在android的textview中显示json数据的结果 [英] display result of json data in textview of android

问题描述

在android应用中显示数据时遇到一个问题.这是我的php.我现在进行了编辑以缩短代码.

I am having one problem while displaying data in android app.This is my php.I have edited now to shorten the code.

if($imei_no!="" && $app_auid!="")
{
        //chk query
        $chk_query=$db1->query("select * from user where imei_number='$imei_no'");
        $chk_cnt=$chk_query->rowCount();
        if($chk_cnt>0)
        {
            $search_qry=$db1->query("select * from candidate_reception cr,candidate_counseling cc ,candidate_admission ca where cr.candidate_id=cc.candidate_id and cr.candidate_id=ca.candidate_id and cr.auid=ca.auid and cr.counseling_id=cc.counseling_id and cr.auid='$app_auid'");
            $ser_count=$search_qry->rowCount();
            if($ser_count>0)
            {
                $stud_row=$search_qry->fetch(PDO::FETCH_OBJ);

                $candidate_id=$stud_row->candidate_id;
                $auid = $stud_row->auid;
                $usn_no = $stud_row->usn_no;
                $sx = $stud_row->candidate_sex


                //image path    
                  if($inst_id==1){ $img_path = "1"; }
                  if($inst_id==2){ $img_path = "2"; }

                $result=array();    
                array_push($result, array('name' => $candidate_name,'img_path'=>$img_path));

            echo json_encode(array("can_data"=>$result));
            echo "Success"          
            }
            else
            {
                echo "Auid not Found.";
            }       
        }
        else
        {
            echo "Invalid IMEI Number.!!!";
        }
}
?>

这是我的Java代码:

and this is my java code:

  private static final String LOGIN_URL = "http://xxx.xxx.x.x/mobile_app/data_push.php";
    private EditText editTextUserName;
    private Button buttonLogin;

    TelephonyManager tel;
    TextView imei;
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_imei__val);

        editTextUserName = (EditText) findViewById(R.id.editTextAUID);
        buttonLogin = (Button) findViewById(R.id.btn);
        buttonLogin.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View view) {
                login();
            }
        });

        tel = (TelephonyManager) getSystemService(Context.TELEPHONY_SERVICE);
        imei = (TextView) findViewById(R.id.textView2);
        imei.setText(tel.getDeviceId().toString());

    }

    private void login(){
        String username = editTextUserName.getText().toString().trim();
        String imei_no = imei.getText().toString().trim();
        userLogin(username,imei_no);
    }

    private void userLogin(final String username,final String imei_no){
        class UserLoginClass extends AsyncTask<String,Void,String> {
            ProgressDialog loading;
            @Override
            protected void onPreExecute() {
                super.onPreExecute();
                loading = ProgressDialog.show(IMEI_Val.this,"Please Wait",null,true,true);
            }

            @Override
            protected void onPostExecute(String s) {
                super.onPostExecute(s);
                loading.dismiss();
                if(s=="Success"){

                    Intent intent = new Intent(IMEI_Val.this, CapturePhoto.class);
                   // intent.putExtra(USER_NAME,username);
                    startActivity(intent);
                }

               else{
                   Toast.makeText(IMEI_Val.this, s, Toast.LENGTH_LONG).show();
                    Log.d("Tag Name", "Log Message");
                }
            }

            @Override
            protected String doInBackground(String... params) {
                HashMap<String,String> data = new HashMap<>();
                data.put("auid",params[0]);
                data.put("imei",params[1]);
                RegisterUserClass ruc = new RegisterUserClass();
                String result = ruc.sendPostRequest(LOGIN_URL,data);
                return result;
            }
        }
        UserLoginClass ulc = new UserLoginClass();
        ulc.execute(username,imei_no);
    }

}

我的问题是当单击登录按钮时，在有效的用户输入后获取带有学生详细信息的json结果.如果用户输入有效，我想转到下一个活动并显示其他详细信息，例如姓名，大学等.我怎样才能做到这一点? 输入有效的输入后，这里将得到json结果作为响应.即使输入正确的regno，也无法进入下一个活动并在文本框中显示其他详细信息.

My problem is when am clicking on Login button, am getting json result in toast with student details after valid user inputs.I want to goto next activity if the the user inputs are valid and display other details like name,college etc.How can I do that? Here am getting json result as response when I enter valid inputs.I could not go to next activity and display other details in text box even though am entering correct regno.

推荐答案

@ user6588225

@user6588225

所以您向Json发送成功消息

so your sending Json with Success message

所以不要使用==尝试使用contains

so rather than using == try to use contains

所以在您的onPostExecute(String s)

@Override
protected void onPostExecute(String s) {
     super.onPostExecute(s);
     loading.dismiss();
     if(s.contains("Success")){
         Intent intent = new Intent(IMEI_Val.this, CapturePhoto.class);
         // intent.putExtra(USER_NAME,username);
         startActivity(intent);
     }
     else{
       Toast.makeText(IMEI_Val.this, s, Toast.LENGTH_LONG).show();
       Log.d("Tag Name", "Log Message");
     }
}

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