Android未连接到数据库以检查用户名和密码

时间:2018-09-28 19:23:51

标签: php android database android-studio xampp

所以这是我在一个教程视频中看到的代码,在该视频中,该应用程序运行正常,甚至显示了成功登录Toast。但是,当我尝试复制它时,即使我在数据库中输入了正确的用户名和密码,它也对我不起作用。 (仅供参考,我尝试使用用户的LAN ip以及WIFI IP在XAMPP上进行尝试,并且还尝试在000webhost.com上托管php脚本,但仍然无法在任何一种方法中工作)

我的ANDROID代码:

package com.example.devarsh.googlemapsbustracker;


import android.app.Dialog;
import android.content.Intent;
import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;
import android.util.Log;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.Toast;

import com.android.volley.AuthFailureError;
import com.android.volley.Request;
import com.android.volley.RequestQueue;
import com.android.volley.Response;
import com.android.volley.VolleyError;
import com.android.volley.toolbox.StringRequest;
import com.android.volley.toolbox.Volley;
import com.google.android.gms.common.ConnectionResult;
import com.google.android.gms.common.GoogleApiAvailability;

import java.util.HashMap;
import java.util.Map;

public class MainActivity extends AppCompatActivity {

private static final String TAG = "MainActivity";
private static final int ERROR_DIALOG_REQUEST = 9001;
private EditText User_name, Pass_word;
private Button btnlogin;
private static String URL_LOGIN = 
"http://192.168.50.127/Android/login.php";


//vars
EditText enroll;


@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    btnlogin = findViewById(R.id.button2);
    User_name = findViewById(R.id.username);
    Pass_word = findViewById(R.id.password);

    btnlogin.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            Login();
        }
    });


    isServicesOK();
}

private void Login(){

    String URL = 
"imagebustracker.000webhostapp.com/Android/parul/login.php";
    RequestQueue requestQueue = Volley.newRequestQueue(this);
    StringRequest stringRequest = new StringRequest(Request.Method.POST, URL, new Response.Listener<String>() {
        @Override
        public void onResponse(String response) {

            if(response.trim().equals("success")){
                Toast.makeText(getApplicationContext(), "Login Successfull", Toast.LENGTH_LONG).show();
                Intent intent = new Intent(getApplicationContext(), Selection_Activity.class);
                startActivity(intent);
            }else{
                Toast.makeText(getApplicationContext(), "Login Unsuccessfull", Toast.LENGTH_LONG).show();
            }

        }
    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {
            Toast.makeText(getApplicationContext(), "Login Unsuccessfull", Toast.LENGTH_LONG).show();
        }
    }) {
        @Override
        protected Map<String, String> getParams() throws AuthFailureError {

            Map<String, String> params = new HashMap<>();
            params.put("UserName", User_name.getText().toString().trim());
            params.put("PassWord", Pass_word.getText().toString().trim());

            return params;
        }
    };

    requestQueue.add(stringRequest);

}


public void isServicesOK() {
    Log.d(TAG, "isServicesOK: checking google services version");

    int available = GoogleApiAvailability.getInstance().isGooglePlayServicesAvailable(this);

    if(available == ConnectionResult.SUCCESS){
        //everything is fine and user can make map request
        Log.d(TAG, "isServicesOK: Google Play Services is working");
    }
    else if(GoogleApiAvailability.getInstance().isUserResolvableError(available)){
        //there is an error but we can fix it
        Log.d(TAG, "isServicesOK: An error is there but we can fix it");
        Dialog dialog = GoogleApiAvailability.getInstance().getErrorDialog(this, available, ERROR_DIALOG_REQUEST);
        dialog.show();
    }else{
        Toast.makeText(this, "We cant make maps request", Toast.LENGTH_SHORT).show();
    }
}

}

在Android代码中,我刚刚添加了一个内容,如果response.trim()。equals(“ success”)然后它将启动一个新活动,但它会继续举杯登录失败。

感谢所有为找到我的问题的答案而付出的努力。

非常感谢。

1 个答案:

答案 0 :(得分:0)

首先,测试您的PHP脚本。因此,像这样

硬编码用户名和密码变量
$username = "myUsername"; 
$password = "myPassword";

确保myUsername和myPassword是数据库中实际存在的数据。从那里,在您的Web浏览器中调用相同的URL,然后查看响应。是成功还是错误。

另一件事值得注意,当您拨打电话时,请在Android应用中阅读响应代码。响应代码将明确确定它是否正在与服务通信。但就您而言,我建议您打印出Volleyerror。

另外,向我提供您用于本教程的链接。有时对代码有双眼会很好。

正如您提到的,您会遇到“ Login Unsuccessful”吐司,这意味着它正在触发onErrorResponse函数。因此,将函数更改为以下内容:-

new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {
            Toast.makeText(getApplicationContext(), "Login Unsuccessfull", Toast.LENGTH_LONG).show();
            Log.e(TAG, error); //or
            System.println("VolleyError Start: " + error + " : End VolleyError");
        } 

请分享错误。如果您有任何问题,请告诉我。