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09/17/07 11:17pm

Idorafan

Anonymous

I've searched these forums up and down... I see lots of similar problems, but none of the fixes ive tried seem to work...

I have the typical SQL error, see error.log for details

and then

Mon 17 Sep 2007 21:49:25 GMT+4 - 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 know its a problem with my user groups...

All group list provide results except for logged in users and all users (from the admin > group manager > memberlists)

Any ideas what changes I need to make to my gl_group_assignments table?

Or is the solution something completely different?

Thanks so much for everyones help!!!

09/17/07 11:51pm

Status: offline

mevans

Forum User

Full Member

Registered: 02/08/04

Posts: 393

Have you looked at this topic? Near the end you will see a small script you can create on your site that should get you going again.

Thanks!
Mark

09/18/07 12:02am

Idorafan

Anonymous

mevans , Thank you for your help, but I'm still having problems - after editing, uploading, and running that script, i get the following error:

Parse error: syntax error, unexpected T_VARIABLE in /home/downtown/public_html/fixdb.php on line 14

Any other ideas?

09/18/07 12:07am

Status: offline

mevans

Forum User

Full Member

Registered: 02/08/04

Posts: 393

Make sure you read the post following my code, there was a small error that is documented. If you are still having problems, cut and paste your fixdb.php here (place it between [ code ] blocks so it will format properly and we'll find the error.

09/18/07 12:12am

Idorafan

Anonymous

lol, please don't think of me as a php moron...

the quote was For those who want to use that last code snippet be sure to move
the semicolon outside the double quotes on line 14 ($sql = ...)

the line is

Text Formatted Code
$result = @mysql_query($sql,$db);

semicolon = ; is currently outside of ($sql=)

Where am I going wrong???

09/18/07 12:21am

Status: offline

mevans

Forum User

Full Member

Registered: 02/08/04

Posts: 393

Well, it is actually the line before that one that is wrong. Here is the corrected version:

Text Formatted Code

<?php

    $hostname = 'localhost';

    $user          = 'dbuserhere';

    $pass         = 'dbpasswdhere';

    $dbname   = 'dbnamehere';

    // connect to the database

    $db = mysql_connect($hostname,$user,$pass) or die('Cannot connect to DB server');

    // Set the database

    @mysql_select_db($dbname) or die('error selecting database');

    $sql = "insert into gl_group_assignments (ug_main_grp_id, ug_uid) values (2,1)";

    $result = @mysql_query($sql,$db);

    echo 'Group assignment database updated - try site again...';

?>

Give this one a try and let's see if it works a little better.

Thanks!
Mark

09/18/07 12:34am

Idorafan

Anonymous

the script ran just fin, but when I go to my group manager and try to list both the "all users" member list and the "logged in users" member list, I get a page cannot be displayed error....

The list shows me (admin) for all other group lists...

For reference:

all users = grp_id=2
logged in users = grp_id=13

09/18/07 09:01am

Status: offline

mevans

Forum User

Full Member

Registered: 02/08/04

Posts: 393

I would recommend looking at your Geeklog error.log and possibly your web servers error log for clues.

I'm not sure what else to tell you without a little more info.

Thanks!
Mark