php - 记录未从android插入到mysql数据库表中

我正在尝试将数据从android应用程序插入到mysql数据库表中。
但不知何故没有插入​​数据。。我认为android和mysql数据库未连接。我没有从android方面得到任何错误。但是在PHP方面...如果我运行url ..我会收到通知。未定义的变量。但是这些变量在android中。

这是我的android边码：

   package com.example.insertintophp;

import java.util.ArrayList;

import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
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.os.Bundle;
import android.app.Activity;
import android.util.Log;
import android.view.Menu;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.EditText;

public class MainActivity extends Activity {

    EditText name,city,email,contact,msg;
    Button insert;

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

        name =(EditText)findViewById(R.id.txtname);
        city =(EditText)findViewById(R.id.txtcity);
        email =(EditText)findViewById(R.id.txteid);
        contact =(EditText)findViewById(R.id.txtno);
        msg =(EditText)findViewById(R.id.txtmsg);
        insert=(Button)findViewById(R.id.btninsert);

        insert.setOnClickListener(new OnClickListener() {

            @Override
            public void onClick(View arg0)
            {
                // TODO Auto-generated method stub
                String nm = name.getText().toString();
                String ct = city.getText().toString();
                String emailid = email.getText().toString();
                String no = contact.getText().toString();
                String str = msg.getText().toString();
                insertRecords(nm, ct, emailid, no, str);

            }
            private void insertRecords(String nm,String ct,String emailid,String no,String str)
            {
                 ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(5);
                    nameValuePairs.add(new BasicNameValuePair("name", nm));
                    nameValuePairs.add(new BasicNameValuePair("city",ct));
                    nameValuePairs.add(new BasicNameValuePair("email",emailid));
                    nameValuePairs.add(new BasicNameValuePair("contact",no));
                    nameValuePairs.add(new BasicNameValuePair("msg",str));
                    sendData(nameValuePairs);
            }
            private void sendData(ArrayList<NameValuePair> data)
            {
                try
                {
                    HttpClient httpclient = new DefaultHttpClient();
                    HttpPost httppost = new HttpPost("http://10.0.2.2:8080/demo/insertData.php");
                    httppost.setEntity(new UrlEncodedFormEntity(data));
                    HttpResponse response = httpclient.execute(httppost);

                }
                catch (Exception e) {
                    // TODO: handle exception
                    Log.e("log_tag", "Error:  "+e.toString());
                }
            }
        });

    }


}

这是我的php辅助代码：

  <?php

$icon = mysql_connect("localhost","root","");
if(!$icon)
{
die('Could not connect : ' . mysql_error());
}
mysql_select_db("db_best_voyage", $icon)or die("database selection error");

echo json_encode($data);
$name=$_POST['name'];
$city=$_POST['city'];
$email = $_POST['email'];
$contact = $_POST['contact'];
$msg = $_POST['msg'];

mysql_query("INSERT INTO ctable (name,city,email,contact,msg) VALUES ('".$name."', '".$city."', '".$email."', '".$contact."', '".$msg."')");
mysql_close($icon);

?>

最佳答案

这是因为

1)您尚未声明/初始化任何$ data变量，但您正在将其编码为json并回显。

2)此外，在访问post参数并将它们插入数据库后，您将回显jsonencoded数据，如下所示。

    $name=$_POST['name'];
    $city=$_POST['city'];
    $email = $_POST['email'];
    $contact = $_POST['contact'];
    $msg = $_POST['msg'];

    mysql_query("INSERT INTO ctable (name,city,email,contact,msg) VALUES ('".$name."', '".$city."', '".$email."', '".$contact."', '".$msg."')");

echo json_encode($data);

确保为$ data提供一些数据，以确保您成功插入了记录。

编辑：

首先从浏览器进行测试：

<?php

$output = array();

$icon = mysql_connect("localhost","root","");
if(!$icon)
{
die('Could not connect : ' . mysql_error());
}
mysql_select_db("db_best_voyage", $icon)or die("database selection error");

echo json_encode($data);
$name= "name1";
$city= "city1";
$email = "email1";
$contact = "contact1";
$msg = "msg1";

$insert1 = mysql_query("INSERT INTO ctable (name,city,email,contact,msg) VALUES ('".$name."', '".$city."', '".$email."', '".$contact."', '".$msg."')");
mysql_close($icon);

if($insert1)
{
        $output["success"] = 1;
    $output["message"] = "Successfully inserted";

    // echoing JSON response
    echo json_encode($output);
}
else
{
     $output["success"] = 0;
    $output["message"] = "insertion failed......";

    // echoing JSON response
    echo json_encode($output);
}


?>