帮助 我在使用 PHP 时收到一条错误消息“警告：mysql_num_rows()：提供的参数不是有效的 MySQL 结果资源...”

Question

I'm trying to run a query in MySql database but I have the same warning everytime. Can someone help.

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX/public_html/login.php on line 279 The Username you supplied does not exist!

I have highlighted line 279 with ++++++++++++$num = mysql_num_rows($res);++++++++++++

Please let me know how I can rectify the problem.

<?php
  //If the user has submitted the form
  if($_POST['submit']) {
    //protect the posted value then store them to variables
    $username = protect($_POST['username']);
    $password = protect($_POST['password']);

    //Check if the username or password boxes were not filled in
    if(!$username || !$password) {
      //if not display an error message
      echo "<center>You need to fill in a <b>Username</b> and a <b>Password</b>!</center>";
    } else {
      //if the were continue checking

      //select all rows from the table where the username matches the one entered by the user
      $res = mysql_query("SELECT * 
                            FROM `users` 
                           WHERE `username` = '".$username."'");
      ++++++++++++++$num = mysql_num_rows($res);++++++++++++++++++

      //check if there was not a match
      if($num == 0) {
        //if not display an error message
        echo "<center>The <b>Username</b> you supplied does not exist!</center>";
      } else {
        //if there was a match continue checking

        //select all rows where the username and password match the ones submitted by the user
        $res = mysql_query("SELECT * FROM `users` 
                             WHERE `username` = '".$username."' 
                               AND `password` = '".$password."'");
        $num = mysql_num_rows($res);

        //check if there was not a match
        if($num == 0) {
          //if not display error message
          echo "<center>The <b>Password</b> you supplied does not match the one for that username!</center>";
        } else {
                    //if there was continue checking

                    //split all fields fom the correct row into an associative array
                    $row = mysql_fetch_assoc($res);

                    //check to see if the user has not activated their account yet
                    if($row['active'] != 1){
                        //if not display error message
                        echo "<center>You have not yet <b>Activated</b> your account!</center>";
                    }else{
                        //if they have log them in

                        //set the login session storing there id - we use this to see if they are logged in or not
                        $_SESSION['uid'] = $row['id'];
                        //show message
                        echo "<center>You have successfully logged in!</center>";

                        //update the online field to 50 seconds into the future
                        $time = date('U')+50;
                        mysql_query("UPDATE `users` SET `online` = '".$time."' WHERE `id` = '".$_SESSION['uid']."'");

                        //redirect them to the usersonline page
                        header('Location: usersOnline.php');
                    }
                }
            }
        }
    }

    ?>

Answer 1

I am not sure what the protect function is doing but here is what you can do:

$username = mysql_real_escape_string($_POST['username']);
$password = mysql_real_escape_string($_POST['password']);

and in your query, for debugging use this to get the correct error

$res = mysql_query("SELECT * 
                  FROM `users` 
                  WHERE 
                     `username` = '$username' 
                  AND 
                     `password` = '$password'") or die('Error: '.mysql_error());

Answer 2

It is possible that $username contains special characters? you should check in the mysql log if your query is executed properly from the database point of view.

Answer 3

Probably your query fails. mysql_query() returns false , if an error occurs. Use mysql_error() to identify the real problem.

$res = mysql_query("SELECT * FROM `users` WHERE `username` = '".$username."'");
if ($res === false) {
    echo mysql_errno() . ': ' . mysql_error();
    exit;
}
$num = mysql_num_rows($res);

Answer 4

As the documentation of mysql_query() states:

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.

So just check, whether you have within the connected database this table, listed columns and your query is ok - it may be just the error of your query.

There is the example within the documentation (see the link at the beginning), how to see what actually happened (see Example #1; replace with your own query):

$result = mysql_query('SELECT * WHERE 1=1');
if (!$result) {
    die('Invalid query: ' . mysql_error());
}