Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given i解决办法

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given i
$sql = "select count(uid) as user from user_table where uid='".$this->username."'and password='".md5($this->password,self::USERCONST)."'";
$query = $this->database->setSQL($sql);
if($row = $this->database->select_array($query)){
echo $row["user"];
有问题 查询的结果只有一条记录为什么 不能读呢 如果user==1就证明有这个用户 1以上就重复了
他的输出结果是:
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in F:\wamp\www\user\DB\MySql.php on line 27
请求助

PHP code

database = new MySql();        $this->username = $u;        $this->password = $p;    }    function register(){        $pmd5 = md5($this->password.self::USERCONST);        $sql = "insert into user_table (username,password) values('".$this->username."','".$pmd5."')";        $database->setSQL($sql);    }    function logout(){        echo "注销";    }    function login(){        echo "登录";    }    function usercheck(){                                                                                    //密码加常量        $sql = "select count(uid) as user from user_table where uid='".$this->username."'and password='".md5($this->password,self::USERCONST)."'";        $query = $this->database->setSQL($sql);        if($row = $this->database->select_array($query)){            echo $row["user"];        }    }}?>

PHP code

host = "localhost";        $this->username = "root";        $this->password = "";        $this->database = "bkqs";        $this->mysqlconnection();     }     function mysqlconnection(){        $connection = mysql_connect($this->host,$this->username,$this->password) or die ("连接数据库失败");        mysql_select_db($this->database,$connection) or die ("打开数据库失败");        mysql_query("set names 'GBK'");     }     function setSQL($sql){        return mysql_query($sql);     }     function select_array($query){         return mysql_fetch_array($query);     }     function select_object($query){         return mysql_fetch_object($query);     }     function close(){         return mysql_close();     } }?>

------解决方案--------------------
sql语句执行失败了。可以在你的setSQL()方法中加上错误提示。or die(mysql_error());就知道错在哪了。

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