Question

当我将值插入数据库时，代码工作正常。我使用

时代码有效

select * from tblLogin

但是当我从android代码中获取$ _POST值并尝试运行

时

SELECT * FROM tblLogin WHERE username='$username' AND   password='$password'

声明它不能正常工作。

PHP

<?php

if($_SERVER["REQUEST_METHOD"] == "POST")
{

    $username = $_POST['username'];
    $password = $_POST['password'];

    $db = mysql_connect("localhost", "root", "");
    mysql_select_db("ntu", $db);

    $sql = mysql_query("SELECT * FROM tblLogin WHERE username='$username' AND password='$password'");

    $a = array(); 
    $index = 0;

    while($row = mysql_fetch_assoc($sql)) 
    {
        $a[$index] = $row;
        $index++;
    }

    echo json_encode(array("students"=>$a));

    mysql_close($db);
}

?>

安卓代码

 package com.example.digi.studentapp;

import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.util.Log;
import android.view.Menu; 
import android.view.MenuItem;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
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.JsonObjectRequest;
import com.android.volley.toolbox.StringRequest;
 import com.android.volley.toolbox.Volley;

import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

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

public class MainActivity extends AppCompatActivity {

EditText username,password;
Button login,register;
RequestQueue requestQueue;
//String insert = "http://192.168.1.104/android/insert.php";
String checklogin = "http://192.168.1.104/android/login.php";
TextView result;

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

    username = (EditText) findViewById(R.id.editText);
    password = (EditText) findViewById(R.id.editText2);

    login = (Button) findViewById(R.id.button);
    register = (Button) findViewById(R.id.button2);
    result = (TextView) findViewById(R.id.textView);

    requestQueue = Volley.newRequestQueue(getApplicationContext());


    login.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {


            JsonObjectRequest jsonObjectRequest = new    JsonObjectRequest(Request.Method.POST,
                    checklogin, new Response.Listener<JSONObject>() {
                public void onResponse(JSONObject response) {
                    try {
                            JSONArray students = response.getJSONArray("students");
                            for (int i = 0; i < students.length(); i++) {
                                JSONObject tblLogin = students.getJSONObject(i);

                                String id = tblLogin.getString("id");
                                String username = tblLogin.getString("username");
                                String password = tblLogin.getString("password");

                                result.append(id+" "+username+" "+password+"\n");

                            }

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

                }
            }, new Response.ErrorListener() {

                public void onErrorResponse(VolleyError error) {

                }

            });
            requestQueue.add(jsonObjectRequest);
        }
    });


  }


}

Answer 1

我不是php的专家，但username='$username'似乎错了！试试username='.$username.'

对您要在请求中使用的变量执行相同的操作。

Answer 2

使用以下命令调试您的SQL查询：

// Check result
// This shows the actual query sent to MySQL, and the error. Useful for debugging.

$query = "SELECT * FROM tblLogin WHERE username='".$username."' AND password='".$password."'";

$sql = mysql_query($query);

if (!$sql) 
{
    $message  = 'Invalid query: ' . mysql_error() . "\n";
    $message .= 'Whole query: ' . $query;
    die($message);
}

请！停止使用mysql_query，它已弃用，您的方法会使您的应用程序易受攻击

<强> EDITED 您可以使用Postman

之类的内容在应用外部发出POST请求来测试结果

将json数据从android代码传递到php文件时，“从哪里选择”语句不起作用

2 个答案: