2

目前正在做一个需要管理面板的网站,我有一个 php 问题,在正确插入用户名和密码的值时,它似乎没有得到行数。这是php代码:

admin_login.php

<?php 
session_start();
if (isset($_SESSION["manager"])) {
    header("location:index.php"); 
    exit();
}
?>
<?php 
if (isset($_POST["username"]) && isset($_POST["password"])) {
    $manager = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["username"]); 
    $password = preg_replace('#[^A-Za-z0-9]#i', '', $_POST["password"]); 
    // Connect to the MySQL database  
    include "../storescripts/connect_to_mysql.php"; 
    $sql = mysql_query("SELECT id FROM admin WHERE username='$manager' AND password='$password' LIMIT 1"); // query the person
    $existCount = mysql_num_rows($sql); // count the row nums
    if ($existCount == 1) { // evaluate the count
         while($row = mysql_fetch_array($sql)){ 
             $id = $row["id"];
         }
         $_SESSION["id"] = $id;
         $_SESSION["manager"] = $manager;
         $_SESSION["password"] = $password;
         header("location: index.php");
         exit();
    } else {
        echo 'That information is incorrect, try again <a href="index.php">Click Here</a>';
        exit();
    }
}

连接到 sql db 工作正常,我做了一个 echo 以确保它工作正常

这是 index.php php 代码:

<?php 
session_start();
if (!isset($_SESSION["manager"])) {
    header("location: admin_login.php"); 
    exit();
}

?>
<?php 
$managerID = preg_replace('#[^0-9]#i', '', $_SESSION["id"]); 
$manager = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["manager"]); 
$password = preg_replace('#[^A-Za-z0-9]#i', '', $_SESSION["password"]); 
include "../storescripts/connect_to_mysql.php"; 
$sql = mysql_query("SELECT * FROM admin WHERE id='$managerID' AND username='$manager' AND password='$password' LIMIT 1");
$existCount = mysql_num_rows($sql);
 if(!$existCount == 0){
     echo "Your login session data is not on record in the database.";
     exit();
}
?>

我确实相信这是 $existCount 中的错误并且它没有得到计数?

干杯

4

2 回答 2

0

我相信您正在从developphp 获取代码。通过使用上面的代码,如果您的密码中有一些字符不是字母数字,可能会导致过滤器将它们删除并与数据库不匹配并将 $existCount 返回为零。

于 2014-08-16T04:54:09.833 回答
0

您的问题可能与最后的 if 语句有关。改变:

if(!$existCount == 0){
     echo "Your login session data is not on record in the database.";
     exit();
}

至:

if($existCount == 0){
     echo "Your login session data is not on record in the database.";
     exit();
}

请注意 missing !,它以您不想要的方式否定表达式。

于 2013-07-10T23:05:55.550 回答