我的登錄jquery沒有正確地調用php文件

Question

嘿即時嘗試編寫代碼的登錄系統，這是我第一次，我用了一個在線教程。最初的Javascript工作正常，輸入空白時顯示錯誤。但是，當我在輸入中輸入文本並單擊登錄時，沒有任何反應。在我看來，它不是調用PHP文件，但我看不到任何理由，因爲它被正確引用？任何幫助將不勝感激！

<?php 
    include ("Includes/dbConnect.php"); 
    $u=$_POST['u']; 
    $p=$_POST['p']; 

    //Strip slashes 
    $u = stripslashes($u); 
    $p = stripslashes($p); 

    //Strip tags 
    $u = strip_tags($u); 
    $p = strip_tags($p); 


    $check = mysqli_query("SELECT * FROM Login WHERE login_user ='$u' AND login_pass='$p'")  or die(mysqli_error()); 
    $check = mysqli_num_rows($check); 
    if($check !=="0"){ 
    $results = mysqli_query("SELECT * FROM Login WHERE login_user = '$u'") or die(mysqli_error()); 
     while ($row = mysqli_fetch_assoc($results)) { 
      $login_user=$row['login_user']; 
      session_register('login_user'); 
      $_SESSION['login_user'] = $Login_user; 
      echo "1"; 
    } 
    } 
?> 

，這裏是JavaScript的

<script type="text/javascript"> 
    $(document).ready(function() { 
     $(".sign_b_btn").live("click", function() { 
      var u = $("#u").val(); 
      var p = $("#p").val(); 

      if (u == "") { 
       $("#u").css("border-color", "red"); 
       $("#un").css("color", "red"); 
       $(".error").show().html("Please enter your username!"); 
       $("#p").css("border-color", "#606060"); 
       $("#up").css("color", "#333333"); 
      } else if (p == "") { 
       $("#u").css("border-color", "#606060"); 
       $("#un").css("color", "#606060"); 
       $(".error").show().html("Please enter your Password!"); 
       $("#p").css("border-color", "red"); 
       $("#up").css("color", "red"); 
      } else { 

       dataString = 'u=' + u + '&p=' + p; 
       $.ajax({ 
        type: "POST", 
        url: "Ajax/login_php.php", 
        data: dataString, 
        cache: false, 
        success: function (html) { 
         if (html == "") { 
          $(".error").show().html("The username or password you entered is incorrect!"); 
          $("#p,#u").css("border-color", "red"); 
          $("#up,#un").css("color", "red"); 
         } else if (html == "1") { 
          $(".error").fadeOut(1000); 
          $("#u").css("border-color", "#606060"); 
          $("#un").css("color", "#333333"); 
          $("#p").css("border-color", "#606060"); 
          $("#up").css("color", "#333333"); 
          $(".center").animate({ 
           opacity: 0.25, 
           left: '+=900', 
           height: 'toggle' 
          }, 5000, function() { 
           $(".done").slideDown(200).html("Welcome " + u); 
           setTimeout(function() { 
            var u = $("#u").val(""); 
            var p = $("#p").val(""); 
            window.location = "index.php"; 
           }, 5000); 
          }); 
         } 
        } 
       }); 
      } 
     }); 
    }); 
</script> 

Answer 1

回答您的問題會話：

編輯你的PHP代碼，使你的腳本開始了與調用在session_start（）;

如：

<?php session_start(); 
//rest of your php code 

更換線路：

session_register('login_user'); 
$_SESSION['login_user'] = $Login_user; 

有：

$_SESSION['login_user'] = $Login_user; 

你都拿到你，因爲函數了session_register提到的錯誤（）;已在PHP 5.3.0版中被棄用，並從PHP版本5.4.0中刪除。 $ _SESSION是在session_start（）之後爲您提供的全局數組。呼叫。

快樂編程:)

Answer 2

感謝您的幫助球員，真的幫我弄明白了。完成代碼，如果任何人感興趣：

<?php 

include ("../Includes/dbConnect.php"); 
$u=$_POST['u']; 
$p=$_POST['p']; 

//Strip slashes 
$u = stripslashes($u); 
$p = stripslashes($p); 

//Strip tags 
$u = strip_tags($u); 
$p = strip_tags($p); 



$query = "SELECT * FROM Login WHERE login_user ='$u' AND login_pass='$p'"; 
$check = mysqli_query($cxn,$query) or die("Couldn't execute query!"); 
$check = mysqli_num_rows($check); 
if($check !=="0"){ 
$results = mysqli_query($cxn,"SELECT * FROM Login WHERE login_user = '$u' AND login_pass='$p'") or die("Couldn't execute query2!"); 
    while ($row = mysqli_fetch_assoc($results)) { 
     $username= $u; 
     session_register('Login'); 
     $_SESSION['user'] = $username; 
     echo "1"; 
} 
} 


?>