Exception in PHP Script
/controller.php
PHP Version: 5.4.4-14+deb7u14;  Zend Engine Version: 2.4.0;  Qcodo Version: 0.3.42 (Qcodo Beta 3)
Application: Apache;  Server Name: www.crosswordpuzzles.net
HTTP User Agent: CCBot/2.0 (http://commoncrawl.org/faq/)
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 1
Exception Type:   QMySqliDatabaseException

Source File:   /www/sites/www.crosswordpuzzles.net/includes/controller/solve-crosswords.php     Line:   49

Line 44:            }
Line 45:        }
Line 46:        public static function getCategoryLinkName($id=null){
Line 47:          $objDatabase = Category::GetDatabase();
Line 48:          $strQuery="SELECT `url`,`short_name` FROM `category` WHERE `category_id`=$id";
Line 49:          $objDbResult = $objDatabase->Query($strQuery);
Line 50:          $r=$objDbResult->FetchArray();
Line 51:          return $r;
Line 52:        }
Line 53:      }
Line 54:      //$crossword=

Database Error Number:  1064

Query:   Show/Hide


Call Stack:

#0 /www/sites/www.crosswordpuzzles.net/includes/controller/solve-crosswords.php(49): QMySqli5Database->Query('SELECT `url`,`s...')
#1 /www/sites/www.crosswordpuzzles.net/includes/controller/solve-crosswords.php(110): Crosswords::getCategoryLinkName(NULL)
#2 /www/sites/www.crosswordpuzzles.net/htdocs/controller.php(89): require('/www/sites/www....')
#3 {main}

Variable Dump:   Show/Hide



Exception Report Generated:  Thursday, October 23 2014, 12:56:02 AM