Exception in PHP Script
/articolo.php		 PHP Version: 5.2.10;  Zend Engine Version: 2.2.0;  Qcodo Version: 0.3.32 (Qcodo Beta 3)
Application: Apache/2.2.3 (CentOS);  Server Name: www.adrenalinik.com
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 4
Exception Type:   QMySqliDatabaseException

Source File:   /home/sites/nextmove/produzione.adrenalinik.com/web/articolo.php     Line:   277 

Line 272:    	            SELECT DISTINCT nome, tipo FROM blocchi_sc_keywords
Line 273:    	            LEFT JOIN blocchi_sc ON	blocchi_sc_keywords.blocchi_sc_id = blocchi_sc.id
Line 274:    	            WHERE
Line 275:    	            blocchi_sc.sottocategoria = '.$this->curSubcat->Id.'
Line 276:    	            ';
Line 277:    	            $objResult = $objDatabase->Query($strCstmQuery);
Line 278:    	            $arrObjKeywords = BlocchiScKeywords::InstantiateDbResult($objResult);
Line 279:    				$this->arrKeywords=array();
Line 280:    				foreach($arrObjKeywords as $objKeywords){
Line 281:    					//$arrKeywords[]=$objKeywords->Tipo.'|'.$objKeywords->Nome;
Line 282:    					$this->arrKeywords[]=$objKeywords->Nome;

Database Error Number:  1064

Query:   Show/Hide


Call Stack: 

#0 /home/sites/nextmove/produzione.adrenalinik.com/web/articolo.php(277): QMySqli5Database->Query('???            ...')
#1 /home/sites/nextmove/produzione.adrenalinik.com/web/qcodo/includes/qcodo/_core/qform/QFormBase.class.php(275): StaticaForm->Form_Create()
#2 /home/sites/nextmove/produzione.adrenalinik.com/web/articolo.php(794): QFormBase::Run('StaticaForm', 'articolo.php.in...')
#3 {main}

Variable Dump:   Show/Hide


Exception Report Generated:  Wednesday, September 8 2010, 4:45:07 AM