Question

我有以下疑问 -

SELECT COUNT(capture_id) as count_captures 
  FROM captures 
 WHERE user_id = 9

...返回5

SELECT COUNT(id) as count_items 
  FROM items 
 WHERE creator_user_id = 9

...返回22

我尝试了以下查询 -

   SELECT COUNT(capture_id) as count_captures, 
          COUNT(items.id) as count_items 
     FROM captures 
LEFT JOIN items ON captures.user_id = items.creator_user_id 
    WHERE user_id = 9

...但它返回两列，其中110为值。我想要一列5列，另一列22块。我做错了什么？

Answer 1

我的下意识是一个子查询：

select count(capture_id) as count_captures, 
    (select count(id) as count_items
         from items i where i.creator_user_id = captures.user_id) as count_items 
from captures 
where user_id = 9

我不确定你能做些什么来避免这种情况。你看到了预期的（通常是期望的行为）。

当然，如果你知道两者中的ID都不会重复，你可以使用distinct：

SELECT COUNT( DISTINCT capture_id) as count_captures, 
      COUNT( DISTINCT items.id) as count_items 
FROM captures 
LEFT JOIN items ON captures.user_id = items.creator_user_id 
    WHERE user_id = 9

Answer 2

LEFT JOIN返回左表中的每一行，右表中的每一行都与结果匹配。因为你的所有id都是相同的，它产生了表的笛卡尔积。（5 * 22 = 110）。

预计会发生这种情况。

Answer 3

您可以随时结合结果（警告，未经测试）：

SELECT SUM(sub.count_captures), SUM(sub.count_items)
FROM (SELECT COUNT(capture_id) as count_captures, 0 as count_items 
from captures where user_id = 9
UNION
SELECT 0 as count_captures, count(id) as count_items
from items where creator_user = 9) sub

Answer 4

将两个（看似不相关的）查询合并为一个的另一种方法：

SELECT
    ( SELECT COUNT(capture_id) 
        FROM captures 
       WHERE user_id = 9
    )
    AS count_captures 

  , ( SELECT COUNT(id) 
        FROM items 
       WHERE creator_user_id = 9
    )
    AS count_items

在这些情况下，确实不需要子查询或JOIN。虽然优化器可能足够聪明，但我不会试图混淆他。

左连接乘法值

4 个答案: