我是PHP新手并开发一个登录表单.请在下面找到我使用的代码.我想要登录work.please帮助我.
config.php文件
<?php
$dbUser="root";
$dbPassword="";
$dbName="forsitelogin";
$dbHost="localHost";
$dbConnection= mysql_connect($dbHost, $dbUser, $dbPassword);
if($dbConnection)
{
mysql_select_db($dbName);
print("Sucessfully connected to database");
}
else
die("<strong>Cound not connect to database </strong> ");
?>
的index.php
<?php
require 'config.php';
require 'thems\login.html';
?>
的login.php
<?php
require 'thems\login.html';
require 'config.php';
$query=mysqli_query($dbConnection,"SELECT * FROM users WHERE email= v AND password=123");
if(mysql_num_rows($query))
{
die("login sucessfully");
}
else {
die("Incorrect password or email");
}
?>
thems/login.html的
<html>
<head>
<title></title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
</head>
<body>
<div>
<form action=".\login.php" method="post">
Email: <input type="text" name="email"><br>
password: <input type="text" name="pass"><br>
<input type="submit" id="Submit_button" >
</form>
</div>
</body>
</html>
警告:mysql_query()期望参数1为字符串,资源在第22行的C:\ Program Files\xampp\htdocs\forsiteSystem\login.php中给出
Login.php应该如下.
<?php
require 'config.php';
require 'thems\login.html';
$sql="SELECT * FROM users WHERE email='".$_POST['email']."' AND password='".$_POST['pass']."'";
$query=mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($query)){
die("login sucessfully");
}
else{
die("Incorrect password or email");
}
?>
index.php应该如下
<?php
require_once('config.php');
include('thems\login.html');
?>
您可以尝试以正确的方式学习php.
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