Maria DB Update and replace

Question

I'm getting error message when run this:

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 #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?

Thanks

Answer 1

I obtained the same error with the expression:

UPDATE catalog_product_entity_tex
     SET value = REPLACE (value, 'yyyyy', 'xxxxx')
     WHERE value LIKE '%yyyyy%';

The solution was using alias:

UPDATE catalog_product_entity_tex c
     SET c.value = REPLACE (c.value, 'yyyyy', 'xxxxx')
     WHERE c.value LIKE '%yyyyy%';

Answer 2

replace is wrong. Here is correct way:

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:

UPDATE
   catalog_product_entity_text 
SET 
   value = REPLACE (value, 'xxxxx', 'zzzzz')
WHERE
   value LIKE 'yyyyy';

Answer 3

Replace takes three arguments, not two. Presumably, you want:

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.