如何判断返回的是JSON-simple(Java)的JSONObject还是JSONArray?

问题描述：

我正在获得服务，有时会找回这样的东西:

I am hitting a service and sometimes getting back something like this:

{ "param1": "value1", "param2": "value2" }

有时会得到这样的回报:

and sometimes getting return like this:

[{ "param1": "value1", "param2": "value2" },{ "param1": "value1", "param2": "value2" }]

我怎么知道我要去哪?当我执行getClass()时，它们两个都评估为字符串，但是如果我尝试这样做:

How do I tell which I'm getting? Both of them evaluate to a String when I do getClass() but if I try to do this:

json = (JSONObject) new JSONParser().parse(result);

在第二种情况下，我得到了例外

on the second case I get an exception

org.json.simple.JSONArray cannot be cast to org.json.simple.JSONObject

如何避免这种情况?我只想知道如何检查我回来了. (第一种情况有时会包含[]，因此我无法索引of，并且我希望使用一种更清洁的方法，而不仅仅是检查第一个字符.

How to avoid this? I would just like to know how to check which I'm getting back. (The first case will sometimes have [] in it so I can't do index of and I'd like a cleaner way than just checking the first character.

一定有某种方法可以检查这一点吗?

There has got to be some sort of method that checks this?

答

简单的Java:

Object obj = new JSONParser().parse(result); 
if (obj instanceof JSONObject) {
    JSONObject jo = (JSONObject) obj;
} else {
    JSONArray ja = (JSONArray) obj;
}

如果您想避免解析错误类型的JSON的开销，还可以测试(声称的)JSON以[还是{开头.但是要注意领先的空格.

You could also test if the (purported) JSON starts with a [ or a { if you wanted to avoid the overhead of parsing the wrong kind of JSON. But be careful with leading whitespace.

如何判断返回的是JSON-simple(Java)的JSONObject还是JSONArray?

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