Question

<form name='form-login' action="<?php header("location: /vardhman/administrator/check.php");?>"  method="post">
    <span class="fontawesome-user"></span>
      <input type="text" id="username" name="username" placeholder="Username">

    <span class="fontawesome-lock"></span>
      <input type="password" id="password" name="password" placeholder="Password">

    <input type="submit" value="Login"></form>

此表格和

这是check.php

$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="mlm"; // Database name 
$tbl_name="administrator"; // Table name 


// Connect to server and select databse.
$conn= mysqli_connect("$host", "$username", "$password")or die("cannot connect"); 

mysqli_select_db($conn,"$db_name")or die("cannot select DB");

// username and password sent from form 
if(isset($_POST['username'])) { 
    // check if the username has been set

$myusername=$_POST['username']; 
}
else{
echo "nothing is passed";   
}

IT只显示没有传递。我不知道是什么问题。请帮助我，我无法传递价值。提前谢谢。

Answer 1

您的代码中存在错误。您不应该在action属性中使用header（），只需将路径写为纯文本

正确的代码：

<form name='form-login' action="/vardhman/administrator/check.php"  method="post">

发现另一个错误：
您不应该将变量包装在引号内。

// Connect to server and select databse.
$conn= mysqli_connect("$host", "$username", "$password")or die("cannot connect"); 

mysqli_select_db($conn,"$db_name")or die("cannot select DB");

正确的代码：

// Connect to server and select databse.
$conn= mysqli_connect($host, $username, $password)or die("cannot connect"); 

mysqli_select_db($conn, $db_name)or die("cannot select DB");

php post方法没有工作请求和get方法无能为力

1 个答案: