20

我想读这JSON行，但因为它开头JSONArray我有点困惑 

 "abridged_cast": [
            {
                "name": "Jeff Bridges",
                "id": "162655890",
                "characters": [
                    "Jack Prescott"
                ]
            },
            {
                "name": "Charles Grodin",
                "id": "162662571",
                "characters": [
                    "Fred Wilson"
                ]
            },
            {
                "name": "Jessica Lange",
                "id": "162653068",
                "characters": [
                    "Dwan"
                ]
            },
            {
                "name": "John Randolph",
                "id": "162691889",
                "characters": [
                    "Capt. Ross"
                ]
            },
            {
                "name": "Rene Auberjonois",
                "id": "162718328",
                "characters": [
                    "Bagley"
                ]
            }
        ],

我只需要使用“名称”并将所有内容保存为一个字符串。（字符串值将是：Jeff Bridges、Charles Grodin、Jessica Lange、John Randolph、Rene Auberjonois）。

这是我的代码：

try {
        //JSON is the JSON code above

        JSONObject jsonResponse = new JSONObject(JSON);
        JSONArray movies = jsonResponse.getJSONArray("characters");
        String hey = movies.toString();


    } catch (JSONException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
4

3 回答 3

67

如果您在“名称”之后，为什么您的代码片段看起来像是在尝试获取“字符”？

无论如何，这与任何其他类似列表或数组的操作没有什么不同：您只需要遍历数据集并获取您感兴趣的信息。检索所有名称应该看起来像这样：

List<String> allNames = new ArrayList<String>();

JSONArray cast = jsonResponse.getJSONArray("abridged_cast");
for (int i=0; i<cast.length(); i++) {
    JSONObject actor = cast.getJSONObject(i);
    String name = actor.getString("name");
    allNames.add(name);
}

（直接在浏览器中输入，因此未经测试）。

于 2013-06-16T19:32:34.670 回答
9

getJSONArray(attrname) 将从给定属性名称的对象中为您获取一个数组，在您的情况下发生的情况是

{"abridged_cast":["name": blah...]}
^ its trying to search for a value "characters"

但是您需要进入数组，然后搜索“字符”

尝试这个

String json="{'abridged_cast':[{'name':'JeffBridges','id':'162655890','characters':['JackPrescott']},{'name':'CharlesGrodin','id':'162662571','characters':['FredWilson']},{'name':'JessicaLange','id':'162653068','characters':['Dwan']},{'name':'JohnRandolph','id':'162691889','characters':['Capt.Ross']},{'name':'ReneAuberjonois','id':'162718328','characters':['Bagley']}]}";

    JSONObject jsonResponse;
    try {
        ArrayList<String> temp = new ArrayList<String>();
        jsonResponse = new JSONObject(json);
        JSONArray movies = jsonResponse.getJSONArray("abridged_cast");
        for(int i=0;i<movies.length();i++){
            JSONObject movie = movies.getJSONObject(i);
            JSONArray characters = movie.getJSONArray("characters");
            for(int j=0;j<characters.length();j++){
                temp.add(characters.getString(j));
            }
        }
        Toast.makeText(this, "Json: "+temp, Toast.LENGTH_LONG).show();
    } catch (JSONException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

检查了它 ：）

于 2013-06-16T19:52:54.350 回答
1

这是一个更好的方法。希望这可以帮助

protected void onPostExecute(String result) {
                Log.v(TAG + " result);


                if (!result.equals("")) {

                    // Set up variables for API Call
                    ArrayList<String> list = new ArrayList<String>();

                    try {
                        JSONArray jsonArray = new JSONArray(result);

                        for (int i = 0; i < jsonArray.length(); i++) {

                            list.add(jsonArray.get(i).toString());

                        }//end for
                    } catch (JSONException e) {
                        Log.e(TAG, "onPostExecute > Try > JSONException => " + e);
                        e.printStackTrace();
                    }


                    adapter = new ArrayAdapter<String>(ListViewData.this, android.R.layout.simple_list_item_1, android.R.id.text1, list);
                    listView.setAdapter(adapter);
                    listView.setOnItemClickListener(new OnItemClickListener() {
                        @Override
                        public void onItemClick(AdapterView<?> parent, View view, int position, long id) {

                            // ListView Clicked item index
                            int itemPosition = position;

                            // ListView Clicked item value
                            String itemValue = (String) listView.getItemAtPosition(position);

                            // Show Alert
                            Toast.makeText( ListViewData.this, "Position :" + itemPosition + "  ListItem : " + itemValue, Toast.LENGTH_LONG).show();
                        }
                    });

                    adapter.notifyDataSetChanged();
...
于 2014-12-03T07:25:06.720 回答