Exception in PHP Script
/web/recursosenlaces.php
PHP Version: 5.2.11;  Zend Engine Version: 2.2.0;  Qcubed Version: 1.0.0 RC2 (QCubed 1.0 RC2)
Application: Apache/2.2.13 (Unix) mod_ssl/2.2.13 OpenSSL/0.9.8g PHP/5.2.11 with Suhosin-Patch;  Server Name: www.idi.es
HTTP User Agent: CCBot/1.0 (+http://www.commoncrawl.org/bot.html)
MySqli Error: 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 3
Exception Type:   QMySqliDatabaseException

Source File:   /usr/home/idi0/www/web/recursosenlaces.php     Line:   90

Line 85:    			
Line 86:    			$sql2 = 'Select TI.nom_tema_idi from temas as T
Line 87:       					inner join temas_idi as TI on T.id_tema=TI.id_tema
Line 88:    					where TI.ididioma = '.$_SESSION['ididioma'];
Line 89:    	    	$this->objDatabase = QApplication::$Database[1];
Line 90:    	    	$objDbResult2 = $this->objDatabase->Query($sql2);
Line 91:    	    		$objData = array();	    		
Line 92:    	    		while ($mixRow = $objDbResult2->FetchArray()) {
Line 93:    					$objData[($mixRow['nom_tema_idi'])]=($mixRow['nom_tema_idi']);
Line 94:    				}
Line 95:    			//print_r ($objData);

Database Error Number:  1064

Query:   Show/Hide


Call Stack:

#0 /usr/home/idi0/www/web/recursosenlaces.php(90): QMySqli5Database->Query('Select TI.nom_t...')
#1 /usr/home/idi0/www/includes/qcodo/_core/qform/QFormBase.class.php(276): RecursosenlacesForm->Form_Create()
#2 /usr/home/idi0/www/web/recursosenlaces.php(271): QFormBase::Run('Recursosenlaces...')
#3 {main}

Variable Dump:   Show/Hide



Exception Report Generated:  Friday, September 3 2010, 11:14:32 PM