java使用params和Body生成HttpPost

Question

我需要用Java复制一篇邮递员文章。 通常，我必须在URL中仅使用参数创建一个HttpPost，因此很容易构建：

ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>();
postParameters.add(new BasicNameValuePair("username", username));
post.setEntity(new UrlEncodedFormEntity(postParameters, Consts.UTF_8));

但是如果我有一篇像下图这样的帖子，URL和Body中都有参数，我该怎么办呢？？ 现在我正在制作HttpPost，如下所示：

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
}

但是我如何将“filename”和“filedata”参数与URL中的参数放在一起呢？ 实际上我在用org。Apache库，但我也可以考虑其他库。

感谢所有愿意帮忙的人

Answer 1

我决定这样做：

1. 放置POST URL头参数
2. 将文件名和文件数据添加为多个部件

这里是代码

private boolean uploadQueue(String username, String password, String filename, byte[] fileData)
{
    HttpClient client = HttpClientBuilder.create().build();
    String URL = "http://post.here.com:8080/";
    HttpPost post = new HttpPost(URL +"?username="+username+"&password="password);

    try
    {
        MultipartEntityBuilder entityBuilder = MultipartEntityBuilder.create();
        entityBuilder.addBinaryBody("filedata", fileData, ContentType.DEFAULT_BINARY, filename);
        entityBuilder.addTextBody("filename", filename);

        post.setEntity(entityBuilder.build());

        HttpResponse httpResponse = client.execute(post);

        if (httpResponse.getStatusLine().getStatusCode() == HttpStatus.SC_OK)
        {
            logger.info(EntityUtils.toString(httpResponse.getEntity()));

            return true;
        } 
        else
        {
            logger.info(EntityUtils.toString(httpResponse.getEntity()));

            return false;
        }
    }
    catch (Exception e)
    {
        logger.error("Error during Updload Queue phase:"+e.getMessage());
    }

    return false;
}

Answer 2

我认为this问题和this问题都是关于类似的问题，并且都有很好的答案

我建议您使用this库，因为它维护良好，如果您愿意，使用起来也很简单

Answer 3

您可以使用下面的代码在POST方法调用中以“application/x-www-form-urlencoded”的形式传递主体参数

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());
   }
}