如何从android中的json对象呈现和选择联系人？

Posted 2023-04-15

【中文标题】如何从android中的json对象呈现和选择联系人？

【英文标题】：How to render and select contacts from json object in android?

【发布时间】：2015-10-10 00:25:26

【问题描述】：

我从服务器收到响应为 json，其中包含名称和联系号码键及其值。我想将姓名和联系电话详细信息与复选框一起显示为一列，以便用户可以从这些联系人中选择多个，并且可以通过单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\"]"

我现在将联系人和姓名放在单独的列表中。我应该怎么做才能把复选框放在那里？并选择多个联系人并将其转发到服务器？

 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);
        

【参考方案1】：

请为您的数据创建一个 Java pojo 类。例如： 类人 字符串标识； 字符串名称； boolean is selected;// 知道这个联系人是否被选中

将所有 Person 对象添加到 Arraylist

如果复选框被选中，则将 isSelected 设置为 true，反之亦然，对于位于该位置的 Person 对象。

然后您可以将您的联系人发送到服务器。

我希望这会有所帮助。

【参考方案2】：

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：可能很多人都知道 SparseArray 对象，但其他人可能不知道。优点是具有 ArrayList 和 HashMap 行为，并且比 HashMap 更有效（除非您有数百个项目）。在这种情况下是一个不错的选择，因为您只有 13 个项目。 希望对您有所帮助！