创建与智能感知配合使用的 JavaScript 函数

我想利用 Visual Studio Intellisense 因此我已阅读：

http://msdn.microsoft.com/en-us/library/bb514138.aspx http://msdn.microsoft.com/en-us/library/bb514138.aspx

无论如何，为什么我没有得到智能感知：

function Customer() {
    this.Id = 0;
    this.FirstName = "";
    this.LastName = "";
}

function Test() {
    /// <returns type="Customer"></returns>

    var c = new Object();
    $.ajax({
        async: false,
        dataType: "json",
        url: "Ajax/GetCustomer.aspx",
        success: function (foundCustomer) {
            // I know that found customer is of type Customer
            c = foundCustomer;
        }
    });

    return c;
}

var x = Test();
x. // No intellicense! why?

我如何告诉 Visual Studio 知道该函数将返回 TYPE 的对象Customer？例如，如果我将函数 Test 替换为：function Test(){ return new Customer(); }那么智能就会发挥作用。

Edit

我的最终目标是：

function Customer() {
    this.Id = 0;
    this.FirstName = "";
    this.LastName = "";
}

Object.prototype.CastToCustomer = function(){
    /// <returns type="Customer"></returns>
    return this;
}

$.ajax({
    async: false,
    dataType: "json",
    url: "Ajax/GetCustomer.aspx",
    success: function (foundCustomer) {

        foundCustomer = foundCustomer.CastToCustomer();
        foundCustomer.// Intellicense does not work :(
    }
});

我得到了很多 json 对象，我想使用这个辅助函数来转换它们。

临时解决方案：

这就是我最终所做的：

function Customer() {
    this.Id = 0;
    this.FirstName = "";
    this.LastName = "";
}

$.ajax({
    async: false,
    dataType: "json",
    url: "Ajax/GetCustomer.aspx",
    success: function (foundCustomer) {
        // this will never be true on real browser. It is always true in visual studio (visual studio will now think that found customer is of type Customer ;)
        if (document.URL.length == 0) foundCustomer = new Customer();

        foundCustomer.// Intellisense works!!!!
    }
});

您正在将返回值初始化为Object所以它是 Intellisense 的基础。如果将其初始化为空Customer，那么 Intellisense 将检测到它正在返回Customer

function Test() {

    var c = new Customer(); //only for Intellisense

    $.ajax({...});

    return c;
}

Test(). //Customer members now appear

您还可以使用/// <param />指定参数类型：

$.ajax({
    ...

        success: function (foundCustomer) {
           /// <param name='foundCustomer' type='Customer' />
           foundCustomer. //Customer members appear            
        }
});

最后，你的document.URL.length也可以在cast方法中使用trick：

Object.prototype.CastToCustomer = function() {

  var c = new Customer();
  if (document.URL.length) c = this;
  return c;

}