The Ultimate Guide to High Availability with MariaDB

Use of UPDATE & SELECT operators

Hi - I'm sure this has been asked before but I've tried looking!

It's been 20 years since I last used SQL (DB2 and an IBM Mainframe), but all I'm trying to do is change a string of text within a cell.

Background - I'm just changing URL address on a wordpress related DB.

So in MYSQL it's UPDATE wp_posts SET post_content = replace(post_content, 'http://www.example.com', 'http://localhost/test-site');

Using MariaDB syntax I've tried: UPDATE `wp_posts` SET `post_content`= REPLACE (`post_content`,'http://www.holidaycottagesloirevalley.com','http://localhost/hclv'); and after some reading I've tried REPLACE INTO `wp_posts` VALUE ('http://www.holidaycottagesloirevalley.com','http://localhost/hclv');

No joy and I've run out of ideas (strength!)

Can anyone point me in the right direction please?

Thanks in advance,

Jon.

Answer Answered by Jon Lethbridge in this comment.

Hi - Thanks for the reponse.

When simulating the querey I was getting the following...

  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

However, with you confirming the SQL I actually ran the SQL and it worked fine. So it must be an issue with the simulator running within XAMPP which is what I am using for my local environments.

So I think I'm sorted - although a little confused.

Thanks for your quick response.

Jon.

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