我试图做POST与HttpURLConnection(我需要使用这种方式,不能使用HttpPost),我想添加参数的连接,如

post.setEntity(new UrlEncodedFormEntity(nvp));

在哪里

nvp = new ArrayList<NameValuePair>();

有一些数据存储在。我找不到一种方法如何添加这个数组列表到我的HttpURLConnection在这里:

HttpsURLConnection https = (HttpsURLConnection) url.openConnection();
https.setHostnameVerifier(DO_NOT_VERIFY);
http = https;
http.setRequestMethod("POST");
http.setDoInput(true);
http.setDoOutput(true);

出现这种尴尬的https和http组合的原因是不需要验证证书。不过,这不是问题,它可以很好地发布服务器。但我需要它来张贴论点。

什么好主意吗?


重复的免责声明:

早在2012年,我还不知道如何在HTTP POST请求中插入参数。我一直在使用NameValuePair,因为它在教程中。这个问题可能看起来像一个重复的问题,然而,我2012年自己读了另一个问题,它没有使用NameValuePair。事实上,这并没有解决我的问题。


当前回答

通过使用org.apache.http.client.HttpClient,你也可以用下面更容易读懂的方式轻松做到这一点。

HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://www.yoursite.com/script.php");

在try catch内可以插入

// Add your data
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
nameValuePairs.add(new BasicNameValuePair("id", "12345"));
nameValuePairs.add(new BasicNameValuePair("stringdata", "AndDev is Cool!"));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

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

其他回答

参数到HttpURLConnection使用POST使用NameValuePair输出

        try {
        URL url = new URL("https://yourUrl.com");
        HttpURLConnection conn = (HttpURLConnection) url.openConnection();

        conn.setUseCaches(false);
        conn.setDoInput(true);
        conn.setDoOutput(true);

        conn.setRequestMethod("POST");

        conn.setRequestProperty("Content-Type", "application/json");

        JSONObject data = new JSONObject();
        data.put("key1", "value1");
        data.put("key2", "value2");

        OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
        wr.write(data.toString());
        wr.flush();
        wr.close();

        BufferedReader in = new BufferedReader(new InputStreamReader(conn.getInputStream()));
        String inputLine;
        StringBuffer response = new StringBuffer();

        while ((inputLine = in.readLine()) != null) {
            response.append(inputLine);
        }
        in.close();
        System.out.println(response.toString());
    } catch (MalformedURLException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    } catch (JSONException e) {
        e.printStackTrace();
    }

一个解决方案是让你自己的params字符串。

这是我在我的最新项目中使用的实际方法。你需要将args从hashtable更改为namevaluepair:

private static String getPostParamString(Hashtable<String, String> params) {
    if(params.size() == 0)
        return "";

    StringBuffer buf = new StringBuffer();
    Enumeration<String> keys = params.keys();
    while(keys.hasMoreElements()) {
        buf.append(buf.length() == 0 ? "" : "&");
        String key = keys.nextElement();
        buf.append(key).append("=").append(params.get(key));
    }
    return buf.toString();
}

发布参数:

OutputStreamWriter writer = new OutputStreamWriter(conn.getOutputStream());
writer.write(getPostParamString(req.getPostParams()));

AsyncTask通过POST方法将数据作为JSONObect发送

public class PostMethodDemo extends AsyncTask<String , Void ,String> {
        String server_response;

        @Override
        protected String doInBackground(String... strings) {
            URL url;
            HttpURLConnection urlConnection = null;

            try {
                url = new URL(strings[0]);
                urlConnection = (HttpURLConnection) url.openConnection();
                urlConnection.setDoOutput(true);
                urlConnection.setDoInput(true);
                urlConnection.setRequestMethod("POST");

                DataOutputStream wr = new DataOutputStream(urlConnection.getOutputStream ());

                try {
                    JSONObject obj = new JSONObject();
                    obj.put("key1" , "value1");
                    obj.put("key2" , "value2");

                    wr.writeBytes(obj.toString());
                    Log.e("JSON Input", obj.toString());
                    wr.flush();
                    wr.close();
                } catch (JSONException ex) {
                    ex.printStackTrace();
                }
                urlConnection.connect();

                int responseCode = urlConnection.getResponseCode();

                if(responseCode == HttpURLConnection.HTTP_OK){
                    server_response = readStream(urlConnection.getInputStream());
                }

            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return null;
        }

        @Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            Log.e("Response", "" + server_response);
        }
    }

    public static String readStream(InputStream in) {
        BufferedReader reader = null;
        StringBuffer response = new StringBuffer();
        try {
            reader = new BufferedReader(new InputStreamReader(in));
            String line = "";
            while ((line = reader.readLine()) != null) {
                response.append(line);
            }
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            if (reader != null) {
                try {
                    reader.close();
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        }
        return response.toString();
    }

我想我找到了你需要的东西。它可能会帮助其他人。

你可以使用UrlEncodedFormEntity.writeTo(OutputStream)方法。

UrlEncodedFormEntity formEntity = new UrlEncodedFormEntity(nvp); 
http.connect();

OutputStream output = null;
try {
  output = http.getOutputStream();    
  formEntity.writeTo(output);
} finally {
 if (output != null) try { output.close(); } catch (IOException ioe) {}
}

接受的回答在以下位置抛出ProtocolException:

OutputStream os = conn.getOutputStream();

因为它没有为URLConnection对象启用输出。解决方案应该包括:

conn.setDoOutput(真正的);

让它起作用。