通过POST方法轻松发送HTTP参数

我成功地使用这段代码通过GET方法发送带有一些参数的HTTP请求

void sendRequest(String request)
{
    // i.e.: request = "http://example.com/index.php?param1=a&param2=b&param3=c";
    URL url = new URL(request); 
    HttpURLConnection connection = (HttpURLConnection) url.openConnection();           
    connection.setDoOutput(true); 
    connection.setInstanceFollowRedirects(false); 
    connection.setRequestMethod("GET"); 
    connection.setRequestProperty("Content-Type", "text/plain"); 
    connection.setRequestProperty("charset", "utf-8");
    connection.connect();
}

现在我可能需要通过POST方法发送参数(即param1, param2, param3)，因为它们非常长。我想添加一个额外的参数，该方法(即字符串httpMethod)。

我怎样才能尽可能少地更改上面的代码，以便能够通过GET或POST发送参数?

我希望情况会有所改变

connection.setRequestMethod("GET");

connection.setRequestMethod("POST");

可以完成这个任务，但是参数仍然通过GET方法发送。

HttpURLConnection有任何方法可以帮助吗? 有什么有用的Java构造吗?

任何帮助都将不胜感激。

当前回答

我强烈推荐构建在apache http api上的http-request。

对于你的案例，你可以看到例子:

private static final HttpRequest<String.class> HTTP_REQUEST = 
      HttpRequestBuilder.createPost("http://example.com/index.php", String.class)
           .responseDeserializer(ResponseDeserializer.ignorableDeserializer())
           .build();

public void sendRequest(String request){
     String parameters = request.split("\\?")[1];
     ResponseHandler<String> responseHandler = 
            HTTP_REQUEST.executeWithQuery(parameters);

   System.out.println(responseHandler.getStatusCode());
   System.out.println(responseHandler.get()); //prints response body
}

如果您对响应体不感兴趣

private static final HttpRequest<?> HTTP_REQUEST = 
     HttpRequestBuilder.createPost("http://example.com/index.php").build();

public void sendRequest(String request){
     ResponseHandler<String> responseHandler = 
           HTTP_REQUEST.executeWithQuery(parameters);
}

对于一般发送post请求与HTTP -request:阅读文档，看到我的答案HTTP post请求与JSON字符串在JAVA，发送HTTP post请求在JAVA, HTTP post使用JSON在JAVA

2017-09-24 22:23:30

其他回答

在GET请求中，参数作为URL的一部分发送。

在POST请求中，参数作为请求体发送，位于请求头之后。

要使用HttpURLConnection进行POST，需要在打开连接后将参数写入连接。

这段代码应该让你开始:

String urlParameters  = "param1=a&param2=b&param3=c";
byte[] postData       = urlParameters.getBytes( StandardCharsets.UTF_8 );
int    postDataLength = postData.length;
String request        = "http://example.com/index.php";
URL    url            = new URL( request );
HttpURLConnection conn= (HttpURLConnection) url.openConnection();           
conn.setDoOutput( true );
conn.setInstanceFollowRedirects( false );
conn.setRequestMethod( "POST" );
conn.setRequestProperty( "Content-Type", "application/x-www-form-urlencoded"); 
conn.setRequestProperty( "charset", "utf-8");
conn.setRequestProperty( "Content-Length", Integer.toString( postDataLength ));
conn.setUseCaches( false );
try( DataOutputStream wr = new DataOutputStream( conn.getOutputStream())) {
   wr.write( postData );
}

2010-11-17 15:40:04

你好，请使用这个类来改善你的张贴方法

public static JSONObject doPostRequest(HashMap<String, String> data, String url) {

    try {
        RequestBody requestBody;
        MultipartBuilder mBuilder = new MultipartBuilder().type(MultipartBuilder.FORM);

        if (data != null) {


            for (String key : data.keySet()) {
                String value = data.get(key);
                Utility.printLog("Key Values", key + "-----------------" + value);

                mBuilder.addFormDataPart(key, value);

            }
        } else {
            mBuilder.addFormDataPart("temp", "temp");
        }
        requestBody = mBuilder.build();


        Request request = new Request.Builder()
                .url(url)
                .post(requestBody)
                .build();

        OkHttpClient client = new OkHttpClient();
        Response response = client.newCall(request).execute();
        String responseBody = response.body().string();
        Utility.printLog("URL", url);
        Utility.printLog("Response", responseBody);
        return new JSONObject(responseBody);

    } catch (UnknownHostException | UnsupportedEncodingException e) {

        JSONObject jsonObject=new JSONObject();

        try {
            jsonObject.put("status","false");
            jsonObject.put("message",e.getLocalizedMessage());
        } catch (JSONException e1) {
            e1.printStackTrace();
        }
        Log.e(TAG, "Error: " + e.getLocalizedMessage());
    } catch (Exception e) {
        e.printStackTrace();
        JSONObject jsonObject=new JSONObject();

        try {
            jsonObject.put("status","false");
            jsonObject.put("message",e.getLocalizedMessage());
        } catch (JSONException e1) {
            e1.printStackTrace();
        }
        Log.e(TAG, "Other Error: " + e.getLocalizedMessage());
    }
    return null;
}

2016-07-22 11:54:19

我强烈推荐构建在apache http api上的http-request。

对于你的案例，你可以看到例子:

private static final HttpRequest<String.class> HTTP_REQUEST = 
      HttpRequestBuilder.createPost("http://example.com/index.php", String.class)
           .responseDeserializer(ResponseDeserializer.ignorableDeserializer())
           .build();

public void sendRequest(String request){
     String parameters = request.split("\\?")[1];
     ResponseHandler<String> responseHandler = 
            HTTP_REQUEST.executeWithQuery(parameters);

   System.out.println(responseHandler.getStatusCode());
   System.out.println(responseHandler.get()); //prints response body
}

如果您对响应体不感兴趣

private static final HttpRequest<?> HTTP_REQUEST = 
     HttpRequestBuilder.createPost("http://example.com/index.php").build();

public void sendRequest(String request){
     ResponseHandler<String> responseHandler = 
           HTTP_REQUEST.executeWithQuery(parameters);
}

对于一般发送post请求与HTTP -request:阅读文档，看到我的答案HTTP post请求与JSON字符串在JAVA，发送HTTP post请求在JAVA, HTTP post使用JSON在JAVA

2017-09-24 22:23:30

我看到一些其他的答案给出了另一种选择，我个人认为直觉上你在做正确的事情;)不好意思，在devoxx上，有几个演讲者一直在抱怨这种事情。

这就是为什么我个人使用Apache的HTTPClient/HttpCore库来做这类工作，我发现它们的API比Java的原生HTTP支持更容易使用。当然是YMMV了!

2010-11-17 15:40:37

这里我发送jsonobject作为参数//jsonobject={"name":"lucifer"，"pass":"abc"}//serverUrl = "http://192.168.100.12/testing" //主机=192.168.100.12

  public static String getJson(String serverUrl,String host,String jsonobject){

    StringBuilder sb = new StringBuilder();

    String http = serverUrl;

    HttpURLConnection urlConnection = null;
    try {
        URL url = new URL(http);
        urlConnection = (HttpURLConnection) url.openConnection();
        urlConnection.setDoOutput(true);
        urlConnection.setRequestMethod("POST");
        urlConnection.setUseCaches(false);
        urlConnection.setConnectTimeout(50000);
        urlConnection.setReadTimeout(50000);
        urlConnection.setRequestProperty("Content-Type", "application/json");
        urlConnection.setRequestProperty("Host", host);
        urlConnection.connect();
        //You Can also Create JSONObject here 
        OutputStreamWriter out = new OutputStreamWriter(urlConnection.getOutputStream());
        out.write(jsonobject);// here i sent the parameter
        out.close();
        int HttpResult = urlConnection.getResponseCode();
        if (HttpResult == HttpURLConnection.HTTP_OK) {
            BufferedReader br = new BufferedReader(new InputStreamReader(
                    urlConnection.getInputStream(), "utf-8"));
            String line = null;
            while ((line = br.readLine()) != null) {
                sb.append(line + "\n");
            }
            br.close();
            Log.e("new Test", "" + sb.toString());
            return sb.toString();
        } else {
            Log.e(" ", "" + urlConnection.getResponseMessage());
        }
    } catch (MalformedURLException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    } catch (JSONException e) {
        e.printStackTrace();
    } finally {
        if (urlConnection != null)
            urlConnection.disconnect();
    }
    return null;
}

2016-07-18 10:12:24

通过POST方法轻松发送HTTP参数

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