关于javascript中回调函数(callback)中this指针的有关问题

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();

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();

------解决方案--------------------

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();