查询表里边 id 为 16 的 content 然后，更新 id 大于 16 的 content 为查出来的 content

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1.使用 sql 只查询一次 2.我自己的 sql: UPDATE guideline_news SET content = ( SELECT content FROM guideline_news WHERE id = 16 ) WHERE id > 16 备注：这个 sql 报错：[Err] 1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'SELECT content FROM guideline_news WHERE id=16 WHERE id>16' at line 1

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5 replies • 2016-12-28 17:17:25 +08:00

kokdemo

Dec 28, 2016

建议用一个临时变量先存一下 SELECT content FROM guideline_news WHERE id = 16

然后再更新

harborM

Dec 28, 2016

mysql 不能直接 set select ，需要使用 inner join

harborM

Dec 28, 2016

UPDATE guideline_news a inner join ( SELECT content FROM guideline_news WHERE id = 16 ) b SET a.content =b.content WHERE id > 16;试试这条语句;我在工作中也遇到过,百度就有解决方案： http://www.3lian.com/edu/2014/05-13/147149.html

q397064399

Dec 28, 2016

http://yonglailizhi.iteye.com/blog/1090210

wesley

Dec 28, 2016

update guideline_news a, guideline_news b set a.content=b.content
where a.id>16 and b.id=16