0

Hey guys, so I've been having an errorwith Php/MySQL for a long time with something I'm trying to do. I posted it in the PHP section before I found this subforum, so I'm just going to place the link to it below. Was hoping maybe I'd have a better chance of getting an answer if I posted it here too.

https://www.daniweb.com/web-development/php/threads/483610/phpmysql-issue-access-denied-for-user-vangua01localhost-using-pass

Basically, I'm trying to use a Php script to receive information from Second Life, the script should connect and inserts the information to my database, but, I'm getting the error in the title.

A, SHOW GRANTS for CURRENT_USER; returns:

Grants for vangua01@localhost
GRANT USAGE ON . TO 'vangua01'@'localhost' IDENTIFIED BY PASSWORD '57c63ca170dc2c10'
GRANT ALL PRIVILEGES ON vangua01\_access\_list.* TO 'vangua01'@'localhost'
GRANT ALL PRIVILEGES ON vangua01\_registration.* TO 'vangua01'@'localhost'
GRANT ALL PRIVILEGES ON vangua01\_%.* TO 'vangua01'@'localhost'

Since I'm having this issue, I tried to create a new user through the SQL query box in PhpMyAdmin (all the others were created through my web host's control panel) using:

CREATE USER 'accesslist_admin'@'localhost' IDENTIFIED by 'n391fnbgfjf0';

And I get:

#1227 - Access denied; you need (at least one of) the CREATE USER privilege(s) for this operation

I don't understand what's going on..

Edited by pritaeas: Closed. Do NOT cross-post.

2
Contributors
4
Replies
18
Views
3 Years
Discussion Span
Last Post by Parker_1
0

Thanks!

-

Talked to support at my web host, SiteGround.com, they say that non-dedicated servers, as in, shared hosting, which is what I have currently, don't have the ability to create users from the SQL query box, and have to do it from the database control panel. But they claim that they have all the same permissions as when creating from inside the database.

Being that I have another database using the same style connect for my website registration and login, that works perfectly, I just plugged it's connection information into this script and ended up with the same error. This leads me to believe that the problem is with my Php script and not MySQL..