Question

此查询的工作原理与我预期的一样，但我试图避免使用“not in”：

select * from subscribers ls join emailopens eo
    on ls.subscriberid=eo.subscriberid
    where ls.listid=1381 
    and opentime  > 1458864000
    and emailaddress not IN (select emailaddress from subscribers where listid=1384)

我试过这个：

select * from subscribers ls join emailopens eo
    on ls.subscriberid=eo.subscriberid left join subscribers ls2
    on ls.emailaddress=ls2.emailaddress
    where ls.listid=1381 
    and opentime  > 1458864000
    and ls2.listid=1384

问题是它只返回listid=1381和listid=1384中具有相同电子邮件的行。我想，因为我使用左连接它应该返回所有行where listid=1381，然后在1384中同一封电子邮件的行上将有数据，但如果1381中有电子邮件1}}而非1384，会显示null列中的ls2.*。

Answer 1

将ls2.listid放入WHERE有效地将LEFT JOIN变为INNER JOIN。尝试在listid中保留ON约束，然后，为了产生NOT IN的效果，通过询问ls2.emailaddress IS NULL：<的记录，使其成为反连接/ p>

select * from subscribers ls join emailopens eo
    on ls.subscriberid=eo.subscriberid left join subscribers ls2
    on ls.emailaddress=ls2.emailaddress and ls2.listid = 1384
    where ls.listid=1381 
    and opentime  > 1458864000
    and ls2.emailaddress is null

使用JOIN而不是NOT IN消除记录

1 个答案: