2017-02-25 83 views
0

我想讓它變成當我的ajax代碼沒有返回錯誤時,我得到了模態。我所有的代碼都可以正常工作,只需要知道如何在沒有錯誤時進行模式彈出。我有一個Ajax成功:函數,但我不知道這段代碼如何工作,因爲目前,它只有在點擊按鈕時才起作用。PHP - 成功註冊後顯示SweetAlert模態彈出窗口(AJAX)

我使用SweetAlert JS文件

我的代碼:

按鈕被按下時,工作模式:

<script> 
    !function(e){ "use strict"; 
    var t = function(){}; 
    t.prototype.init = function() { 
     e("#sa-success").click(function() { 
     swal("Registration Successful","Your registration was successful! Check your email and click on your activation link to be able to access your account.","success") 
     } 
     ) 
    },e.SweetAlert=new t,e.SweetAlert.Constructor=t}(window.jQuery),function(e){"use strict";e.SweetAlert.init()}(window.jQuery); 
</script> 

我的AJAX註冊碼我想讓msg =''的模式彈出:

<script> 
     $(document).ready(function() { 

    jQuery.validator.addMethod("noSpace", function(value, element) { 
    return value.indexOf(" ") < 0 && value != ""; 
    }, "Spaces are not allowed"); 

      $("#registered_form").submit(function() { 

       if ($("#registered_form").valid()) { 
        var data1 = $('#registered_form').serialize(); 
        $.ajax({ 
         type: "POST", 
         url: "inc/pgs/register.php", 
         data: data1, 
         success: function(msg) { 
          console.log(msg); 
          if(msg == '') { 
           //setTimeout("window.location.href='login.php';",4000); 
          } else { 
           $("#result").html('<div class="alert alert-danger"><button type="button" class="close"></button>' + msg +'</div>').slideDown(100); 
           window.setTimeout(function() 
           { 
            $(".alert").fadeTo(500, 0).slideUp(500, function(){ 
             $(this).remove(); 
            }); 
           }, 3000); 
          } 
         } 
        }); 
       } 
       return false; 
      }); 
     }); 
    </script> 

回答

0

全部排序。我不得不看SweetAlert手冊,看看它是如何工作的。我應該檢查過。

我不得不使用:

swal("Registration Success!", "Your registration was successful! Check your email and click on your activation link to be able to access your account.", "success");