如何显示模式，而不是警告

Question

这是模式弹出我希望当用户选择上的一个按钮，他没有登录显示。

从我的javascript函数我叫它如下：

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，当按钮被点击时，没有任何显示。

Answer 1

替换：

  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."); 
      } 

为：

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