Question

我正在尝试使用phpMyadmin替换wordpress帖子内容中的一些句子这是我的更新查询

UPDATE `wp_posts` SET `post_content` = REPLACE (`post_content`,  'Example Routes:',  'Routes:') WHERE `post_content` LIKE '%Example Routes:%';

但它说“0行受影响”。

我运行了一个具有相同条件的select查询，以确保有resutls并且我有很多resutls 这是选择查询

SELECT * FROM `wp_posts` WHERE `post_content` LIKE '%Example Routes:%';

请注意，两个查询条件都与我更新的条件相同“LIKE'％示例路由：％'”，更新为“REPLACE（post_content，'示例路由：'，'路线：'）”

知道为什么更新会给我0行影响吗？

Answer 1

尝试删除SELECT a.Record_id, a.Reference_id,V.Name FROM TableA AS a inner join ( SELECT Doc_id id, NameB Name,'B' Type from TableB UNION ALL SELECT Doc_id ID, NameC name,'C' Type From TableC ) AS V On V.ID=Reference_id AND Type=Param和左括号之间的空格：

REPLACE

默认情况下，不允许在函数名称和括号之间包含空格：http://dev.mysql.com/doc/refman/5.7/en/functions.html

Answer 2

After a lot of search and working around I figured it out, the select query give results because the LIKE operator is not case sensitive but Replace is, so I needed to replace 'Example Routes:' to 'Example routes:' in the replace function which is a bit tricky.

I ended up using this query which works for me

UPDATE `wp_posts` SET `post_content` = REPLACE(`post_content`,'Example routes:','Routes:')
WHERE `post_content` LIKE '%Example Routes:%';

为什么更新查询在phpmyadmin中没有按预期工作？

2 个答案: