java - 从 JSON 字符串中提取单个元素

Question

我知道可能有人会问这个问题，但我无法得到具体答案。我有一个 json 字符串:

GraphObject{graphObjectClass=GraphPlace, state={"id":"268367713308665","category":"School","location":{"state":"","zip":"","longitude":32.631482614136,"latitude":0.24519375867929,"country":"Uganda","city":"Kampala","street":"PO BOX 28493"},"category_list":[{"id":"365182493518892","name":"School"}],"name":"Little Wonders Montessori"}}
GraphObject{graphObjectClass=GraphPlace, state={"id":"142442605900768","category":"Education","location":{"state":"","zip":"","longitude":32.606752647018,"latitude":0.28746491511647,"country":"Uganda","city":"Kampala","street":"Heritage\/ Kiwafu Stage, Wheeling Zone, Gaba Road, Kansanga"},"category_list":[{"id":"151676848220295","name":"Education"}],"name":"Wisdomgate Pre-School, Kansanga."}}

当我查询附近的地点时，我从 Facebook 获得了这些信息，我想提取纬度、经度、类别和名称。这就是我正在尝试的

public void onCompleted(List<GraphPlace> places, Response response) {
    //Toast.makeText(getActivity(), (CharSequence) places.listIterator(), Toast.LENGTH_LONG).show();
    ListIterator<GraphPlace> x =    places.listIterator();
    while(x.hasNext()){
        String f = x.next().toString();
        Toast.makeText(getActivity(), f, Toast.LENGTH_LONG).show();
        mytext.setText(f);
        try {
            JSONObject json = new JSONObject(f);
            JSONArray array = json.getJSONArray( "GraphPlace" );
            Toast.makeText(getActivity(), array.length(), Toast.LENGTH_LONG).show();
        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }
}   
});
request.executeAsync(); 

但是，我似乎无法获得入口点，无论我如何尝试，我都无法获得任何东西，感谢您提前提供的帮助。

Answer 1

如果您的 JSON 字符串如下所示:

{"id":"268367713308665","category":"School","location":{"state":"","zip":"","longitude":32.631482614136,"latitude":0.24519375867929,"country":"Uganda","city":"Kampala","street":"PO BOX 28493"},"category_list":[{"id":"365182493518892","name":"School"}],"name":"Little Wonders Montessori"}}

要获取纬度和经度，请执行以下操作:

        JSONObject json = new JSONObject(yourJSONVariableHere);
        JSONObject location = json.getJSONObject( "Location" );
        String lat = location.getString("latitude");
        String long = location.getString("longitude");

对于 lat 和 long 变量，您可以根据 JSON 的值分配 int 或 long。如果更改变量类型，则需要对相应的变量值执行 getInt() 或 getLong 操作。

要记住的其他事情是，您有一个嵌套对象，其中包含 lat 和 long，因此您必须获取它，然后解析它。如果您有嵌套数组或其他对象，情况也是如此。目前您正在使用 getJSONArray，但 JSON 中的数组具有 [array, array] 或方括号，其中 JSON 对象具有 {object, object} 大括号。

希望这有帮助。