Question

我需要用Java复制Postman POST。通常我只需要在URL中创建一个只有params的HttpPost，因此很容易构建：

HttpClient client = HttpClientBuilder.create().build();
HttpPost post = new HttpPost("someUrls.com/upload");
ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>();
postParameters.add(new BasicNameValuePair("username", username));
postParameters.add(new BasicNameValuePair("password", password));
postParameters.add(new BasicNameValuePair("owner", owner));
postParameters.add(new BasicNameValuePair("destination", destination));
try{
    post.setEntity(new UrlEncodedFormEntity(postParameters, Consts.UTF_8));
    HttpResponse httpResponse = client.execute(post);
    //Do something
}catch (Exception e){
    //Do something
}

但是，如果我有一个像下面的图像那样在URL和身体中有Params的POST，我该怎么办？现在我正在制作像这样的HttpPost：

protected function createRequest(
    $method,
    $content,
    $uri = '/test',
    $server = ['CONTENT_TYPE' => 'application/json'],
    $parameters = [],
    $cookies = [],
    $files = []
) {
    $request = new \Illuminate\Http\Request;
    return $request->createFromBase(
        \Symfony\Component\HttpFoundation\Request::create(
            $uri,
            $method,
            $parameters,
            $cookies,
            $files,
            $server,
            $content
        )
    );
}

但我如何把＆＃34; filename＆＃34;和＆＃34; filedata＆＃34;身体中的参数与URL中的参数一起？实际上我使用org.Apache库，但我也可以考虑其他库。

感谢任何有帮助的人！

Answer 1

你可以使用下面的代码在POST方法调用中将body参数作为“application / x-www-form-urlencoded”传递

package han.code.development;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.net.URL;

import javax.net.ssl.HttpsURLConnection;

public class HttpPost
{
    public String getDatafromPost()
    {
        BufferedReader br=null;
        String outputData;
        try
        {
            String urlString="https://www.google.com"; //you can replace that with your URL
            URL url=new URL(urlString);
            HttpsURLConnection connection=(HttpsURLConnection) url.openConnection();
            connection.setRequestMethod("POST");
            connection.addRequestProperty("Content-Type", "application/x-www-form-urlencoded");
            connection.addRequestProperty("Authorization", "Replace with your token"); // if you have any accessToken to authorization, just replace
            connection.setDoOutput(true);
            String data="filename=file1&filedata=asdf1234qwer6789";

            PrintWriter out;
            if((data!=null)) 
            {
                 out = new PrintWriter(connection.getOutputStream());
                 out.println(data);
                 out.close();
            }
            System.out.println(connection.getResponseCode()+" "+connection.getResponseMessage());
            br=new BufferedReader(new InputStreamReader(connection.getInputStream()));
            StringBuilder sb=new StringBuilder();
            String str=br.readLine();
            while(str!=null)
            {
                sb.append(str);
                str=br.readLine();
            }

            outputData=sb.toString();
            return outputData;
       }
       catch(Exception e)
       {
           e.printStackTrace();
       }
       return null;
   }
   public static void main(String[] args)
   {
       HttpPost post=new HttpPost();
       System.out.println(post.getDatafromPost());
   }
}

Answer 2

我认为this问题和this问题是关于类似问题的，并且都有很好的答案。

我建议使用this库，因为它维护得很好，如果你愿意，也很容易使用。

Answer 3

我已经解决了这个问题：

放置POST URL标题参数;
将MultipartEntity添加为文件名和filedata。

这里的代码......

async def on_message(self, message):
    print(message)
    while True:
        try:
            #print('ws_connections: ', self.ws_connection, self.ws_connection.stream.socket)
            _fut = self.write_message(self.users[self].request_data())
        except tornado.iostream.StreamClosedError as e:
            print('StreamClosedError:', e)
            break
        except tornado.websocket.WebSocketClosedError as e:
            print('WebSocketClosedError:', e)
            break
        except KeyError as e:
            print('KeyError:', e)
            break
        await gen.sleep(1)

用params和Body制作一个HttpPost

3 个答案: