Question

我正在尝试从Android应用程序执行一个非常简单的POST到将更新数据库的php脚本。不幸的是，这给了我第52行的调试器错误（eclipse）。下面是代码：

package com.example.testhttppost;

import java.io.IOException;
import java.util.ArrayList;
import java.util.List;

import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.message.BasicNameValuePair;

import android.app.Activity;
import android.os.Bundle;
import android.view.Menu;
import android.view.View;

public class MainActivity extends Activity {

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


    @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;
    }

    public void updateDiscountTable(View view)
    {
        // Create a new HttpClient and Post Header
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://www.test.com/jsonpost.php");

        try {
            // Add your data
            List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
            nameValuePairs.add(new BasicNameValuePair("id", "12345"));
            nameValuePairs.add(new BasicNameValuePair("shop", "ZARA"));
            nameValuePairs.add(new BasicNameValuePair("discount", "20%"));
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

            // Execute HTTP Post Request
            HttpResponse response = httpclient.execute(httppost);    //Line 52

        } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
        } catch (IOException e) {
            // TODO Auto-generated catch block
        }
    }
}

，PHP脚本是：

<?php
    // PHP variable to store the host address
     $db_host  = "localhost";
     // PHP variable to store the username
     $db_uid  = "dsdsdsv_android";
     // PHP variable to store the password
     $db_pass = "test1234";
     // PHP variable to store the Database name  
     $db_name  = "dsdsdsv_android"; 
            // PHP variable to store the result of the PHP function 'mysql_connect()' which establishes the PHP & MySQL connection  
     $db_con = mysql_connect($db_host,$db_uid,$db_pass) or die('could not connect');
     mysql_select_db($db_name);

     $shopId = $_POST['id'];
     $shopName = $_POST['shop'];
     $discount = $_POST['discount'];

     mysql_query("insert into discounts(id, shop, discount) values ($shopId, $shopName, $discount)");
//  mysql_query("insert into discounts(id, shop, discount)values(121, 'sadsdas','dasdasdsa')");

?>

应用程序的界面只是一个按钮。此按钮与上面定义的updateDiscountTable方法链接。谢谢，期待回复。

Answer 1

您可能正在主应用程序线程上执行网络操作。您需要将其移动到后台线程。您可以使用Asynctask。

您需要Keep your app responsive。

发布导致错误的数据

1 个答案: