幫助 我在使用 PHP 時收到一條錯誤消息“警告：mysql_num_rows()：提供的參數不是有效的 MySQL 結果資源...”

Question

我正在嘗試在 MySql 數據庫中運行查詢，但每次都收到相同的警告。 有人可以幫忙嗎。

警告：mysql_num_rows()：在第 279 行提供的參數不是有效的 MySQL 結果資源/XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX/public_html/login.php 您提供的用戶名不存在！

我用 ++++++++++++$num = mysql_num_rows($res);++++++++++++ 突出顯示了第 279 行

請讓我知道如何解決這個問題。

<?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

我不確定保護 function 正在做什么，但您可以執行以下操作：

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

並在您的查詢中，用於調試使用它來獲得正確的錯誤

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

Answer 2

$username 可能包含特殊字符嗎？ 如果您的查詢從數據庫的角度正確執行，您應該檢查 mysql 日志。

Answer 3

可能您的查詢失敗。 如果發生錯誤， mysql_query()返回false 。 使用mysql_error()來確定真正的問題。

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

Answer 4

正如 mysql_query() 的文檔所述：

對於 SELECT、SHOW、DESCRIBE、EXPLAIN 和其他返回結果集的語句，mysql_query() 成功時返回資源，錯誤時返回 FALSE。

因此，只需檢查您在連接的數據庫中是否有此表、列出的列以及您的查詢是否正常 - 這可能只是您的查詢錯誤。

文檔中有示例（請參閱開頭的鏈接），如何查看實際發生的情況（請參閱示例 #1；替換為您自己的查詢）：

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