كود PHP:
Database error in vBulletin 3.0.3:
Invalid SQL:
INSERT INTO setting
(varname, grouptitle, value, defaultvalue, optioncode, displayorder,
advanced, volatile)
VALUES
(
'cpstylefolder',
'admincp',
'vBulletin_3_Default',
'vBulletin_3_Default',
'cpstylefolder',
10,
0,
1
),
(
'timeoutcontrolpanel',
'admincp',
'0',
'0',
'yesno',
20,
0,
1
),
(
'adminquickstats',
'admincp',
'0',
'0',
'yesno',
30,
0,
1
),
(
'cp_collapse_forums',
'admincp',
'0',
'0',
'<select name=\\\"setting[$setting[varname]]\\\" tabindex=\\\"1\\\"
class=\\\"bginput\\\">
<option value=\\\"0\\\" \" .
iif($setting[\'value\']==0,\'selected=\"selected\"\') .
\">$vbphrase[default]</option>
<option value=\\\"1\\\" \" .
iif($setting[\'value\']==1,\'selected=\"selected\"\') .
\">$vbphrase[collapsible]</option>
<option value=\\\"2\\\" \" .
iif($setting[\'value\']==2,\'selected=\"selected\"\') .
\">$vbphrase[single]</option>
</select>',
40,
0,
1
),
(
'cp_usereditcolumns',
'admincp',
'2',
'2',
'<select name=\\\"setting[$setting[varname]]\\\" tabindex=\\\"1\\\"
class=\\\"bginput\\\">
<option value=\\\"1\\\" \" .
iif($setting[\'value\']==1,\'selected=\"selected\"\') . \">1</option>
<option value=\\\"2\\\" \" .
iif($setting[\'value\']==2,\'selected=\"selected\"\') . \">2</option>
</select>',
50,
0,
1
)
mysql error: Duplicate entry 'timeoutcontrolpanel' for key 1
mysql error number: 1062
Date: Monday 28th of February 2005 03:14:19 PM
Script: http://www.*****.net/vb/install/upgrade15.php?step=1
Referer: http://www.*****.net/vb/install/upgrade15.php
IP Address:
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