从安卓应用程序插入到mysql


Inserting into mysql from android app

我是安卓开发的新手。作为一个更大项目的一部分,我想将数据从安卓设备插入网络服务器。所以我做了一些研究和文章,比如来自androidhive的文章和来自codeproject的这篇文章,对于尝试开发一个插入到mysql数据库的测试应用程序非常有帮助,该数据库驻留在远程Web服务器上。

这是我的安卓代码

ConnectivityManager connMgr = (ConnectivityManager)getSystemService(Context.CONNECTIVITY_SERVICE);
    NetworkInfo networkInfo = connMgr.getActiveNetworkInfo();
    if (networkInfo != null && networkInfo.isConnected()){
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://xxxxxxxx.in/installment.php");
        try{
            List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
            nameValuePairs.add(new BasicNameValuePair("name", editTextCustomer.getText().toString()));
            nameValuePairs.add(new BasicNameValuePair("amount", editTextAmount.getText().toString()));
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpclient.execute(httppost);
            HttpEntity entity = response.getEntity();
            InputStream is = entity.getContent();
            Log.i("postData", response.getStatusLine().toString());
        }catch(Exception e){
            Log.e("log_tag", "Error in http connection"+e.toString());
         }
     }
    else { 
        Toast.makeText(PayBillActivity.this, "Internet Access, Denied!!", Toast.LENGTH_LONG).show();
    } 

这是 php 代码

<?php
/*
 * Following code will create a new product row
 * All product details are read from HTTP Post Request
 */
// array for JSON response
$response = array();
// check for required fields
if (isset($_POST['name']) && isset($_POST['amount'])) {
   $name = $_POST['name'];
   $amount = $_POST['amount'];
   $con=mysqli_connect("localhost","db_user","passwd","db_name");
    // Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
$sql="INSERT INTO installment (name, amount) VALUES ('$_POST[name]','$_POST[amount]')";
if (!mysqli_query($con,$sql))
  {
  die('Error: ' . mysqli_error($con));
  }
echo "1 record added";
    // check if row inserted or not
    if ($sql) {
        // successfully inserted into database
        $response["success"] = 1;
        $response["message"] = "Installment made successfully";
        // echoing JSON response
        echo json_encode($response);
    } else {
        // failed to insert row
        echo $result;
        $response["success"] = 0;
        $response["message"] = "Oops! An error occurred.";
        // echoing JSON response
        echo json_encode($response);
    }
} else {
    // required field is missing
    //$amount = 1000;
    //echo $amount;
    $response["success"] = 0;
    $response["message"] = "Required field(s) is missing";
    // echoing JSON response
    echo json_encode($response);
}
?>

当我运行该应用程序时,我收到此"NetworkOnMainThreadException"异常,因此没有添加任何行。但它与HTML POST完美配合。

谁能告诉我代码中的问题在哪里?提前感谢!:)

我想

如果你花时间把这个问题发布到谷歌上,你可能会得到一些很好的答案...... 只是为了完成这个问题

有两个选项可供您使用,您可以添加一行代码并允许在主线程上进行网络操作,但它对您的应用程序以及编码风格非常非常不利。

StrictMode.ThreadPolicy

policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();

StrictMode.setThreadPolicy(policy);

更长的选项是重新设计代码,以便在单独的线程中执行网络操作。 这对应用程序都有好处,您将学习如何在多线程程序上工作。

我认为你不应该使用那么严格或手动将其从main中删除。

只需使用一个智能的预制库,它就会为你制作一切!

下载 : http://loopj.com/android-async-http/

注意:这个库甚至使用 gzip 来压缩请求:)