当我运行的代码没有语法或任何错误实际上OCUR但问题是,SQL查询似乎没有被阅读的用户名和密码:PHP和MySQL登录错误存在的
这里是我所得到的当我运行的代码,并尝试登录:
That information is incorrect, try again Click Here
这里是PHP
<?php
if(isset($_POST["username"]) && isset($_POST["password"])){
$manager = preg_replace('#[^A-Za-z0-9]#i','',$_SESSION["username"]);
$password = preg_replace('#[^A-Za-z0-9]#i','',$_SESSION["password"]);
include"db_connection.php";
$sql = mysql_query("SELECT id FROM admin WHERE username='$manager' AND password='$password'");
$existCount = mysql_num_rows($sql);
if ($existCount ==1){
while($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();
}
}
?>
下面是HTML:
<html>
<head>
<title> Admin Login Page</title>
<head>
<body>
<div align="center" id="mainWrapper">
<div id="pageContent"><br/>
<div align="left" style="margin-left:24px;">
<h2> Please log in To manage the store</h2>
<form id="form1" name="form1" method="post" action="admin_login.php">
User Name: <br/>
<input name="username" type="text" id="username" size="40"/>
<br/></br>
Password: <br/>
<input name="password" type="password" id="password" size="40">
<br/>
<br/>
<br/>
<br/>
<br/>
<label>
<input type="submit" name="button" id="button" value="LogIn">
</label>
</form>
</body>
</html>
感谢我固定的@jan的方式,但我仍然得到同样的错误 – user3311898
我们对我们的方式:-)。你得到的错误究竟是什么? – jan
:)呃,错误是当我尝试登录它给了我'这个信息是不正确的,当再次输入用户名和密码时,请再次点击这里'@jan – user3311898