2016-04-23 69 views
0

這是模式彈出我希望當用戶選擇上的一個按鈕,他沒有登錄顯示。如何顯示模式,而不是警告

從我的javascript函數我叫它如下:

function addToWishlistFun(tourid) { 
     if (document.getElementById("wishlist" + tourid).value == "added") 
     { 
      $.post('@Url.Action("RemoveFromWishlist", "Home")', { id: tourid }, function (data) { 
       if (data) { 
        var elementid = "wishlist" + tourid; 
        document.getElementById(elementid).innerHTML = "Add to wishlist" + "<i class=\"fa fa-heart-o\"></i>"; 
        document.getElementById(elementid).value = "notadded"; 
       } 
       else { 
        //$("#testPopup").toggle(); 
        document.getElementById("#must-be-logged-in") 
        // alert("Tour could not be removed from wishlist. Please make sure that you are logged in."); 
       } 
      }); 
     } 

但是,當按鈕被點擊時,沒有任何顯示。

回答

0

替換:

  else { 
       //$("#testPopup").toggle(); 
       document.getElementById("#must-be-logged-in") 
       // alert("Tour could not be removed from wishlist. Please make sure that you are logged in."); 
      } 

爲:

  else { 
       $("#must-be-logged-in").toggle(); 
       // alert("Tour could not be removed from wishlist. Please make sure that you are logged in."); 
      } 
+0

嘗試過,但仍然沒有運氣 – Techworld

+0

嘗試$( '#必須待登錄')模態( '秀'); –

+0

仍然不起作用:( – Techworld