2

我有一个像表单一样工作的应用程序，它需要四个字段并验证信息以确保没有输入无效字符。这四个字段存储在变量中：

  • 电话
  • 姓名
  • 电子邮件

  • 评论

    现在我想将表单数据（在这四个字段中输入并存储到变量中的任何内容）提交到一个 url（将使用http://www.test.com），但我不知道该怎么做. 我想我正在寻找一种叫做 HttpURLConnection 的东西，但我不确定如何指定发送哪个变量。下面的代码是我从网站http://developer.android.com/reference/java/net/HttpURLConnection.html找到的

    private class UploadFilesTask extends AsyncTask<URL, Integer, Long>{
protected Long doInBackground(URL... urls) {

    try {
        HttpClient http = new DefaultHttpClient();
        HttpPost post = new HttpPost("http://www.test.com");

        List<NameValuePair> data = new ArrayList<NameValuePair>();
        data.add(new BasicNameValuePair("phone", "value"));
        data.add(new BasicNameValuePair("name", "value"));
        data.add(new BasicNameValuePair("email", "value"));
        data.add(new BasicNameValuePair("comments", "value"));
        post.setEntity(new UrlEncodedFormEntity(data));

        HttpResponse response = http.execute(post);
        // do something with the response
    }
    catch (ClientProtocolException e) {
        // do something
        finish();
    }
    catch (IOException e) {
        // do something
        finish();
    }

    }

    }

任何帮助将不胜感激，谢谢！

4

4 回答 4

5

将表单数据发送到服务器的最简单方法是使用 HttpClient 和 HttpPost。

尝试这样的事情：

try {
    HttpClient http = new DefaultHttpClient();
    HttpPost   post = new HttpPost("http://www.example.com/process");

    List<NameValuePair> data = new ArrayList<NameValuePair>();
    data.add(new BasicNameValuePair("phone", "value");
    data.add(new BasicNameValuePair("name", "value");
    data.add(new BasicNameValuePair("email", "value");
    data.add(new BasicNameValuePair("comments", "value");
    post.setEntity(new UrlEncodedFormEntity(data));

    HttpResponse response = http.execute(post);
    // do something with the response
}
catch (ClientProtocolException e) {
    // do something
}
catch (IOException e) {
    // do something
}

请注意，您需要在 AsyncTask 中执行此操作，这样您就不会锁定等待网络操作完成的 UI 线程。

编辑

这是一个简单的快速示例，说明它在 AsyncTask 中的外观。

public class SendTask extends AsyncTask<Void, Void, Boolean> {

    String responseString;

    @Override
    protected Boolean doInBackground(Void... params) {
        try {

            HttpClient http = new DefaultHttpClient();
            HttpPost post = new HttpPost("http://www.test.com");

            List<NameValuePair> data = new ArrayList<NameValuePair>();
            data.add(new BasicNameValuePair("phone", "value"));
            data.add(new BasicNameValuePair("name", "value"));
            data.add(new BasicNameValuePair("email", "value"));
            data.add(new BasicNameValuePair("comments", "value"));
            post.setEntity(new UrlEncodedFormEntity(data));

            HttpResponse response = http.execute(post);
            responseString = new BasicResponseHandler().
                                 handleResponse(response); // Basic handler
            return true;
        }
        catch (ClientProtocolException e) {
            // do something useful to recover from the exception
            // Note: there may not be anything useful to do here
        }
        catch (IOException e) {
            // do something useful to recover from the exception
            // Note: there may not be anything useful to do here
        }           
        return false;
    }
    @Override
    protected void onPostExecute(Boolean success) {
        // TODO: Do something more useful here
        if (success) {
            Toast.makeText(MainActivity.this, "Success: " + responseString, Toast.LENGTH_LONG).show();
        } else {
            Toast.makeText(MainActivity.this, "Failed", Toast.LENGTH_LONG).show();
        }
    }
}
于 2012-07-23T14:27:52.330 回答
1

将字段作为参数添加到 url。

String phone="phone=234432";
String name="name=John Smith";
String email="email=test@email.com"; 

Url = new URL("http://www.test.com?"+phone+"&"+name+"&"+email);

未经测试，但它应该在 GET Http 请求中工作。

于 2012-07-23T14:26:57.813 回答
0

//使用xml请求

String url="http://www.test.com";
String xmlRequest="<message xmlns=\"http://test.com/schemas/ma\"><header></header>"+
"<body><phone>9944556622</phone><name>Mytest name</name><email>mytest@email.com</email>"+"<comments>hay this seems nice</comments></body></message>"

HttpPost httppost = new HttpPost(url);
    StringEntity se = new StringEntity(xmlRequest,HTTP.UTF_8);


httppost.setHeader("Content-Type","application/xml");
httppost.setHeader("Accept","application/xml");

httppost.setEntity(se);

BasicHttpResponse httpResponse = (BasicHttpResponse) httpclient.execute(httppost);

   StatusLine statusLine = httpResponse.getStatusLine();
于 2012-07-23T15:41:03.507 回答
0 
    HttpClient httpClient = new DefaultHttpClient();
                    Log.i("sampath","line 139");
                    HttpPost httpPost = new HttpPost(url);

                    BasicNameValuePair usernameBasicNameValuePair = new BasicNameValuePair("textMessage", msg);

                    List<NameValuePair> nameValuePairList = new ArrayList<NameValuePair>();
                    nameValuePairList.add(usernameBasicNameValuePair);

                    try {

                        UrlEncodedFormEntity urlEncodedFormEntity = new UrlEncodedFormEntity(nameValuePairList);


                        httpPost.setEntity(urlEncodedFormEntity);

                        try {

                            HttpResponse httpResponse = httpClient.execute(httpPost);


                            InputStream inputStream = httpResponse.getEntity().getContent();

                            InputStreamReader inputStreamReader = new InputStreamReader(inputStream);

                            BufferedReader bufferedReader = new BufferedReader(inputStreamReader);

                            StringBuilder stringBuilder = new StringBuilder();

                            String bufferedStrChunk = null;

                            while((bufferedStrChunk = bufferedReader.readLine()) != null){
                                stringBuilder.append(bufferedStrChunk);
                            }

                            return stringBuilder.toString();

                        } catch (ClientProtocolException cpe) {
                            System.out.println("Firstption caz of HttpResponese :" + cpe);
                            cpe.printStackTrace();
                        } catch (IOException ioe) {
                            System.out.println("Secondption caz of HttpResponse :" + ioe);
                            ioe.printStackTrace();
                        }

                    } catch (UnsupportedEncodingException uee) {
                        System.out.println("Anption given because of UrlEncodedFormEntity argument :" + uee);
                        uee.printStackTrace();
                    }

//---------------------------------------------------------------------------
Here is the php file

<?php
$message = $_POST['textMessage'];
echo 'working fine'.$message;
$filename="androidmessages.html";
file_put_contents($filename,"Entered Text is:".$message."<br />",FILE_APPEND);
?>
于 2015-07-21T15:25:50.563 回答