Php 有人能帮我解决这个错误吗；警告：无法修改标题信息-标题已由“发送”；

这是登录表单…它说它不能修改标题信息，我尝试了所有方法来修复，但我不能…有人能帮我修改代码吗…提前谢谢

---

如果查询失败，您的脚本将因
而死亡或死亡（mysql\u错误）
在$getuser之后不会输出任何内容

将您的代码修改为

 include('movieshub/includes/config.php');
    if ($getuser = mysqli_query($con,$login)) { // check if the query succeeded running
      $count = mysqli_num_rows($getuser);
      if ($count == 0 ) {

             echo "<SCRIPT>alert('$failpopup');</SCRIPT>";
             header("location:index.php");

      } else  {

        while($row = mysqli_fetch_array($getuser)) 
        { //output data }
          echo "<SCRIPT>alert('$loginpopup');</SCRIPT>";
          header("location:home.php");
        }
    }
 } else {
echo "query failed to run";
 }

include（'movieshub/includes/config.php'）；
如果（$getuser=mysqli_query（$con，$login））{//检查查询是否成功运行
$count=mysqli\u num\u行（$getuser）；
如果（$count==0）{
回显“警报（'$failpopup'）；”；
标题（“location:index.php”）；
}否则{
while（$row=mysqli\u fetch\u数组（$getuser））
{//输出数据}
回显“警报（'$loginpopup'）；”；
标题（“location:home.php”）；
}
}
}否则{
echo“查询运行失败”；
}
请尝试以下代码：

 <?php

        //if your are using wamp then let $servername,$username and $password be same as below otherwise change them.

        $servername = "localhost"; //insert your severname at the place of localhost
        $username = "root"; //insert your username at the place of root
        $password = "";  //insert your password at the place of ""

        // Create connection
        $con = mysqli_connect($servername, $username, $password); 

        //select database
        mysqli_select_db($con,"test"); //here enter your database name at the place of test

        // Check connection
        if (!$con) {
            die("Connection failed: " . mysqli_connect_error());
        }

       $loginpopup = 'Login Success';
       $failpopup = 'Wrong Username or Password';




    if (isset($_POST['submitlogin']))
    {
        $user=$_POST["user"];
        $pass=$_POST["pass"];

       $login = "SELECT * from admin where username=$user AND password=$pass";

            $getuser = mysqli_query($con,$login);

            $row=mysqli_affected_rows($con);
              if($row>1)
              {
                echo "<SCRIPT>alert('$loginpopup');</SCRIPT>";
                  header("location:home.php");


               }

             else
                {
                  echo "<SCRIPT>alert('$failpopup');</SCRIPT>";
                     header("location:index.php");
                }

    }
    ?>

显示错误报告（E_ALL）；ini设置（“显示错误”，1）；您能在这里跳过全部错误吗？警告：编写您自己的访问控制层并不容易，而且有很多机会使它严重出错。这有很多危险，因为你没有。此代码允许任何人从您的站点获取任何内容。不要编写自己的身份验证系统。任何类似的密码都有一个内置的。警告：至少要遵循并且永远不要将密码存储为纯文本。给您提供密码的人相信您会安全地存储密码。警告通知中有答案xD。警告：mysqli_num_rows（）期望参数1是mysqli_result，在第204行的C:\wamp\www\eserver\index.php中给出布尔值……仍然是一个错误，这就是我们需要的echo$login；在最后一个else语句SELECT*from admin，其中username=dan和password=dan应该是SELECT*from admin，其中username='dan'和password='dan'起作用了，先生谢谢…但是这里还有一个关于此标题的错误。警告：无法修改标题信息-标题已由发送（输出从C:\wamp\www\eserver\index.php:165开始）在C:\wamp\www\eserver\index.php第208行，您面临什么？我已经解决了问题，先生，另一个问题弹出…警告：无法修改标题信息-标题已由发送（输出从C:\wamp\www\eserver\index.php:166开始）在第216行的C:\wamp\www\eserver\index.php中，当出现此警告时？当我已成功登录并显示此错误时，只需转到您的问题，您将获得编辑选项，然后在那里更新您的问题
 <?php

        //if your are using wamp then let $servername,$username and $password be same as below otherwise change them.

        $servername = "localhost"; //insert your severname at the place of localhost
        $username = "root"; //insert your username at the place of root
        $password = "";  //insert your password at the place of ""

        // Create connection
        $con = mysqli_connect($servername, $username, $password); 

        //select database
        mysqli_select_db($con,"test"); //here enter your database name at the place of test

        // Check connection
        if (!$con) {
            die("Connection failed: " . mysqli_connect_error());
        }

       $loginpopup = 'Login Success';
       $failpopup = 'Wrong Username or Password';




    if (isset($_POST['submitlogin']))
    {
        $user=$_POST["user"];
        $pass=$_POST["pass"];

       $login = "SELECT * from admin where username=$user AND password=$pass";

            $getuser = mysqli_query($con,$login);

            $row=mysqli_affected_rows($con);
              if($row>1)
              {
                echo "<SCRIPT>alert('$loginpopup');</SCRIPT>";
                  header("location:home.php");


               }

             else
                {
                  echo "<SCRIPT>alert('$failpopup');</SCRIPT>";
                     header("location:index.php");
                }

    }
    ?>