13

我正在尝试在 java 中编写一个宁静的 Web 服务，它将采用一些字符串参数和一个二进制文件 (pdf) 参数。

我知道如何处理字符串，但我被二进制文件挂断了。有什么想法/例子吗？

这是我到目前为止所拥有的

@GET
@ConsumeMime("multipart/form-data")
@ProduceMime("text/plain")
@Path("submit/{client_id}/{doc_id}/{html}/{password}")
public Response submit(@PathParam("client_id") String clientID,
                   @PathParam("doc_id") String docID,
                   @PathParam("html") String html,
                   @PathParam("password") String password,
                   @PathParam("pdf") File pdf) {
  return Response.ok("true").build();
}

由于我已经发布了这个答案的链接已被删除，所以这是我的实现。

@POST
@Consumes(MediaType.MULTIPART_FORM_DATA)
@Produces(MediaType.TEXT_PLAIN)
@Path("submit")
public Response submit(@FormDataParam("clientID") String clientID,
                   @FormDataParam("html") String html,
                   @FormDataParam("pdf") InputStream pdfStream) {

    try {
        byte[] pdfByteArray = DocUtils.convertInputStreamToByteArrary(pdfStream);
    } catch (Exception ex) {
        return Response.status(600).entity(ex.getMessage()).build();
    }
}


...

public static byte[] convertInputStreamToByteArrary(InputStream in) throws IOException {
    ByteArrayOutputStream out = new ByteArrayOutputStream();
    final int BUF_SIZE = 1024;
    byte[] buffer = new byte[BUF_SIZE];
    int bytesRead = -1;
    while ((bytesRead = in.read(buffer)) > -1) {
        out.write(buffer, 0, bytesRead);
    }
    in.close();
    byte[] byteArray = out.toByteArray();
    return byteArray;
}
4

3 回答 3

5 
@POST
@Consumes(MediaType.MULTIPART_FORM_DATA)
@Produces(MediaType.TEXT_PLAIN)
@Path("submit")
public Response submit(@FormDataParam("clientID") String clientID,
                   @FormDataParam("html") String html,
                   @FormDataParam("pdf") InputStream pdfStream) {

    try {
        byte[] pdfByteArray = DocUtils.convertInputStreamToByteArrary(pdfStream);
    } catch (Exception ex) {
        return Response.status(600).entity(ex.getMessage()).build();
    }
}


...

public static byte[] convertInputStreamToByteArrary(InputStream in) throws IOException {
    ByteArrayOutputStream out = new ByteArrayOutputStream();
    final int BUF_SIZE = 1024;
    byte[] buffer = new byte[BUF_SIZE];
    int bytesRead = -1;
    while ((bytesRead = in.read(buffer)) > -1) {
        out.write(buffer, 0, bytesRead);
    }
    in.close();
    byte[] byteArray = out.toByteArray();
    return byteArray;
}
于 2015-03-16T23:06:57.233 回答
2

您可以将二进制附件存储在请求的正文中。或者，在此处查看此邮件列表存档：

http://markmail.org/message/dvl6qrzdqstrdtfk

它建议使用 Commons FileUpload 来获取文件并适当地上传。

此处使用 MIME 多部分 API 的另一种选择：

http://n2.nabble.com/File-upload-with-Jersey-td2377844.html

于 2009-07-20T23:40:16.383 回答
1

使用 jersey restful web 服务上传文件的示例程序

需要 Jar 文件（从 Apache 站点下载）：commons-fileupload.jar、commons-io.jar

package com.sms.web;

import java.io.File;
import java.util.Iterator;
import java.util.List;

import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.core.Context;
import javax.servlet.http.HttpServletRequest;

import org.apache.commons.fileupload.FileItem;
import org.apache.commons.fileupload.FileUploadException;
import org.apache.commons.fileupload.disk.DiskFileItemFactory;
import org.apache.commons.fileupload.servlet.ServletFileUpload;


@Path("/UploadTest")
public class UploadData {


    @POST 
    // public String upload(@Context HttpServletRequest request, @PathParam("myfile") String fileName) throws Exception {
    public String upload(@Context HttpServletRequest request) throws Exception {

        String response = "none";

        if (ServletFileUpload.isMultipartContent(request)) { 

            response="got file in request";

            // Create a factory for disk-based file items 
            DiskFileItemFactory  fileItemFactory = new DiskFileItemFactory();

            String path = request.getRealPath("") + File.separatorChar + "publishFiles" + File.separatorChar;

            // File f = new File(path + "myfile.txt");
            // File tmpDir = new File("c:\\tmp");

            File destinationDir = new File(path);


            // Set the size threshold, above which content will be stored on disk.
            // fileItemFactory.setSizeThreshold(1*1024*1024); //1 MB

            // Set the temporary directory to store the uploaded files of size above threshold.
            // fileItemFactory.setRepository(tmpDir);

            // Create a new file upload handler             
            ServletFileUpload uploadHandler = new ServletFileUpload(fileItemFactory);

            try {
                /*
                 * Parse the request
                 */
                List items = uploadHandler.parseRequest(request);
                Iterator itr = items.iterator();

                while(itr.hasNext()) {
                    FileItem item = (FileItem) itr.next();
                    /*
                     * Handle Form Fields.
                     */
                    if(item.isFormField()) {
                        response += "<BR>" + "Field Name = "+item.getFieldName()+", Value = "+item.getString();
                    } else {
                        //Handle Uploaded files.
                        response += "<BR>" + "File Field Name = "+item.getFieldName()+
                            ", File Name = "+item.getName()+
                            ", Content type = "+item.getContentType()+
                            ", File Size = "+item.getSize();
                        /*
                         * Write file to the ultimate location.
                         */
                        File file = new File(destinationDir,item.getName());
                        item.write(file);
                    }
                }
            }catch(FileUploadException ex) {
                response += "Error encountered while parsing the request " + ex;
            } catch(Exception ex) {
                response += "Error encountered while uploading file " + ex;
            }
        } 

        return response;

        }
}
于 2010-06-16T13:35:36.923 回答