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

[miaoxu1com]

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

[kokdemo]

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]

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

[harborM]

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]

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

[wesley]

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