Question

我试图在不使用URL的情况下在android中读取简单的JSON数据。

以下是我的以下JSON数据。任何人都可以帮我阅读以下数据吗？

{"_id":"550ac135a2c513115c6cfa32","UserID":195,"LoginName":"Asdfa","EmailAddress":"adifa1@gmail.com","Password":"Head","Gender":1,"DateOfBirthString":"5-21-1988","DateOfBirth":"1988-05-21T00:00:00","DOB":"21/05/1988","CreatedDate":"2015-03-19T12:29:40.313Z","Status":"P","UserAge":26,"PreffredMaxAge":99,"PreffredMinAge":21,"ProfilePhoto":null,"IsUserOnline":false,"IsUserHomo":false,"ZipCode":"","IsUserSubscribed":false,"ProfileCompletionPoints":-1.0,"LastOnlineTime":null}

Answer 1

请为所有属性创建一个具有setter gatter的用户的模型类（ID名称，照片等）。我只为id创建。

public class AppUser{
private string id,

public void getId(){
return id;
}

public void setId(String id){
this.id = id;
}

并为AppUser类型创建一个数组列表。

ArrayList<AppUser> userlist = new ArrayList<AppUser>();

现在解析数据：

String serverData = "your server response in json";

JSONObject jsonObj = new JSONObject(serverdata);

                        String id= jsonObj .getString("_id");
                        String userID = phone.getString("UserID");
                        ................//parse other attribute 
AppUser user = new AppUser();
user.setId(id);
................// for other attribute 

//save into arraylist

userlist.add(user);

有关详细信息，请参阅以下教程：

http://mrbool.com/how-to-use-json-to-parse-data-into-android-application/28944

http://androidexample.com/JSON_Parsing_-_Android_Example/index.php?view=article_discription&aid=71&aaid=95

http://www.vogella.com/tutorials/AndroidJSON/article.html

http://www.androidhive.info/2012/01/android-json-parsing-tutorial/

Answer 2

看看这个......简单

网址myURL =新网址（“您的网址”）;

URLConnection tc = myURL.openConnection（）;

使用带有obj“tc”

的InputStreamReader在//读取行中的BufferedReader

 String line;
while ((line = in.readLine()) != null) {
JSONArray ja = new JSONArray(line);

    for (int i = 0; i < ja.length(); i++) {
    JSONObject jo = (JSONObject) ja.get(i);
    listItems.add(jo.getString("text")); // add in list view
    }

Answer 3

创建一个新的JSONObject并将json字符串作为参数传递。

    try {
        json_object = new JSONObject(json_str);
    } catch (JSONException) {
        //thrown when it cannot parse the json
      ;
    }

在android中读取简单的json数据

3 个答案: