概述

以下curl命令在 java中的等价物是什么： curl -X POST -F "file=@$File_PATH" 我想用Java执行的请求是： curl -X POST -F 'file=@file_path' http://localhost/files/ 我在努力： httpClient httpClient = new DefaulthttpClient();

以下curl命令在 java中的等价物是什么：

curl -X POST -F "file=@$File_PATH"

我想用Java执行的请求是：

curl -X POST -F 'file=@file_path' http://localhost/files/

我在努力：

httpClient httpClient = new DefaulthttpClient();        

    httpPost httpPost = new httpPost(_URL);

    File file = new File(PATH);

            multipartentity mpEntity = new multipartentity();
        ContentBody cbFile = new FileBody(file,"bin");
        mpEntity.addPart("userfile",cbFilE);

        httpPost.setEntity(mpEntity);

    httpResponse response = httpClient.execute(httpPost);
    InputStream instream = response.getEntity().getContent();

解决方法

我昨天碰到了这个问题.这是一个使用Apache http库的解决方案.

package curldashf;

import java.io.File;
import java.io.IOException;
import org.apache.commons.io.FileUtils;
import org.apache.http.httpResponse;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.fluent.request;
import org.apache.http.entity.mime.multipartentity;
import org.apache.http.entity.mime.content.byteArrayBody;
import org.apache.http.util.EntityUtils;

public class CurlDashF
{
    public static void main(String[] args) throws ClientProtocolException,IOException
    {
        String filePath = "file_path";
        String url = "http://localhost/files";
        File file = new File(filePath);
        multipartentity entity = new multipartentity();
        entity.addPart("file",new FileBody(filE));
        httpResponse returnResponse = request.Post(url)
            .body(entity)
            .execute().returnResponse();
        System.out.println("Response status: " + returnResponse.getStatusLine().getStatusCode());
        System.out.println(EntityUtils.toString(returnResponse.getEntity()));
    }
}