关于javascript中回调函数(callback)中this指针的问题

JScript code

function Test() {    this.sClassName = "Test";}Test.prototype.show = function(0 {        alert(this.sClassName);    // 正常显示Test    var callback = function() {        alert(this.sClassName); // 显示undefined，如何才能显示Test中的this.sClassName?    };        $.ajax({        success: callback    });    callback = null;};var test = new Test();test.show();

JScript code

function Test() {    this.sClassName = "Test";}Test.prototype.show = function(0 {        alert(this.sClassName);    // 正常显示Test    var cn = this.sClassName;    var callback = function() {        alert(cn); // 这样写就可以    };        $.ajax({        success: callback    });    callback = null;};var test = new Test();test.show();
[解决办法]
要理顺上下文：
function Test() {
   this.sClassName = "Test";
}

Test.prototype.show = function(0 {
  
   alert(this.sClassName);    // 正常显示Test
   var This = this;
   var callback = function() {
       alert(This.sClassName); // 显示undefined，如何才能显示Test中的this.sClassName?
   };
  
   $.ajax({
       success: callback
   });

   callback = null;
};

var test = new Test();
test.show();