Question

我正在尝试更新＆＃34;％2017-08-01％＆＃34;之间的金额和＆＃34;％2017年8月7日％＆＃34;到下一个日期的金额＆＃34;％2017-08-07％＆＃34; e.auto =＆＃34; true＆＃34;在交易表中。但是，mysql正在返回

1064 - 您的SQL语法出错;查看与MariaDB服务器版本对应的手册，以获取正确的语法靠近＆＃39; FROM user as u JOIN e_wallet as e ON e.id = u.ewallet JOIN transacti＆＃39;在第1行

update t set t.amount = (
                          select t.amount 
                          FROM user as u 
                          JOIN e_wallet as e ON e.id = u.ewallet 
                          JOIN transaction as t on t.user = u.id 
                          where e.auto="true" 
                                and t.create_date like "%2017-08-07%"
                        ) 
FROM user as u 
JOIN e_wallet as e ON e.id = u.ewallet 
JOIN transaction as t on t.user = u.id 
where e.auto="true" 
      and t.create_date BETWEEN "%2017-08-01%" and "%2017-08-07%"

Answer 1

如果忽略子查询，则UPDATE查询如下所示：

UPDATE t
SET t.amount = ...
FROM user AS u ...
JOIN e_wallet AS e ...

您在FROM查询中不能拥有UPDATE条款，而且导致语法错误的原因是什么。语法是：

UPDATE tablename AS t1
JOIN tablename2 AS t2 ON ...
JOIN tablename3 AS t3 ON ...
SET ...
WHERE ...

所以它应该是：

UPDATE transaction AS t
JOIN user AS u ON u.id = t.user
JOIN e_wallet AS e ON e.id = u.ewallet
SET t.amount = (
  select t.amount 
  FROM user as u 
  JOIN e_wallet as e ON e.id = u.ewallet 
  JOIN transaction as t on t.user = u.id 
  where e.auto="true" 
        and t.create_date like "%2017-08-07%"
) 
WHERE e.auto = "true"
AND t.create_date BETWEEN "2017-08-01" and "2017-08-07"

这应解决语法错误，但我不确定它会做你想要的。子查询与主查询不相关，因此它将返回符合条件的所有事务的数量。但是当您使用子查询作为表达式时，它必须最多返回一行。您可能只想第二次加入transaction表。

UPDATE transaction AS t
JOIN user AS u ON u.id = t.user
JOIN e_wallet AS e ON e.id = u.ewallet
JOIN transaction AS t1 ON t1.user = t.user
SET t.amount = t1.amount
WHERE e.auto = "true"
AND t.create_date BETWEEN "2017-08-01" and "2017-08-07"
AND t1.create_date LIKE '%2017-08-07%'

从精选MySQL更新

1 个答案: