Question

Sections.class

private void loadSection() {
  final String username = SharedPrefManager.getInstance(getActivity()).getUsername();
final String userSection_URL = Constants.USER_SECTION + "?username=" + username;
  JsonObjectRequest jsonObjectRequest = new JsonObjectRequest(Request.Method.GET, userSection_URL, null, new Response.Listener<JSONObject>() {
        @Override
        public void onResponse(JSONObject response) {
            try {
                JSONArray jsonArray = response.getJSONArray("data");
                for (int i = 0; i < jsonArray.length(); i++) {
                    JSONObject sections = (JSONObject) jsonArray.get(i);
                    sectionList.add(new ListGradeData(
                            sections.getInt("id"),
                            sections.getString("section"),
                            sections.getString("year_level"),
                            sections.getString("school_year")
                    ));

                    //creating adapter object and setting it to recyclerview
                    LatestGradeAdapter adapter = new LatestGradeAdapter(getActivity(), sectionList);
                    recyclerView.setAdapter(adapter);
                }
            } catch (JSONException e) {
                e.printStackTrace();
                Toast.makeText(getContext(), "No Data", Toast.LENGTH_LONG).show();
            }
            // Stopping swipe refresh
            mSwipeRefreshLayout.setRefreshing(false);
        }
    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {
            // Stopping swipe refresh
            mSwipeRefreshLayout.setRefreshing(false);

            headerSection.setVisibility(View.GONE);
            noConnectionLayout.setVisibility(View.VISIBLE);
        }
    });
    //adding our stringrequest to queue
    Volley.newRequestQueue(getActivity().getApplicationContext()).add(jsonObjectRequest);
}

userSection.php

require_once "../include/Constants.php";

$ conn =新的mysqli（DB_HOST，DB_USER，DB_PASSWORD，DB_NAME）;

if（$ _ SERVER ['REQUEST_METHOD'] =='GET'）{ $ uname = trim（$ _ GET ['username']）;

if(isset($uname)){
    $query = "SELECT * FROM students WHERE s_id = '".$uname."' ";
    $tbl = mysqli_query($conn, $query);
    if(mysqli_num_rows($tbl)>0) {
        $row = mysqli_fetch_array($tbl, MYSQLI_ASSOC);
        $uid = $row['id'];
        $stmt = $conn->prepare("SELECT students.id, sections.name AS section_name, years.name AS year_level, CONCAT(academic_years.start, '-', academic_years.end) AS school_year FROM students INNER JOIN section_student ON students.id = section_student.student_id INNER JOIN sections ON section_student.section_id = sections.id INNER JOIN years ON sections.year_id = years.id INNER JOIN academic_years ON sections.academic_year_id = academic_years.id WHERE students.id = ?");
    $stmt->bind_param("s", $uname);
    $stmt->execute();
    $stmt->bind_result($id, $section, $year_level, $school_year);
    $sections = array(); 

    //traversing through all the result 
    while($stmt->fetch()){
      $temp = array();
      $temp['id'] = $id; 
      $temp['section'] = $section; 
      $temp['year_level'] = $year_level; 
      $temp['school_year'] = $school_year; 
      array_push($sections, $temp);
    }
        echo json_encode(array('data' => $sections));
    }else {
        echo json_encode(array('data' => "No Data"));
    }
}

} 问题解决了！

我有一个问题，涉及到检索数据。我想根据当前登录的用户名检索信息。但这是行不通的。我已经尝试显示所有数据，但是它不是基于当前登录的用户。它将显示所有数据。主要问题是如何根据当前登录的用户检索数据。我当前登录的用户基于此代码final String username = SharedPrefManager.getInstance(getActivity()).getUsername();

Answer 1

我建议使用JSONObjectRequest。它将允许您检索对象。

String url = YOUR_OWN_URL + "?username=" + username;
JsonObjectRequest jsonObjectRequest = new JsonObjectRequest(Request.Method.GET, url, null, new Response.Listener<JSONObject>() {
    @Override
    public void onResponse(JSONObject response) {
        try {
            JSONArray jsonArray = response.getJSONArray("data");
            for (int i = 0; i < jsonArray.length(); i++) {
                JSONObject jsonObject = (JSONObject) jsonArray.get(i);
                int id = jsonObject.getInt("id");
                String section = jsonObject.getString("section");
                String year_level = jsonObject.getString("year_level");
                String school_year = jsonObject.getString("school_year");
                // insert your own logic
            }
        } catch (JSONException e) {
        }
    }
}, new Response.ErrorListener() {
    @Override
    public void onErrorResponse(VolleyError error) {
    }
});

然后，在PHP中，我将json_encode部分更改为这样。

echo json_encode(array('data' => $sections));

Answer 2

StringRequest中没有用户名

在您的PHP代码中，您正在username处寻找一个$uname = trim($_GET['username']); url参数。但是您不是从Volley StringRequest发送它的。您的网址由Constants.USER_SECTION定义。如果这是请求的端点，则应使用类似以下内容的参数将其添加到url中：

String url = Constants.USER_SECTION + '?username=' + username; //considering that Constants.USER_SECTION doesn't have any url params
StringRequest stringRequest = new StringRequest(Request.Method.GET, url ,
            new Response.Listener<String>() {
               ...

调试代码

修复上述错误后，请按照以下步骤调试代码：

检查final String username值。如果不正确，请确认将其保存到SharedPreferences的位置。
在StringRequest之前检查url的值。如果错误，请检查Constants.USER_SECTION的定义位置。
从StringRequest中检查response。如果不正确，甚至使用浏览器的REST客户端调试服务器端代码。

如何根据当前在Android上登录的用户检索数据

2 个答案:

StringRequest中没有用户名

调试代码