Question

基本上我有一个主表（帐户）和一个元表（accounts_meta）...元表如下所示：

id | account_id | meta_key | meta_value

我想要做的只是在accounts_meta表中选择没有'referrer_paid'作为行的帐户...

到目前为止，这是我的代码......

SELECT a.* FROM accounts AS a
    LEFT JOIN accounts_meta AS am ON a.id = am.account_id AND am.meta_key != 'referrer_paid'
    WHERE a.account_referrer != ''
    GROUP BY a.id

希望我有意义。我做错了什么？

Answer 1

SELECT * 
  FROM accounts 
 WHERE id NOT IN ( select DISTINCT account_id 
                     from `account_meta_table` 
                    where meta_key != 'referrer_paid'
                  );

Answer 2

来自@lexu的微小变化：

SELECT * 
  FROM accounts 
 WHERE id NOT IN ( select account_id 
                     from `account_meta_table` 
                    where meta_key = 'referrer_paid'
                  );

Answer 3

SELECT a.* FROM accounts AS a
LEFT JOIN accounts_meta AS am ON a.id = am.account_id AND am.meta_key = 'referrer_paid'
WHERE a.account_referrer != ''
  AND am.account_id IS NULL

你不需要group by，因为left-join-is-null不会产生重复的帐户行

编辑： duh，将am.meta_key != 'referrer_paid'更改为am.meta_key = 'referrer_paid'

这就是你想要的。如果连接的行不匹配，则返回NULL，并且只接受NULL行

Answer 4

选择a。* FROM帐户AS a LEFT JOIN accounts_meta AS am ON a.id = am.account_id AND am.meta_key！='referrer_paid' 哪里有ISNULL（am.account_referrer） GROUP BY a.id

如何选择第二个表中没有值的行

4 个答案: