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Php应用程序登录失败

php html login

Php应用程序登录失败,php,html,login,Php,Html,Login,我不熟悉PHP，正在开发一个登录表单。请在下面找到我使用的代码。在这里我想让登录工作。请帮助我 config.php <?php $dbUser="root"; $dbPassword=""; $dbName="forsitelogin"; $dbHost="localHost"; $dbConnection= mysql_connect($dbHost, $dbUser, $dbPassword); if($dbConnection) { mysql_select_db($

我不熟悉PHP，正在开发一个登录表单。请在下面找到我使用的代码。在这里我想让登录工作。请帮助我

config.php

<?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> ");

?>

<?php

        require 'config.php';
        require 'thems\login.html';

     ?>

<?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");
}


?>
index.php

<?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> ");

?>

<?php

        require 'config.php';
        require 'thems\login.html';

     ?>

<?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");
}


?>
login.php

<?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> ");

?>

<?php

        require 'config.php';
        require 'thems\login.html';

     ?>

<?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

    <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>


电子邮件：

密码：
警告：mysql_query（）希望参数1是字符串，即C:\Program Files\xampp\htdocs\forsiteSystem\login.php中在线提供的资源 二十二

如果查看，您将看到
mysql\u query
需要的第一个参数是SQL查询（而您提供的是链接标识符，然后是查询）

您需要将第一个参数作为查询，第二个参数作为连接

mysql_查询（“从表中选择*，$connection）

Login.php应该如下所示

<?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");
}

?>

试试这个


$query=mysqli_query("SELECT * FROM users WHERE email= 'v' AND password='123' ",$dbConnection);


login.php文件


$query=mysql_query("SELECT * FROM users WHERE email='v' AND password='123'",$dbConnection);


你错把“”放在里面了检查一下

您可以更换此线路并避免警告


$query=mysql_query("SELECT * FROM users WHERE email='v' AND password='123'",$dbConnection);



mysql_查询（“从表中选择*，$connection）；它不起作用啊，几分钟前我看到的是mysql\u查询而不是mysql\u查询。很抱歉，答案是针对“mysql_query”$con=mysqli_connect（“localhost”、“user”、“password”、“db”）；mysqli_查询（$con，“从表中选择*）；你能进一步解释一下你的mysql查询（$connection，“SELECT*FROM…”）吗
但文档中规定参数的顺序必须是
mysql\u查询（“SELECT*FROM…”，$connection）