在Android中从具有嵌套json对象的json获取字符串和具有多个json对象的嵌套json数组 [英] Get String from json with nested json object and nested json arrays with multiple json object, in Android

问题描述

我需要以String的形式访问复杂Json中包含的所有单个参数.

I need to access as String all the single parameters contained in a complex Json.

例如String people=...; String idPeople=...;等

for example String people=...; String idPeople=...; etc.

我尝试使用JSONTokeners，就像我尝试搜索类似的问题一样，对于简单的json我没有问题，但是我不知道如何从中正确获取参数:

I have tried to use the JSONTokeners, as I have tried to search for similar question, and for simple json I haven't problem but I don't know how to get the parameters correctly from this:

{"id":1,"error":null,"result":
  {"nPeople":2,
    "people":[
            {"namePeople":"Inca",
             "power":"1235",
             "location":"asdfghjja",
             "idPeople":189,
             "mainItems":"brownGem",
             "verified":false,
             "description":"Lorem impsum bla bla",
             "linkAvatar":"avatar_12.jpg",
             "longitude":16.2434263,
             "latitude":89.355118},

            {"namePeople":"Maya",
             "power":"1235",
             "location":"hcjkjhljhl",
             "idPeople":119,
             "mainItems":"greenstone",
             "verified":false,
             "description":"Lorem impsum bla bla",
             "linkAvatar":"avatar_6.jpg",
             "longitude":16.2434263,
             "latitude":89.3551185}]
    }
}

请注意，数组中对象的数目并不总是2 ...并且可能包含4个或更多的对象

NB the numbers of object in the array people is not always 2... and may contains 4 or more people object

推荐答案

我没有尝试过. 但是我想这可能有用.

I've not tried. But i guess it may work.

    JSONObject obj = new JSONObject(jsonString);
    String id = obj.getString("id");
    String error = obj.getString("error");
    JSONObject result = obj.getJSONObject("result");
    int nPeople = result.getInt("nPeople");
    JSONArray people = result.getJSONArray("people");
    for(int i = 0 ; i < people.length() ; i++){
        JSONObject p = (JSONObject)people.get(i);
        String namePeople = p.getString("namePeople");
        ...
    }

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