Question

问题解决了!!!

我在下面发布我的php文件和android类。问题是当我尝试检查从PHP返回的结果时，它给了我很多垃圾数据。

loginStatus.php

    <?php
$mysql = new mysqli("10.0.0.2", "root", "nishad0963", "cabService");

if($mysql->connect_errno){
    echo "Failed to connect to DB"; 
}   
//echo $mysql->host_info . "\n";

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

$query_select = $mysql->query("SELECT status FROM loginStatus WHERE userName = 'nishad' and password = 'password' ");

//echo $query_select;

while(($row = $query_select->fetch_assoc())!=null);
{
    printf("result = %s", $row["status"]);
}

$mysql->close();

?>

signInActivity.java

package com.example.helloworld;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.UnsupportedEncodingException;
import java.net.MalformedURLException;
import java.net.URL;
import java.net.URLConnection;
import java.net.URLEncoder;

import android.os.AsyncTask;
import android.util.Log;
import android.widget.Toast;

public class signInActivity extends AsyncTask<String, Void, String> {

    protected void onPreExecute() {

    }

    @Override
    protected String doInBackground(String... arg0) {


        //preparing data
        String data = null;
        String userName = (String) arg0[0];
        String password = (String) arg0[1];
        try {
            data = URLEncoder.encode("userName", "UTF-8")+ "=" + URLEncoder.encode(userName, "UTF-8");
        } catch (UnsupportedEncodingException e) {
            Log.e("DATA_ERROR", e.getMessage());
            e.printStackTrace();
        }
        //making connection
        String url_address = "http://10.0.0.2/cabService/loginStatus.php";
        try {
            URL url = new URL(url_address);
            URLConnection connect = url.openConnection();
            connect.setDoOutput(true);
            OutputStreamWriter wr = new OutputStreamWriter(connect.getOutputStream());
            wr.write(data);
            wr.flush();
            Log.d("Progress Status", "It's working");

            BufferedReader br = new BufferedReader(new InputStreamReader(connect.getInputStream()));
            //reading response
            StringBuilder sb = new StringBuilder();
            String Line;
            while((Line = br.readLine())!=null){
                sb.append(Line);
                Log.d("Result Status", Line);
            }
            return sb.toString();

        } catch (MalformedURLException e) {
            Log.e("URL_ERROR", e.getMessage());
            e.printStackTrace();
        } catch (IOException e) {
            Log.e("URL_CONNECTION_ERROR", e.getMessage());
            e.printStackTrace();
        } catch (Exception e) {
            Log.e("ERROR", e.getMessage());
        }




        return null;
    }

    protected void onPostExecute(String result) {

        if(result.equals(1)){
            Log.d("LOGIN_SUCCESS", result);
        }
        else
            Log.d("LOGIN_FAILURE", result);
    }


}

调试的logcat

   01-03 16:42:52.102: D/Status(626): Before Clicking
01-03 16:42:52.452: D/Progress Status(626): It's working
01-03 16:42:52.532: D/Result Status(626): result =

结果状态是返回结果的标记。

希望你能帮忙！任何建议将不胜感激。

我认为错误在于PHP文件。

编辑：

mainActivity.java

package com.example.helloworld;

import android.os.Bundle;
import android.app.Activity;
import android.util.Log;
import android.view.Menu;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;

public class MainActivity extends Activity {

    private EditText userName, passWord;
    private Button loginButton;

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


    userName = (EditText) findViewById(R.id.nameEdit);    
    passWord = (EditText) findViewById(R.id.passwordEdit);

    final String name = userName.getText().toString();
    final String password = passWord.getText().toString();

    loginButton = (Button) findViewById(R.id.loginButton);

    loginButton.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View arg0) {
            Log.d("Status","Before Clicking");
            new signInActivity().doInBackground(name, password);

        }
    });

    }


    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        // Inflate the menu; this adds items to the action bar if it is present.
        getMenuInflater().inflate(R.menu.main, menu);
        return true;
    }

}

查看asyncTask传递的参数。我很确定这不正确。

Answer 1

你的logcat输出说明了它; loginStatus.php的第9行：

Function name must be a string.

您正在使用函数调用语法访问超级全局$_POST，这是一个数组 $_POST('user')代替$_POST['user']。因为php允许基于字符串的函数调用，$myfuncname ="someFunc"; $myfuncname();和$_POST不是字符串而是数组，您会收到此错误消息。

Answer 2

使用以下链接中的脚本完成同样的操作，这应该有所帮助。

How to create Simple Login form using php in android? – Connect php with android.

从php获取结果到android

2 个答案: