如何从android中的json对象渲染和选择联系人?

时间:2015-07-20 18:42:03

标签: android android-listview contactpicker

我从服务器得到一个响应,因为json的名字和联系号码键都带有值。我想显示姓名和联系电话号码的详细信息以及一个复选框作为列,以便用户可以从这些联系人中选择多个,并且可以通过点击按钮将联系人发送到服务器。

我的json文件

{
"users": "[{\"id\":1,\"name\":\"test_name\",\"contact\":\"23456543\",\"gender\":\"F\",\"age\":234,\"city\":\"delhi\",\"state\":\"india\",\"created_at\":\"2015-07-19T17:58:42.000Z\",\"updated_at\":\"2015-07-19T17:58:42.000Z\",\"district\":\"test_district\"},{\"id\":2,\"name\":\"test_name\",\"contact\":\"23456543\",\"gender\":\"F\",\"age\":234,\"city\":\"delhi\",\"state\":\"india\",\"created_at\":\"2015-07-19T17:58:42.000Z\",\"updated_at\":\"2015-07-19T17:58:42.000Z\",\"district\":\"test_district\"},{\"id\":3,\"name\":\"qwrrtt\",\"contact\":\"1234567890\",\"gender\":\"F\",\"age\":12,\"city\":\"Bokaro\",\"state\":\"Jharkhand\",\"created_at\":\"2015-07-19T18:01:16.000Z\",\"updated_at\":\"2015-07-19T18:01:16.000Z\",\"district\":\"Bokaro\"},{\"id\":4,\"name\":\"wetur\",\"contact\":\"1234567890\",\"gender\":\"F\",\"age\":22,\"city\":\"Bokaro\",\"state\":\"Jharkhand\",\"created_at\":\"2015-07-19T18:41:17.000Z\",\"updated_at\":\"2015-07-19T18:41:17.000Z\",\"district\":\"Bokaro\"},{\"id\":5,\"name\":\"tfjko\",\"contact\":\"1234567990\",\"gender\":\"F\",\"age\":34,\"city\":\"Bokaro\",\"state\":\"Jharkhand\",\"created_at\":\"2015-07-19T19:30:09.000Z\",\"updated_at\":\"2015-07-19T19:30:09.000Z\",\"district\":\"Bokaro\"},{\"id\":6,\"name\":\"tfjko\",\"contact\":\"1234567990\",\"gender\":\"F\",\"age\":34,\"city\":\"Bokaro\",\"state\":\"Jharkhand\",\"created_at\":\"2015-07-19T19:30:22.000Z\",\"updated_at\":\"2015-07-19T19:30:22.000Z\",\"district\":\"Bokaro\"},{\"id\":7,\"name\":\"fghjk\",\"contact\":\"4567890123\",\"gender\":\"F\",\"age\":45,\"city\":\"Bokaro\",\"state\":\"Jharkhand\",\"created_at\":\"2015-07-19T19:35:14.000Z\",\"updated_at\":\"2015-07-19T19:35:14.000Z\",\"district\":\"Bokaro\"},{\"id\":8,\"name\":\"cvbnm\",\"contact\":\"7894561203\",\"gender\":\"F\",\"age\":23,\"city\":\"Bokaro\",\"state\":\"Jharkhand\",\"created_at\":\"2015-07-19T19:37:42.000Z\",\"updated_at\":\"2015-07-19T19:37:42.000Z\",\"district\":\"Bokaro\"},{\"id\":9,\"name\":\"tfjko\",\"contact\":\"1234567990\",\"gender\":\"F\",\"age\":34,\"city\":\"Bokaro\",\"state\":\"Jharkhand\",\"created_at\":\"2015-07-19T19:53:14.000Z\",\"updated_at\":\"2015-07-19T19:53:14.000Z\",\"district\":\"Bokaro\"},{\"id\":10,\"name\":\"edgujn\",\"contact\":\"4894521360\",\"gender\":\"F\",\"age\":45,\"city\":\"Bokaro\",\"state\":\"Jharkhand\",\"created_at\":\"2015-07-20T02:45:01.000Z\",\"updated_at\":\"2015-07-20T02:45:01.000Z\",\"district\":\"Bokaro\"},{\"id\":11,\"name\":\"qwert\",\"contact\":\"4568217390\",\"gender\":\"F\",\"age\":45,\"city\":\"Bokaro\",\"state\":\"Jharkhand\",\"created_at\":\"2015-07-20T06:12:57.000Z\",\"updated_at\":\"2015-07-20T06:12:57.000Z\",\"district\":\"Bokaro\"},{\"id\":12,\"name\":\"surbhi\",\"contact\":\"1334567890\",\"gender\":\"F\",\"age\":12,\"city\":\"Bokaro\",\"state\":\"Jharkhand\",\"created_at\":\"2015-07-20T07:17:53.000Z\",\"updated_at\":\"2015-07-20T07:17:53.000Z\",\"district\":\"Bokaro\"},{\"id\":13,\"name\":\"preefu\",\"contact\":\"5641287092\",\"gender\":\"F\",\"age\":56,\"city\":\"Bokaro\",\"state\":\"Jharkhand\",\"created_at\":\"2015-07-20T07:23:54.000Z\",\"updated_at\":\"2015-07-20T07:23:54.000Z\",\"district\":\"Bokaro\"}]"

}

我现在正在将联系人和姓名放在单独的列表中。我该怎么做才能把复选框放在那里?选择并转发多个联系人到服务器?

 JSONObject ob = new JSONObject(strres);
        List<String> allNames = new ArrayList<String>();

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

2 个答案:

答案 0 :(得分:1)

  1. 请为您的数据创建一个Java pojo类。例如:  class Person {  String Id;  字符串名称;  选择布尔值; //知道是否选择了此联系人

    }

  2. 将所有Person对象添加到Arraylist

  3. 如果选中复选框,则为该位置的Person对象设置isSelected为true,反之亦然。
  4. 然后你可以将你的联系人发送到服务器。

    我希望这会有所帮助。

答案 1 :(得分:0)

就像Priya Singhal所说的那样:

public class Actor { //simplified

    private int id;
    private String name;

    public Actor(){}

    public Actor(int id, String name) {
        this.id = id;
        this.name = name;
    }
}
JSONObject json = new JSONObject();
SparseArray<Actor> actors = new SparseArray<Actor>();

JSONArray actorsJSON = null;
try {
    actorsJSON = json.getJSONArray("users");
    for (int i=0; i<actorsJSON.length(); i++){
        JSONObject actorJSON = actorsJSON.getJSONObject(i);
        Actor actor = new Actor(actorJSON.getInt("id"), actorJSON.getString("name"));
        actors.put(id, actor);
    }
 } catch (JSONException e) {
     e.printStackTrace();
 }

Offtopic:可能你们很多人都知道SparseArray对象但可能是其他人没有。优点是具有ArrayList和HashMap行为,并且比HashMap更有效(除非你有数百个项目)。在这种情况下是一个很好的选择,因为你只有13个项目。 希望它有所帮助!