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我正在尝试创建一个可以使用 JSON 将数据发送到 PHP 服务器的 android 应用程序。我遵循了本指南：http ://www.androidhive.info/2012/05/how-to-connect-android-with-php-mysql/这是我的 Android 代码：

import java.util.ArrayList;
import java.util.List;
import org.apache.http.NameValuePair;
import org.apache.http.message.BasicNameValuePair;
import org.json.JSONObject;
import android.app.Activity;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;

public class HelloWorldActivity extends Activity {

    JSONParser jsonParser = new JSONParser();
    EditText userName;

    // url to create new product
    private static String url = "http://192.168.1.35/workspace/sosapp/db_connect.php";    

    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

        // Edit Text
        userName = (EditText) findViewById(R.id.userName);

        // Send button
        Button sendButton = (Button) findViewById(R.id.sendButton);

        // button click event
        sendButton.setOnClickListener(new View.OnClickListener() {

            public void onClick(View view) {
                String name = userName.getText().toString();

                // Building Parameters
                List<NameValuePair> params = new ArrayList<NameValuePair>();
                params.add(new BasicNameValuePair("name", name));

                // getting JSON Object
                JSONObject json = jsonParser.makeHttpRequest(url, "POST", params);

            } 
        }); 

    }
}

至于我的 PHP 代码：

  function insert($var1) {

    // mysql inserting a new row
    $result = mysql_query("INSERT INTO report (name) VALUES ('$var1')");

    // check if row inserted or not
    if ($result) {
        // successfully inserted into database;
    $msg = "Message received.";
    } else {
    $msg = "Not received.";
    }
return ($msg);
}

if (isset($_POST['name'])) {
   $name = $_POST['name']);
   insert($name);   
} else {
   echo "No input.";
}

我一直没有输入。我运行 android 应用程序（在连接到计算机的手机上），然后刷新 PHP 站点，但一直没有输入。android 应用程序不发送数据。有人可以帮忙吗？非常感谢！

4

2 回答 2

0

据我所知，它是本教程的 JSONParser 类。它检查

method == "POST"

这是错误的。将此更改为

method.equals("POST")

看看它是否有效。

于 2013-02-17T19:18:04.277 回答
0

Android Http 仅允许后台运行

import java.util.ArrayList;
import java.util.List;
import org.apache.http.NameValuePair;
import org.apache.http.message.BasicNameValuePair;
import org.json.JSONObject;
import android.app.Activity;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;

public class HelloWorldActivity extends Activity {

    JSONParser jsonParser = new JSONParser();
    EditText userName;

    // url to create new product
    private static String url = "http://192.168.1.35/workspace/sosapp/db_connect.php";    

    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

        // Edit Text
        userName = (EditText) findViewById(R.id.userName);

        // Send button
        Button sendButton = (Button) findViewById(R.id.sendButton);

        // button click event
        sendButton.setOnClickListener(new View.OnClickListener() {

            public void onClick(View view) {
                String name = userName.getText().toString();

                // Building Parameters
                List<NameValuePair> params = new ArrayList<NameValuePair>();
                params.add(new BasicNameValuePair("name", name));

                // getting JSON Object

            Thread httpthread=new Thread(new Runnable() {
            @Override
            public void run() {

          JSONObject json = jsonParser.makeHttpRequest(url, "POST", params);


  runOnUiThread(new Runnable(){
            @Override
             public void run() {
                 // ui working  ex)textview.setText("test");


             }
        });

            }
        });

httpthread.start();
            } 
        }); 

    }
}
于 2016-07-05T15:08:59.617 回答