使用mysql中的select进行更新

我将使用存储在table_b中的变量更新table_a。但是当我尝试更新与选择查询，我有错误，请帮助我。万分感谢。使用mysql中的select进行更新

这2代表的结构：

CREATE TABLE IF NOT EXISTS `table_a` (
    `fk1` int(11) DEFAULT NULL, 
    `avg_100` int(11) DEFAULT NULL, 
    `avg_score` int(11) DEFAULT NULL, 
    `cvg_date` datetime DEFAULT NULL 
) ENGINE=InnoDB DEFAULT CHARSET=latin1; 


CREATE TABLE IF NOT EXISTS `table_b` (
    `fk1` int(11) NOT NULL DEFAULT '0', 
    `avg_100` int(11) DEFAULT NULL, 
    `avg_score` int(11) DEFAULT NULL, 
    `cvg_date` int(11) DEFAULT NULL 
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

，当我尝试执行查询

UPDATE table_a a LEFT JOIN (( 
     SELECT fk1, SUM(avg_100) as avg_100, SUM(avg_score) as avg_score, MAX(cvg_date) as cvg_date 
     FROM table_b 
      GROUP BY fk1 
) AS b1) AS b ON a.fk1= b.fk1 
SET 
    a.avg_score = b.avg_score, 
    a.avg_100 = b.avg_100, 
    a.cvg_date = b.cvg_date

我得到一个错误：

[Err] 1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'b ON a.fk1= b.fk1 
SET 
    a.avg_score = b.avg_score, 
    a.avg_100 = b.avg_100, 
' at line 4

来源

2013-03-27 Mr Jerry

要设置别名“b”和“B1”被选择返回的表。你只需要一个。试试这个查询：

UPDATE table_a a 
LEFT JOIN ( 
    SELECT fk1, SUM(avg_100) as avg_100, SUM(avg_score) as avg_score, MAX(cvg_date) as cvg_date 
    FROM table_b GROUP BY fk1 
) AS b 
ON a.fk1= b.fk1 
SET 
a.avg_score = b.avg_score, 
a.avg_100 = b.avg_100, 
a.cvg_date = b.cvg_date

来源

2013-03-27 04:56:24 Eddie

谢谢@Eddie你是对的。 – 2013-03-27 04:59:51

我认为你有一个语法错误通过在左连接之后添加一个选择。我发现一个堆栈溢出后用更新，也许你可以使用为例来重新安排你的查询： UPDATE multiple tables in MySQL using LEFT JOIN

来源

2013-03-27 04:23:20

解决了这个问题“并称一个只选择左侧后加入”我认为这是正确的，因为我选择结果保存为表 – 2013-03-27 04:33:17

我使用ROWNUM

UPDATE TABLE1 set TABLE1.COLUMN1 = (select T2.COLUMN1 from TABLE1 AS T2 
where T2.PK = TABLE1.PK 
and rownum = 1)

来源

2015-12-17 17:27:29

使用mysql中的select进行更新

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