Question

我有这个查询，它按预期工作

SELECT i0_.address AS address0, COUNT(d1_.id) AS sclr2 
FROM IP i0_ 
LEFT JOIN Date_IP d3_ ON i0_.id = d3_.ip 
LEFT JOIN Date d1_ ON d1_.id = d3_.date 
GROUP BY i0_.address 
ORDER BY sclr2 DESC

此查询也可按预期工作：

SELECT i0_.address AS address0, COUNT(u2_.id) AS sclr3 
FROM IP i0_ 
LEFT JOIN IP_UserAgent i4_ ON i0_.id = i4_.ip 
LEFT JOIN UserAgent u2_ ON u2_.id = i4_.useragent 
GROUP BY i0_.address 
ORDER BY sclr3 DESC

但我如何将这两者合二为一呢？

我试过

SELECT i0_.address AS address0, COUNT(d1_.id) AS sclr2, COUNT(u2_.id) AS sclr3 
FROM IP i0_ 

LEFT JOIN Date_IP d3_ ON i0_.id = d3_.ip 
LEFT JOIN Date d1_ ON d1_.id = d3_.date 

LEFT JOIN IP_UserAgent i4_ ON i0_.id = i4_.ip 
LEFT JOIN UserAgent u2_ ON u2_.id = i4_.useragent 

GROUP BY i0_.address 
ORDER BY sclr2 DESC

在这种情况下，值sclr2是正确的，但sclr3与sclr2相同。我做错了什么？

Answer 1

您可以通过制作子查询来组合这些：

SELECT t1.address0, t1.sclr2, t2.sclr3
FROM (SELECT i0_.address AS address0, COUNT(d1_.id) AS sclr2 
      FROM IP i0_ LEFT JOIN 
           Date_IP d3_ ON i0_.id = d3_.ip LEFT JOIN
           Date d1_
           ON d1_.id = d3_.date 
      GROUP BY i0_.address
     ) t1 JOIN
     (SELECT i0_.address AS address0, COUNT(u2_.id) AS sclr3 
      FROM IP i0_ LEFT JOIN
           IP_UserAgent i4_
           ON i0_.id = i4_.ip LEFT JOIN
           UserAgent u2_
           ON u2_.id = i4_.useragent 
      GROUP BY i0_.address
     ) t2
     on t1.address0 = t2.address0;

如另一个答案所述，count(distinct)也可以。然而，它产生了一个中间表，它是＆＃34;日期＆＃34;的笛卡尔积。和＃34;用户代理＆＃34;。因此，如果有100个日期和100个用户代理，则中间表将有10,000行 - 如果您的数据有很多这样的示例，则处理时间和中间存储要求可能会变得过高。

Answer 2

尝试使用

COUNT(DISTINCT d1_.id) AS sclr2, COUNT(DISTINCT u2_.id)

MySQL在一个查询中多次连接多个计数

2 个答案: