如何在REST API中以编程方式使用文件发布多部分/表单数据

Question

我有一个表格数据的Rest URL，我从我的REST客户端POST MAN成功执行。如何从java程序执行此操作？如何以编程方式传递附加文件。其余呼叫的快照如下：

Answer 1

检查以下代码示例。

import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;

import javax.ws.rs.Consumes;
import javax.ws.rs.FormParam;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response;

import com.sun.jersey.core.header.FormDataContentDisposition;
import com.sun.jersey.multipart.FormDataParam;

@Path("/file")
public class UploadFileService {

@POST
@Path("/upload")
@Consumes(MediaType.MULTIPART_FORM_DATA)
public Response uploadFile(
        @FormDataParam("file") InputStream uploadedInputStream,
        @FormDataParam("file") FormDataContentDisposition fileDetail,
        @FormDataParam("path") String path) {


    // Path format //10.217.14.97/Installables/uploaded/
    System.out.println("path::"+path);
    String uploadedFileLocation = path
            + fileDetail.getFileName();

    // save it
    writeToFile(uploadedInputStream, uploadedFileLocation);

    String output = "File uploaded to : " + uploadedFileLocation;

    return Response.status(200).entity(output).build();

}

// save uploaded file to new location
private void writeToFile(InputStream uploadedInputStream,
        String uploadedFileLocation) {

    try {
        OutputStream out = new FileOutputStream(new File(
                uploadedFileLocation));
        int read = 0;
        byte[] bytes = new byte[1024];

        out = new FileOutputStream(new File(uploadedFileLocation));
        while ((read = uploadedInputStream.read(bytes)) != -1) {
            out.write(bytes, 0, read);
        }
        out.flush();
        out.close();
    } catch (IOException e) {

        e.printStackTrace();
    }

   }