“Curl -F”Java等价物
发布时间:2020-12-14 02:12:23 所属栏目:Linux 来源:网络整理
导读:以下curl命令在 java中的等价物是什么: curl -X POST -F "file=@$File_PATH" 我想用Java执行的请求是: curl -X POST -F 'file=@file_path' http://localhost/files/ 我在努力: HttpClient httpClient = new DefaultHttpClient(); HttpPost httpPost = new
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以下curl命令在
java中的等价物是什么:
curl -X POST -F "file=@$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()));
}
}
根据需要设置filePath和url.如果您使用的是文件以外的其他内容,则可以使用ByteArrayBody,InputStreamBody或StringBody替换FileBody.我的特殊情况需要ByteArrayBody,但上面的代码适用于文件. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |
