PHP的登录代码不起作用

Question

所以我用php的登录页面写了这段代码

<?php 
function SignIn() {
    session_start(); //starting the session for user profile page
    if(!empty($_POST['username'])) //checking the 'user' name which is from Sign-In.html, is it empty or have some text 
    {
        $query = mysql_query("SELECT * FROM register where Username = '$_POST[username]' AND Password = '$_POST[password]'") or die(mysql_error()); 
        $row = mysql_fetch_array($query) or die(mysql_error());
        if(!empty($row['Username']) AND !empty($row['Password'])) { 
            if($password = $row['Password']) {
                $_SESSION['Username'] = $row['Password'];
                echo "welcome    ".$row['Name']; 
            }
            if ($password != $row['Password']) {
                echo "Something went wrong!" ;
            }
        }
    }
}

if(isset($_POST['submit'])) {
        SignIn(); 
}
?>

我的问题是，虽然登录部分可以正常工作（当我输入正确的密码时），当我输入错误的密码时，它不显示消息，而是将其设置为回显，而是显示空白页。很多次但没有结果...有人可以给我一些帮助吗？

Answer 1

您的第一个IF语句将$password设置$password $row['password']值：

if($password = $row['Password'])

当第二个if语句开始时， $password与$row['password'] ，条件等于false：

if($password != $row['Password'])

解决方案是将第一个if语句更改为可比较值：

if($password == $row['Password'])

注意，有两个=符号？ 这告诉PHP检查它们是否都是相同的值。

Answer 2

您似乎错过了一个=号：

 if($password = $row['Password'])

因此，您不是在进行比较，而是进行分配。 我希望这应该是：

 if($password == $row['Password'])

Answer 3

尝试这个

function SignIn() { 
 session_start(); //starting the session for user profile page
  if(!empty($_POST['username'])) //checking the 'user' name which is from Sign-In.html, is it empty or have some text 
  {
   $query = mysql_query("SELECT * FROM register where Username = '$_POST[username]' AND Password = '$_POST[password]'") or die(mysql_error()); 
   $row = mysql_fetch_array($query) or die(mysql_error());
    if(!empty($row['Username']) && !empty($row['Password'])) { 
        if($password == $row['Password'])
        {
            $_SESSION['Username'] = $row['Password'];
             echo "welcome   ".$row['Name']; 

        }
         else
            echo "Something went wrong!" ;
     }

    }

     }

Answer 4

第二个if条件必须在第一个if条件之外。

您还将密码分配给$_SESSION['Username']而不是$row['Username'] 。 我编辑了代码..见下文

<?php 
    function SignIn() {
        session_start(); //starting the session for user profile page
        if(!empty($_POST['username'])) //checking the 'user' name which is from Sign-In.html, is it empty or have some text 
        {
            $query = mysql_query("SELECT * FROM register where Username = '$_POST[username]' AND Password = '$_POST[password]'") or die(mysql_error()); 
            $row = mysql_fetch_array($query) or die(mysql_error());

            if(!empty($row['Username']) AND !empty($row['Password'])) { 
            if($_POST['password'] = $row['Password'])
            $_SESSION['Username'] = $row['Username']; 
            echo "welcome   ".$row['Username']; 
            }

             if ($_POST['password'] != $row['Password']) {
               echo "Something went wrong!" ;
            }
        }
    }

    if(isset($_POST['submit'])) {
            SignIn(); 
    }


    ?>