Geeklog Forums

Hi, sorry to email u directly but I need help, dont we all...???

I changed one of my site moderator accounts to be able to be the "link" admin also..after i did that the below happend..

1064: You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 1. SQL in question: SELECT DISTINCT ft_name FROM gl_access,gl_features WHERE ft_id = acc_ft_id AND acc_grp_id IN ()

I cant access my geeklog system now at all cuz of this error... any work arounds..?

Thanks,

Brandon

The user in question is not a member of any groups. However, a user has to be at least in the All Users and Logged-In Users groups.

Try logging in as someone else (you may have to delete the cookies in your browser first, or use another browser, so Geeklog doesn't try to log you in automatically).

If you don't have another account with sufficient permissions to access the Admin's list of users, see this post for how to create a new Root account.

bye, Dirk

I cant do this because each time i try to goto my site the error comes up, thus not allowing me access.. eg i type www.hello.com and its says "SQL Error. Pls check the logs for details".

Dirk, my site is not accessablee at all. Even after i clear cookies and my cache, if i try to access my site I get the "SQL Error has occured, pls refer to error logg for details".

again, in the error log I have this messege:

Check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 1. SQL in question: SELECT DISTINCT ft_name FROM gl_access,gl_features WHERE ft_id = acc_ft_id AND acc_grp_id IN ()

what way can i get rid of this messege and log into my site...? what do i need to edit in the DB and do i need to edit the lib-secrity file..?

Hmm, that's odd. That would mean that the Anonymous user is also affected.

Try this: Find the place in phpMyAdmin where you can enter SQL requests and type This should list one entry where ug_main_grp_id = 2.

Only(!) if you do not get that entry, do this:

Also check that in your gl_groups table, there is a group with grp_id = 2 and grp_name = "All Users".

bye, Dirk