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Error when login Admin


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koalasoft

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Registered: 03/09/05
Posts: 242
Location:Mexico
When trying to enter my site as administrator appears to me the following message: - -
Text Formatted Code
An SQL error has occurred. Please see error.log for details.

and seeing error.log he shows this to me: -
Text Formatted Code
 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 ') OR bid = '-1') ORDER BY blockorder,title asc' at line 1. SQL in question: SELECT *,UNIX_TIMESTAMP(rdfupdated) AS date FROM gl_blocks WHERE is_enabled = 1 AND onleft = 1 AND (tid = 'homeonly' OR tid = 'all') AND (bid NOT IN (65,67,44,72,43,19,57,64,53,21,75,59,47,41,62,9,51,4,) OR bid = '-1') ORDER BY blockorder,title asc

This happens to me SINGLE when trying to login (user/pass) with the Admin user

How I can fix this? :helpme:
**Cuando el Alumno esta listo, el Maestro Aparece **
::Geeklog support in Spanish::
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koalasoft

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Full Member
Registered: 03/09/05
Posts: 242
Location:Mexico
previously it recovers the last endorsement and it solves the problem, now it returned to me to happen once again the same and it leaves in log error the following thing to me when logeado being like administrator.
Text Formatted Code
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 ') OR bid = '-1') ORDER BY blockorder,title asc' at line 1. SQL in question: SELECT name,owner_id,group_id,perm_group,perm_members,perm_anon FROM gl_blocks WHERE onleft = 1 AND is_enabled = 1 AND (tid = 'all' AND type <> 'layout') AND (bid NOT IN (65,67,44,72,26,19,57,64,53,21,66,75,59,47,49,41,23,62,77,9,51,4,) OR bid = '-1') ORDER BY blockorder,title asc
 


the other users have the same problem when he enters his name of user and password the site and modify his profile when keeping this, single when the administrator enters similarly passes east error... as he could be the error. in summary ALL those that modify their data (profiles) and keep the changes, mark east error to them and they cannot enter but...

... as he could be the error. :helpme:

My System:
Geeklog 1.4.1
PHP version 5.1.6
MySQL version 5.0.27-standard-log
**Cuando el Alumno esta listo, el Maestro Aparece **
::Geeklog support in Spanish::
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jmucchiello

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Registered: 08/29/05
Posts: 985
Quick fix: Execute this SQL using whatever query tools you have
UPDATE gl_userindex SET boxes = '' WHERE uid = ?
Where ? is the broken account's uid.

Alternatively, find this line of code in lib-common.php, it's in the COM_showBlocks() function:
$BOXES = str_replace( ' ', ',', $_USER['boxes'] );
Replace it with
$BOXES = str_replace( ' ', ',', trim($_USER['boxes']) );

The SQL will eliminate the problem. The code change is a piece of duct tape. The real fix is somewhere in usersettings.php but I'm not sure what it is (and fixing it there doesn't mend currently broken users).
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koalasoft

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Registered: 03/09/05
Posts: 242
Location:Mexico
Thanks! .. If it were solved, which I notice is that in the section of profiles, in the options of filtrate of blocks when ing personalize a block returns to send the same error to me.
as this can be avoided?

and on code SQL... how him I can make nonsingle for doing it with a single user, if not to ALL my users, (923 altogether), not to do it of one in one?

Grettings !!
**Cuando el Alumno esta listo, el Maestro Aparece **
::Geeklog support in Spanish::
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jmucchiello

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Registered: 08/29/05
Posts: 985
Quote by: koalasoft

Thanks! .. If it were solved, which I notice is that in the section of profiles, in the options of filtrate of blocks when ing personalize a block returns to send the same error to me.
as this can be avoided?


Which method did you use? SQL or code change?
and on code SQL... how him I can make nonsingle for doing it with a single user, if not to ALL my users, (923 altogether), not to do it of one in one?

Replace uid = ? with 1 = 1
 Quote

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koalasoft

Forum User
Full Member
Registered: 03/09/05
Posts: 242
Location:Mexico
Quote by: jmucchiello

Which method did you use? SQL or code change?


For SQL ..
Quote by: jmucchielloReplace uid = ? with 1 = 1[/p]

uid = ? .. for ..
Text Formatted Code
UPDATE gl_userindex SET boxes = '' WHERE uid = 1 = 1
??

This for my ALL users ?? .. or it says as serious sentence SQL for ALL the users...
**Cuando el Alumno esta listo, el Maestro Aparece **
::Geeklog support in Spanish::
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jmucchiello

Forum User
Full Member
Registered: 08/29/05
Posts: 985
update gl_userindex set boxes = '' where 1 = 1

That will reset every user's boxes preferences. (Thus, they will once again see all left and right side blocks unless they have noboxes = 1.)
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