Question

我正在尝试根据其他列的值替换列值，但我不断收到语法错误。我究竟做错了什么？这就是我累了：

SELECT * FROM `table` WHERE `id_1` =164 and `id_2` = 6
SET `id_1` = (188|164)
WHERE `id_1` = 164 AND `id_2` = 6 
UPDATE `table`

这给了我一个靠近SET的语法。

SELECT * FROM `table` WHERE `id_1` = 164 AND `id_2` = 6
UPDATE `table`
SET `id_1` = 188
WHERE `id_2` = 6

这使我在UPDATE以下的所有内容都出错。

Answer 1

您不能在一个查询中使用UPDATE和SELECT。

单独执行这两个查询：

SELECT * FROM `table` WHERE `id_1` = 164 AND `id_2` = 6;

然后执行：

UPDATE `table`
SET `id_1` = 188
WHERE `id_2` = 6;

Answer 2

你错过了标点符号的结尾';'在你的select语句之后。在执行SELECT之后，必须在开始下一个语句（UPDATE）之前结束该语句：

SELECT * FROM `table` WHERE `id_1` = 164 AND `id_2` = 6; -- semicolon goes here
UPDATE `table`
SET `id_1` = 188
WHERE `id_2` = 6; -- and another semicolon goes here

这将消除您的语法错误。

mysql SET和UPDATE

2 个答案: