php - 从 MySQL 获取数组到 android

Question

我已经编写了一个 php 代码来从 MYSQL 获取数据数组，但现在我正在努力将其返回到我的 android 中并只是为了庆祝它。

这是我的 php 代码:

if($_SERVER['REQUEST_METHOD']=='GET'){

    $user_id  = $_GET['user_id'];

    require_once('dbConnect.php');

    $user_tags = array();

           foreach ($_REQUEST['user_id'] as $key => $val) {
           $user_tags[$key] = filter_var($val, FILTER_SANITIZE_STRING);
            }
            $user_ids = "'" . implode("','", $user_tags) . "'";
            $sql = "SELECT * FROM user_tags WHERE user_id IN ({$user_ids})";

    $r = mysqli_query($con,$sql);

//creating a blank array 
$result = array();

//looping through all the records fetched

while($row = mysqli_fetch_array($r)){

    $result['tags'][] = $row['tags'];   
       }

//Displaying the array in json format 
echo json_encode(array('result'=>$result));

mysqli_close($con);
}

这里是代码的 URL:“http://allwaysready.16mb.com/Cuboid/tagsTest.php?user_id[]=7”

这是 php 代码的结果:{“结果”:{“标签”:[“Pascol”，“PHP”，“Python”]}}

我想在我的 Android 中得到这个“结果”。

这是我的 Android 代码:

 public class Jobs extends AppCompatActivity {

public static String tags;

private String JSON_STRING;
public static final String TAG_JSON_ARRAY="result";
public static final String KEY_USER_TAGS="tags";
public static final String URL_GET_TAG =   "http://allwaysready.16mb.com/Cuboid/tagsTest.php?user_id[]=";
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_jobs);


    getJSON();

    Toast.makeText(getApplicationContext(),tags,Toast.LENGTH_LONG).show();


}

private void showEmployee() {
    JSONObject jsonObject = null;
    ArrayList<HashMap<String, String>> list = new ArrayList<HashMap<String, String>>();
    try {
        jsonObject = new JSONObject(JSON_STRING);
        JSONArray result = jsonObject.getJSONArray(TAG_JSON_ARRAY);

        for (int i = 0; i < result.length(); i++) {
            JSONObject jo = result.getJSONObject(i);
            tags = jo.getString(KEY_USER_TAGS);

            HashMap<String, String> employees = new HashMap<>();
            employees.put(KEY_USER_TAGS, tags);

        }

    } catch (JSONException e) {
        e.printStackTrace();
    }
}

private void getJSON(){
    class GetJSON extends AsyncTask<Void,Void,String>{

        ProgressDialog loading;
        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            loading = ProgressDialog.show(Jobs.this,"Fetching Data","Wait...",false,false);
        }

        @Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            loading.dismiss();
            JSON_STRING = s;
            showEmployee();
        }

        @Override
        protected String doInBackground(Void... params) {
            RequestHandler rh = new RequestHandler();
            String s = rh.sendGetRequest(URL_GET_TAG);
            return s;
        }
    }
    GetJSON gj = new GetJSON();
    gj.execute();
}

但我只得到空白的 toast 。如果我犯了任何错误，请帮助我。

Answer 1

像这样修改你的 showEmployee:

private void showEmployee() {
    JSONObject jsonObject = null;
    ArrayList<HashMap<String, String>> list = new ArrayList<HashMap<String, String>>();
    try {
        jsonObject = new JSONObject(JSON_STRING);
        JSONObject result = jsonObject.getJSONObject(TAG_JSON_ARRAY);
        JSONArray jsonTags = result.getJSONArray(KEY_USER_TAGS);

        for (int i = 0; i < jsonTags.length(); i++) {
            tags = tags + jsonTags.getString(i);

            HashMap<String, String> employees = new HashMap<>();
            employees.put(KEY_USER_TAGS, jsonTags.getString(i));

        }

    } catch (JSONException e) {
        e.printStackTrace();
    }
}

并将标签字符串设置为空字符串(否则第一个对象将为空)

   public static String tags = "";

代码仍然存在一些逻辑缺陷，您的员工 HashMap 将只包含最后一个条目，因为您总是在创建一个新的 HashMap。