javascript - 改进js代码去掉全局变量和函数

Question

我有一段 js 代码，我真的很想改进它，但不确定如何改进。下面的工作版本有一个全局变量和一个单独的函数声明，我认为它可以合并到一个匿名函数中(下面的代码片段:不工作)

如有任何帮助或指点，我们将不胜感激。

工作版本:

var Data = {}; // would like to remove the global variable

function callBack() {
    $.ajax({
        url: "http://host/callBacks/script.js",
        //get and execute Script to process json data below
        dataType: "script"
    });
}


$(document).ready(function() {
    $.ajax({
        type: "Get",
        url: "http://host/data/json/",
        success: function(json) {
            Data = json; // Would like to just call the callback here
            callBack(Data);
        },
        error: function() {
            alert("error");
        }
    });
});


// Script which gets loaded from callBack
(function(Data) {
    var json = $.parseJSON(Data);
    $.each(json, function(i, v) {
        alert(v);
    });
})(Data);

所需代码:不工作

// Error: Length is null or not an object. In jQuery script

var Data = {}; // Ideally would like to remove this from global scope if possible

$(document).ready(function() {
    $.ajax({
        type: "Get",
        url: "http://host//data/json/",
        success: function(Data) {
            $.ajax({
                url: "http://host/callBacks/script.js",
                dataType: "script"
            });
        },
        error: function() {
            alert("error");
        }
    });
// Error: Length is null or not an object. In jQuery script

更新:根据 adeneo 的回答:还是需要global Data = {};因为我想返回的立即执行脚本作为参数接收

var Data = {};

$(document).ready(function() {
    function doAjax() {
        return $.ajax({
            type: "GET",
            url: "http://host/data/json/"
        });
    }

    var XHR = doAjax();
    XHR.done(function(data) {
        Data = data; // <--- If I remove this I get Error:'length' is not or not an object
        $.ajax({
            url: "http://host/callBacks/script.js",
            dataType: "script"
        });
    }).fail(function() {
        alert("error");
    });
});

不过看起来还不错。可能应该提到我正在 IE 8 中对此进行测试。约束。更新标签

Answer 1

$(function() {
    function doAjax() { //put inside function
        return $.ajax({ //return deffered object
            type: "GET",
            url: "/data/json/" //X domain not supported
        });
    }

    var XHR = doAjax();  //do ajax call
    XHR.done(function(data) { // use deffered object
        var json = $.parseJSON(data); //why would you need to get an external script just to parse some JSON?
        $.each(json, function(i, v) {
            alert(v);
        });
    }).fail(function() {
        alert("error");
    });
});