mysql COUNT返回的值多于拉出实际结果时的值

Question

查询

SELECT DISTINCT walls.wall_id FROM walls 
LEFT JOIN wall_categories ON walls.wall_id = wall_categories.wall_id 
LEFT JOIN wall_devices ON walls.wall_id = wall_devices.wall_id
WHERE wall_categories.category_id IN (1,2,3)
AND wall_devices.device_id IN (1,2,3)
AND walls.wall_id>113;

上面的查询返回以下结果：

如果我将COUNT添加到完全相同的查询中，如：

SELECT DISTINCT COUNT(walls.wall_id) FROM walls 
LEFT JOIN wall_categories ON walls.wall_id = wall_categories.wall_id 
LEFT JOIN wall_devices ON walls.wall_id = wall_devices.wall_id
WHERE wall_categories.category_id IN (1,2,3)
AND wall_devices.device_id IN (1,2,3)
AND walls.wall_id>113;

它返回= 14

哪里可以发行？我做错了什么？

Answer 1

你想：

SELECT COUNT(DISTINCT walls.wall_id)

当你写：

SELECT DISTINCT <expr>, <expr>, <expr>, ...

它有效地将查询视为没有DISTINCT修饰符，然后从结果集中删除所有重复项。所以当你写道：

SELECT DISTINCT COUNT(walls.wall_id)

它只是对符合查询其余部分条件的所有wall_id行进行了正常计数。这只返回一个结果，即总行数，然后DISTINCT删除任何重复，这是微不足道的，因为它只是一个结果。

当您将DISTINCT修饰符放在COUNT()中时，它会告诉该函数它应该在计数时删除重复项。

Answer 2

试试这个：

SELECT  COUNT(DISTINCT walls.wall_id) FROM walls 
LEFT JOIN wall_categories ON walls.wall_id = wall_categories.wall_id 
LEFT JOIN wall_devices ON walls.wall_id = wall_devices.wall_id
WHERE wall_categories.category_id IN (1,2,3)
AND wall_devices.device_id IN (1,2,3)
AND walls.wall_id>113;