Question

我正在比较mysql中的两个电子邮件地址列表 - listA和listB（两者都是视图）。最初，我想查找listA中不在listB中的所有电子邮件地址。我成功完成了这个：

$comparison_query = mysql_query("SELECT DISTINCT email_addresses FROM listA WHERE email_addresses NOT IN (SELECT DISTINCT email_addresses FROM listB) ORDER BY email_addresses");

现在，我想查找listA中不在listB中的所有电子邮件地址或异常表。我试过这样做：

$comparison_query = mysql_query("SELECT email_addresses FROM listA WHERE email_addresses NOT IN (SELECT DISTINCT email_addresses FROM ((SELECT email_addresses FROM listB) UNION (SELECT email_addresses FROM exception))) ORDER BY email_addresses");

但是，这不起作用。谁能看到我错在哪里？

谢谢！

Answer 1

我觉得这样的事情可以解决问题：

SELECT email_addresses FROM listA
WHERE email_addresses NOT IN
(SELECT DISTINCT email_addresses FROM FROM listB)
AND email_addresses NOT IN
(SELECT email_addresses FROM exception);

Answer 2

USE AND代替UNION

 mysql_query("SELECT email_addresses FROM listA
        WHERE email_addresses  NOT IN
        ( SELECT DISTINCT email_addresses FROM listB) AND 
         email_addresses  NOT IN
        (SELECT DISTINCT email_addresses FROM exception)) 
        ORDER BY email_addresses");

mysql列表与查询逻辑的比较问题

2 个答案: