Question

我是网络开发的新手，我尝试使用PHP进行登录。 mylogin.php的内容在加载文件index.php之后立即得到处理，我在这里有点困惑。下面的截图是加载索引文件后的响应。

response just after loading the index.php without doing anything

的index.php

  <!DOCTYPE html>
    <html>
    <head>
        <title>
            Welcome     
        </title>
    </head>
    <body>

    <h1> Please login</h1>
    <br>

    <!--
    <form action="" method="POST">
    <label>Username: </label><br>
    <input type="text" name="username" placeholder="Enter username"/><br>
    <label>Password: </label><br>
    <input type="password" name="password" placeholder="Enter password"/><br><br>
    <input type="submit" name="submit" value="Login"/><br><br>
    </form>
    -->

    <form method="post" action="">
    <label>Username:</label><br>
    <input type="text" name="username" placeholder="username" /><br><br>
    <label>Password:</label><br>
    <input type="password" name="password" placeholder="password" />  <br><br>
    <input type="submit" name="submit" value="Login" /> 
    </form>

    </body>
    </html>

    <?php
    include("mylogin.php");
    ?>

mylogin.php

<?php
    include("myconnection.php");

    $error="";
    if(empty($_POST["submit"]))
    {
        if($_POST["username"] =='' || $_POST["password"]=='')
        {
            $error='Please fill the blanks!';

        }else
        {
            $username=$_POST['username'];
            $password=$_POST['password'];

            $sql="SELECT * FROM users WHERE username='$username' AND password='$password'";
            $result=mysqli_query($db,$sql);
            $row=mysqli_fetch_array($result,MYSQLI_ASSOC);

            if(mysqli_num_rows($result)==1)
            {
                $login_user=$_POST["$username"];    
                header("location: myhome.php");

                $error="Connected";

            }
            else
            {

            //$error="Incorrect Username/Password";
            //  echo"failed";

            }

        }
    }
    else
    {

        echo"failed to  submit2";

    }

?>

Answer 1

您遇到这种情况if(empty($_POST["submit"]))时出现问题。你想要这个的负面条件，所以你需要像：

if(!empty($_POST["submit"]))（!）

Answer 2

更改包含mylogin.php文件的方式，如下面的代码所示：并且不存在这个问题。

欢迎

<h1> Please login</h1>
<br>

<!--
<form action="" method="POST">
<label>Username: </label><br>
<input type="text" name="username" placeholder="Enter username"/><br>
<label>Password: </label><br>
<input type="password" name="password" placeholder="Enter password"/><br><br>
<input type="submit" name="submit" value="Login"/><br><br>
</form>
-->

<form method="post" action="">
<label>Username:</label><br>
<input type="text" name="username" placeholder="username" /><br><br>
<label>Password:</label><br>
<input type="password" name="password" placeholder="password" />  <br><br>
<input type="submit" name="submit" value="Login" /> 
</form>

</body>
</html>

<?php
if(isset($_POST["submit"]))
{
    include("mylogin.php");
}
?>

当然，更改if文件的mylogin.php条件，如下所示：

if(isset($_POST["submit"])) OR you can change it 
if(!empty($_POST["submit"]))

如何处理include语法中的内容

2 个答案: