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php - 如何使用我创建的HTML,PHP和MySQL数据库创建登录表单,如下所示。

转载 作者:行者123 更新时间:2023-11-29 21:05:26 25 4
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main_login.php

 <table width="300" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<form name="form1" method="post" action="checklogin.php">
<td>
<table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF">
<tr>
<td colspan="3"><strong>Member Login </strong></td>
</tr>
<tr>
<td width="78">Username</td>
<td width="6">:</td>
<td width="294"><input name="myusername" type="text" id="myusername"></td>
</tr>
<tr>
<td>Password</td>
<td>:</td>
<td><input name="mypassword" type="password" id="mypassword"></td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td><input type="submit" name="Submit" value="Login"></td>
</tr>
</table>
</td>
</form>


checklogin.php

<?php

ob_start();
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="database"; // Database name
$tbl_name="login"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

// Define $myusername and $mypassword
$myusername=$_POST['myusername'];
$mypassword=$_POST['mypassword'];

// To protect MySQL injection (more detail about MySQL injection)
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);
$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);

// If result matched $myusername and $mypassword, table row must be 1 row
if($count==1){

// Register $myusername, $mypassword and redirect to file "login_success.php"
session_register("myusername");
session_register("mypassword");
header("location:login_success.php");
}
else {
echo "Wrong Username or Password";
}
ob_end_flush();
?>


login_success.php

<?php
session_start();
if(!session_is_registered(myusername)){
header("location:login.php");
}
?>

<html>
<body>
Login Successful
</body>
</html>


没用如果用户名和密码正确,则不会发生任何事情。我应该如何纠正?我真的不知道这段代码有什么问题,非常感谢您的帮助!

最佳答案

正如我在评论中所说,您必须在会话中开始会话session_start();
  checklogin.php文件位于顶部。


您需要以一种良好的方式设计表单,不要在表内创建form。在表单内创建一个表。

进行任何查询时,请使用PDO


  不要在密码中使用mysql_real_escape_string

关于php - 如何使用我创建的HTML,PHP和MySQL数据库创建登录表单,如下所示。,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36856117/

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