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");

to

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

似乎你还必须调用connection. getoutputstream ()"至少一次"(以及setDoOutput(true)),以将其视为POST。

所以最低要求的代码是:

    URL url = new URL(urlString);
    HttpURLConnection connection = (HttpURLConnection) url.openConnection();
    //connection.setRequestMethod("POST"); this doesn't seem to do anything at all..so not useful
    connection.setDoOutput(true); // set it to POST...not enough by itself however, also need the getOutputStream call...
    connection.connect();
    connection.getOutputStream().close(); 

令人惊讶的是,你甚至可以在urlString中使用“GET”样式参数。尽管这可能会让事情变得复杂。

显然,你也可以使用NameValuePair。

其他回答

import java.net.*;

public class Demo{

  public static void main(){

       String data = "data=Hello+World!";
       URL url = new URL("http://localhost:8084/WebListenerServer/webListener");
       HttpURLConnection con = (HttpURLConnection) url.openConnection();
       con.setRequestMethod("POST");
       con.setDoOutput(true);
       con.getOutputStream().write(data.getBytes("UTF-8"));
       con.getInputStream();

    }

}

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

我发现HttpURLConnection使用起来非常麻烦。你必须写大量样板,容易出错的代码。我需要一个轻量级的包装为我的Android项目,并提出了一个库,你可以使用:DavidWebb。

上面的例子可以这样写:

Webb webb = Webb.create();
webb.post("http://example.com/index.php")
        .param("param1", "a")
        .param("param2", "b")
        .param("param3", "c")
        .ensureSuccess()
        .asVoid();

您可以在提供的链接上找到备选库的列表。

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 );
}

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;
}