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

试试这个:

HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("your url");
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(3);
nameValuePairs.add(new BasicNameValuePair("user_name", "Name"));
nameValuePairs.add(new BasicNameValuePair("pass","Password" ));
nameValuePairs.add(new BasicNameValuePair("user_email","email" ));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

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

String ret = EntityUtils.toString(response.getEntity());
Log.v("Util response", ret);

您可以根据需要添加任意数量的nameValuePairs。别忘了在名单上提到人数。

其他回答

如果你不需要ArrayList<NameValuePair>作为参数,这是一个更短的解决方案,它使用Uri构建查询字符串。建筑类:

URL url = new URL("http://yoururl.com");
HttpsURLConnection conn = (HttpsURLConnection) url.openConnection();
conn.setReadTimeout(10000);
conn.setConnectTimeout(15000);
conn.setRequestMethod("POST");
conn.setDoInput(true);
conn.setDoOutput(true);

Uri.Builder builder = new Uri.Builder()
        .appendQueryParameter("firstParam", paramValue1)
        .appendQueryParameter("secondParam", paramValue2)
        .appendQueryParameter("thirdParam", paramValue3);
String query = builder.build().getEncodedQuery();

OutputStream os = conn.getOutputStream();
BufferedWriter writer = new BufferedWriter(
            new OutputStreamWriter(os, "UTF-8"));
writer.write(query);
writer.flush();
writer.close();
os.close();

conn.connect();

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();
    }

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);

使用PrintWriter有一个更简单的方法(见这里)

基本上你所需要的是:

// set up URL connection
URL urlToRequest = new URL(urlStr);
HttpURLConnection urlConnection = (HttpURLConnection)urlToRequest.openConnection();
urlConnection.setDoOutput(true);
urlConnection.setRequestMethod("POST");
urlConnection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");

// write out form parameters
String postParamaters = "param1=value1&param2=value2"
urlConnection.setFixedLengthStreamingMode(postParameters.getBytes().length);
PrintWriter out = new PrintWriter(urlConnection.getOutputStream());
out.print(postParameters);
out.close();

// connect
urlConnection.connect();

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) {}
}