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="186b68797b7d357c6d7c7d587d75797174367b7775" rel="noreferrer noopener nofollow">[email protected]</a>|
+---------------------------------------+
这就是我的任务表的样子:
+------------------------------------------------+
| ID | Title | Text | Status |
+------------------------------------------------+
| 1 | Dishes | Kitchen = nasty | incomplete|
| 1 | Library | drop off books | complete |
| 3 | Gym | get swole dude | incomplete|
+------------------------------------------------+
这就是我的提醒表的样子:
+------------------------------------+
| ID | Title | Text |
+------------------------------------+
| 1 | Dishes | Kitchen = nasty |
| 2 | Library | drop off books |
| 1 | Gym | get swole dude |
+------------------------------------+
我希望从上述查询中得到以下结果:
+-------------------------------------------+
| Name | Tasks | Reminders |
+-------------------------------------------+
| John Smith | 2 | 2 |
| Mark Twain | 1 | 0 |
| Elon Musk | 0 | 1 |
+-------------------------------------------+
我实际上得到以下信息:
+-------------------------------------------+
| Name | Tasks | Reminders |
+-------------------------------------------+
| John Smith | 4 | 4 | <---2 tasks x 2 reminders?
| Mark Twain | 1 | 0 |
| Elon Musk | 0 | 1 |
+-------------------------------------------+
最佳答案
您将获得交叉连接，每项任务都有提醒。
尝试
select
users.name,
remindercount,
taskcount
FROM users
LEFT JOIN (select id, count(*) as remindercount from reminders group by id) reminders on users.id = reminders.id
LEFT JOIN (select id, count(*) as taskcount from tasks group by id) tasks on users.id = tasks.id
关于MySQL Count函数将两列的结果相乘(应该分别返回每列的计数)，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/51888523/