将JSONECODE（PHP）中的值存储到responseHandler中，并将其存储到变量中

我找了很多东西，但都没找到。这就是我的问题

我已成功使用responseHandler制作登录表单。代码如下：

  public void login()
  {
        httpClient=new DefaultHttpClient();
        httpPost= new HttpPost("http://studentstracking.hol.es/loginDatabase2.php");
        nameValuePairs = new ArrayList<NameValuePair>(2);
        nameValuePairs.add(new BasicNameValuePair("username",txtId.getText().toString().trim()));
        nameValuePairs.add(new BasicNameValuePair("password",txtPassword.getText().toString().trim()));

        httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        response=httpClient.execute(httpPost);

        ResponseHandler<String> responseHandler = new BasicResponseHandler();
        final String response = httpClient.execute(httpPost, responseHandler);
        System.out.println("Response : " + response);

        runOnUiThread(new Runnable() {
            public void run() {
                tv.setText("Response from PHP : " + response);
                dialog.dismiss();
            }
        });

        if(response.equalsIgnoreCase("User Found")){
            runOnUiThread(new Runnable() {
                public void run() {
                    Toast.makeText(LoginActivity.this,"Login Success", Toast.LENGTH_SHORT).show();
                }
            });

            startActivity(new Intent(LoginActivity.this, student_activity.class));
        }else{
            showAlert();
        }}

但这次我想执行PHP并从中获取值，然后将其存储到变量中，或者将其存储到SQLite中

这是PHP代码：

 $username = $_POST['username'];
 $password = $_POST['password'];
 $query_search = "select * from user2 where kode_user2 = '".$username."' AND password = '".$password."'";
 $query_exec = mysqli_query($conn, $query_search) or die(mysqli_error($conn));
 $rows = mysqli_num_rows($query_exec);
 if($rows == 0) { 
     echo "No Such User Found"; 
 }
 else  {
     echo "User Found"; 
 }

 $query_search1 = "SELECT `schedule`.`kode_schedule`,`schedule`.`kode_matkul`,`matakuliah`.`title`,`schedule`.`ruang`,`kelas`.`latitude`,`kelas`.`longitude`,`kelas`.`lantai`,`schedule`.`hari`,`schedule`.`jam` FROM schedule LEFT JOIN `u466318196_track`.`matakuliah` ON `schedule`.`kode_matkul` = `matakuliah`.`kode_matkul` LEFT JOIN `u466318196_track`.`kelas` ON `schedule`.`ruang` = `kelas`.`ruang` WHERE(( kode_schedule = 'sch0001'))";
 $query_exec1 = mysqli_query($conn, $query_search1) or die(mysqli_error($conn));


  while ($data2 = mysqli_fetch_assoc($query_exec1))
 {
     $response['length'][] = $data2;
 }

 print json_encode($response);

这个json的结果是：

{
   "length":
   [
      {
          "kode_schedule":"sch0001",
          "kode_matkul":"TIB01",
          "title":"Basis Data",
          "ruang":"501",
          "latitude":"-6.18861653446272",
          "longitude":"106.801122526546",
          "lantai":"5",
          "hari":"Selasa",
          "jam":"17:45:00"
      },          
      {
          "kode_schedule":"sch0001",
          "kode_matkul":"TIB02",
          "title":"Pemrograman Berorientasi Objek",
          "ruang":"LABB",
          "latitude":"-6.18864706134991",
          "longitude":"106.801161122636",
          "lantai":"5",
          "hari":"Selasa",
          "jam":"19:30:00"
       }
    ]
 }

所以我想将“kode_schedule”、“kode_matkul”等值存储到变量中，比如String，或者可能是带有数组的String

---

请帮助我。

您将在响应字符串中获得结果

   JSONObject mainObject=new JSONObject(response);

        for(int i: 0 and 1){
        mainObject.getJSONArray("length").getJSONObject(i).getString("kode_schedule");
        mainObject.getJSONArray("length").getJSONObject(i).getString("kode_matkul");
}

---

使用json方法进行相应的访问。他们有很多学习json解析的教程。

use。我将把json解析为java对象，我对github感到困惑。你能在聊天中做些什么吗？（编辑）你能直接从我的第一栏代码中给我完整的代码吗？我也是一个初学者，但这可能会对你有所帮助。我已经编辑了答案