javascript - 如何在jquery ajax成功回调函数中传递上下文

Question

var Box = function(){
    this.parm = {name:"rajakvk",year:2010};
    Box.prototype.jspCall = function() {
        $.ajax({
            type: "post",
            url: "some url",
            success: this.exeSuccess,
            error: this.exeError,
            complete: this.exeComplete
        });
    }
    this.exeSuccess = function(){
        alert(this.parm.name);
    }
}

我没有在 exeSuccess 方法中获取 Box 对象。如何在 exeSuccess 方法中传递 Box 对象？

Answer 1

使用 context option ，像这样:

    $.ajax({
        context: this,
        type: "post",
        url: "some url",
        success: this.exeSuccess,
        error: this.exeError,
        complete: this.exeComplete
    });

context 选项决定了调用回调的上下文......因此它决定了 this 在该函数中指的是什么。