MySQL操作实施成功，但没有返回值

MySQL操作执行成功，但没有返回值？
在MySQL console中执行：

SQL code

<!--Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/-->mysql>update `1001` set finish=0 where sid=94664;Query OK,1 row affected (0.07 sec)Rows matched:1 changed:1 warnings:0

php中的语句：

PHP code

<!--Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/-->$tid="`".$_COOKIE['uid']."`";$which="finish=1";$where="sid=".$sid;$db=new Mysql("localhost","root","","members_check","","UTF8");$db->connect();$result=$db->update($tid,$which,$where);echo $result;    //if($result){echo 0;}else{echo 1;}

js那里返回来的就是空，什么都没有。用下面那个if语句返回不了0。但是数据库里全部操作成功。

附上封装的函数：

PHP code

<!--Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/-->public function query($sql) {     if ($sql == "") {     echo "SQL语句为空";     }     $this->sql = $sql;     $result = mysql_query($this->sql, $this->conn) or die('faill'.mysql_error());     $this->result = $result;     return $this->result;}public function update($table, $mod_content, $condition, $url = '') {     $result=$this->query("UPDATE $table SET $mod_content WHERE $condition");     return $result;}

------解决方案--------------------
$result=$db->update($tid,$which,$where);
var_dump($result);
显示 null

显然LZ使用的代码与他贴出的代码是不一样的
继续讨论是无意义的