[英]Maria DB Update and replace
I'm getting error message when run this: 运行此命令时出现错误消息:
UPDATE catalog_product_entity_text
SET value = REPLACE (value, 'xxxxx')
WHERE value LIKE 'yyyyy';
Error: #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 ')' at line 1
错误:
#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 ')' at line 1
#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 ')' at line 1
Any Ideas? 有任何想法吗?
Thanks 谢谢
I obtained the same error with the expression: 我获得了与表达式相同的错误:
UPDATE catalog_product_entity_tex
SET value = REPLACE (value, 'yyyyy', 'xxxxx')
WHERE value LIKE '%yyyyy%';
The solution was using alias: 解决方案是使用别名:
UPDATE catalog_product_entity_tex c
SET c.value = REPLACE (c.value, 'yyyyy', 'xxxxx')
WHERE c.value LIKE '%yyyyy%';
replace
is wrong. replace
是错误的。 Here is correct way: 这是正确的方法:
MariaDB [maison]> select replace('aaaaaa', 'a', 'b');
+-----------------------------+
| replace('aaaaaa', 'a', 'b') |
+-----------------------------+
| bbbbbb |
+-----------------------------+
1 row in set (0.01 sec)
Your query must be something like: 您的查询必须类似于:
UPDATE
catalog_product_entity_text
SET
value = REPLACE (value, 'xxxxx', 'zzzzz')
WHERE
value LIKE 'yyyyy';
Replace takes three arguments, not two. 替换需要三个参数,而不是两个。 Presumably, you want:
大概,您想要:
UPDATE catalog_product_entity_tex
SET value = REPLACE (value, 'yyyyy', 'xxxxx')
WHERE value LIKE '%yyyyy%';
Also note the use of wildcards in the LIKE
pattern. 还要注意在
LIKE
模式中使用通配符。
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