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SQL Problems - seen before! Please Help!


Idorafan

Anonymous
freakingout
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 'Wink' 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!!!
 Quote

Status: offline

mevans

Forum User
Full Member
Registered: 02/08/04
Posts: 393
Location:Texas
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
 Quote

Idorafan

Anonymous
freakingout
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?
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Status: offline

mevans

Forum User
Full Member
Registered: 02/08/04
Posts: 393
Location:Texas
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.
 Quote

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???
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Status: offline

mevans

Forum User
Full Member
Registered: 02/08/04
Posts: 393
Location:Texas
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
 Quote

Idorafan

Anonymous
freakingout
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
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Status: offline

mevans

Forum User
Full Member
Registered: 02/08/04
Posts: 393
Location:Texas
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
 Quote

Idorafan

Anonymous
cheerful
Well, I apparently haven't had the issue since I ran the script (not showing up on the error log) So thanks so much mevans!!!!!
 Quote

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