JSON对象返回undefined？

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var theObject = [{"id":"412","thn_akademik":"20152","id_prodi":"15301","kode_mk":"20038","tanggal":"23\/8\/2016","jam_1":"17:00\t","jam_2":"19:00\t","npm":"07150100190","flag1":"0","flag2":"0"},{"id":"558","thn_akademik":"20152","id_prodi":"15301","kode_mk":"20053","tanggal":"25\/8\/2016","jam_1":"17:00\t","jam_2":"19:00\t","npm":"07150100190","flag1":"0","flag2":"0"},{"id":"704","thn_akademik":"20152","id_prodi":"15301","kode_mk":"20052","tanggal":"30\/8\/2016","jam_1":"17:00\t","jam_2":"19:00\t","npm":"07150100190","flag1":"0","flag2":"0"},{"id":"850","thn_akademik":"20152","id_prodi":"15301","kode_mk":"000047","tanggal":"31\/8\/2016","jam_1":"17:00\t","jam_2":"19:00\t","npm":"07150100190","flag1":"0","flag2":"0"},{"id":"996","thn_akademik":"20152","id_prodi":"15301","kode_mk":"20050","tanggal":"23\/8\/2016","jam_1":"09:00\t","jam_2":"11:00\t","npm":"07150100190","flag1":"0","flag2":"0"},{"id":"1142","thn_akademik":"20152","id_prodi":"15301","kode_mk":"20051","tanggal":"25\/8\/2016","jam_1":"09:00\t","jam_2":"11:00\t","npm":"07150100190","flag1":"0","flag2":"0"},{"id":"1288","thn_akademik":"20152","id_prodi":"15301","kode_mk":"000030","tanggal":"30\/8\/2016","jam_1":"09:00\t","jam_2":"11:00\t","npm":"07150100190","flag1":"0","flag2":"0"}] ; 


for (var i = 0; i < theObject.length; i++) { 
    alert(theObject[i].kode_mk); 
}

，当我在fiddle测试的工作。

但是，当我尝试在我的应用程序中返回未定义。

有人知道是什么让这种情况发生？

来源

2016-08-12 Tanya von Degurechaff

返回？代码中没有返回 - 没有函数可以返回 –

[没有像“JSON对象”那样的东西]（http://benalman.com/news/2010/03/theres-no-这样的事情作为一个JSON /） –

保持良好的斗争@FelixKling - 不幸的是，下一步是“我有一个JSON” –

这是一个的javascript对象NOT一个JSON对象。目前还不清楚你在问什么“但是，当我尝试在我的应用程序中返回undefined。”。什么样的应用程序？

如果您的应用检索JSON，你必须它通过它试图循环之前解析。 E.g

var json = JSON.parse(JSON_HERE);

然后可以循环通过JSON对象并获取kode_mk。

如果您有来自“应用程序”的代码，请提供它。告诉我们您的应用正在做什么以及它如何获得JSON

来源

2016-08-12 02:37:07 James111

谢谢。。。。〜 –

对不起，我总是把我的项目称为应用程序。 –

为什么在小提琴我不需要解析，但在我的代码我需要解析它？ –

JSON对象返回undefined？

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