php – mysql_fetch_array()期望参数1是资源,字符串给出教程

参见英文答案 > mysql\_fetch\_array()/mysql\_fetch\_assoc()/mysql\_fetch\_row()/mysql\_num\_rows etc… expects parameter 1 to be resource or result 32个
我有两个类DBConn和DBQueries. DBQueries在DBConn之后继承.当我要在用户数据库中显示所有用户时,我看到消息：

mysql\_fetch\_array() expects parameter 1 to be resource, string given.

谢谢你的任何建议.

class DBConn /*extends Config*/ {
  public function dbConnection(){
    $db_host = 'localhost';
    $db_login = 'root';
    $db_password = '';
    $db_name = "database";
    $conn = mysql_connect($db_host, $db_login, $db_password);
    $db = mysql_select_db($db_name);
  }  
}

class DBQueries extends DBConn {
  function displayUsers(){
    $this->dbConnection();
    $query = "SELECT * FROM users";
    $result = mysql_query($query);
    while ($row = mysql_fetch_array($query)) {
      echo $row['password'];
    }
  }
}

解决方法:

当你需要传入$result这是你的db结果对象时,你传入$query这是你的sql字符串.

class DBQueries extends DBConn {
    function displayUsers(){
        $this->dbConnection();
        $query = "SELECT * FROM users";
        $result = mysql_query($query);
        while ($row = mysql_fetch_array($result)) {
            echo $row['password'];
        }
    }
}