マルチパート/フォームデータ安らかなリクエスト

Question

私は境界で分割されたmultipart/form-data POSTリクエストを扱っています。

POST .... HTTP/1.1 
. 
. 
. 
---boundary123 
Content-type:application/octet-stream 
content-Disposition: form-data filenale="payload.txt" name="someuniquename" 
[paylaod content](this is in xml format) 
---boundary123 
content-type:application/json 
content-Disposition:form-data name="someuniquname" 
{ID:"999"} 
---boundary123 

このマルチパートリクエストはどのように処理できますか？また、Spring4とRESTを使用してPOSTリクエストを行う前に、データを検証する必要があります。

Answer 1

あなたは以下のコードに従ってください。マルチパートリクエスト）Jersey2で

import org.springframework.web.multipart.MultipartHttpServletRequest;  
@RequestMapping(value = "/your_webservice_path", method = RequestMethod.POST, consumes = "multipart/form-data") 
public Void myController(MultipartHttpServletRequest request) throws Exception 
{ 
Iterator<String> iterator = request.getFileNames(); 
while (iterator.hasNext()) {  
     String str=iterator.next().toString(); 
     try{ 
     //extract the file from request 
     } 
     catch(Exception ex){ 
      logger.error("Error while parsing the Payload ro metadata from request", ex); 
     } 
    } 
} 

：

import org.glassfish.jersey.media.multipart.MultiPart; 
@POST 
@Consumes({MediaType.MULTIPART_FORM_DATA}) 
    @Path("/your_webservice_path") 
    public void myController(MultiPart request) throws Exception { 
    //use request .getBodyParts() for extracting files from Multipart request 
} 

Answer 2

@RequestMapping(value = "/upload", 
    method = RequestMethod.POST, 
    consumes = MediaType.MULTIPART_FORM_DATA_VALUE, 
    produces = MediaType.APPLICATION_JSON_VALUE) 
public @ResponseBody String upload(
    @RequestParam("file") final MultipartFile file, 
    @RequestParam("name") final String name) { 
     // your work here using the file. you can use file.getBytes() 
     // or get an input stream from file using file.getInputStream() 
     // and save it. 
     return "{\"status\" : \"success\"}"; 
    } 

file引数にファイルをアップロードし、name引数にStringパラメータ名を取得します。これは私が春4を（これはあなたがファイル名または総無について知らない場合で使用してマルチパートリクエストを処理し方法です

import java.io.BufferedOutputStream; 
import java.io.File; 
import java.io.FileOutputStream; 
import org.springframework.stereotype.Controller; 
import org.springframework.web.bind.annotation.RequestMapping; 
import org.springframework.web.bind.annotation.RequestMethod; 
import org.springframework.web.bind.annotation.RequestParam; 
import org.springframework.web.bind.annotation.ResponseBody; 
import org.springframework.web.multipart.MultipartFile; 
@Controller 
public class FileUploadController { 
    @RequestMapping(value="/singleUpload") 
    public String singleUpload(){ 
     return "singleUpload"; 
    } 
    @RequestMapping(value="/singleSave", method=RequestMethod.POST) 
    public @ResponseBody String singleSave(@RequestParam("file") MultipartFile file, @RequestParam("desc") String desc){ 
     System.out.println("File Description:"+desc); 
     String fileName = null; 
     if (!file.isEmpty()) { 
      try { 
       fileName = file.getOriginalFilename(); 
       byte[] bytes = file.getBytes(); 
       BufferedOutputStream buffStream = 
         new BufferedOutputStream(new FileOutputStream(new File("F:/cp/" + fileName))); 
       buffStream.write(bytes); 
       buffStream.close(); 
       return "You have successfully uploaded " + fileName; 
      } catch (Exception e) { 
       return "You failed to upload " + fileName + ": " + e.getMessage(); 
      } 
     } else { 
      return "Unable to upload. File is empty."; 
     } 
    } 
    @RequestMapping(value="/multipleUpload") 
    public String multiUpload(){ 
     return "multipleUpload"; 
    } 
    @RequestMapping(value="/multipleSave", method=RequestMethod.POST) 
    public @ResponseBody String multipleSave(@RequestParam("file") MultipartFile[] files){ 
     String fileName = null; 
     String msg = ""; 
     if (files != null && files.length >0) { 
      for(int i =0 ;i< files.length; i++){ 
       try { 
        fileName = files[i].getOriginalFilename(); 
        byte[] bytes = files[i].getBytes(); 
        BufferedOutputStream buffStream = 
          new BufferedOutputStream(new FileOutputStream(new File("F:/cp/" + fileName))); 
        buffStream.write(bytes); 
        buffStream.close(); 
        msg += "You have successfully uploaded " + fileName +"<br/>"; 
       } catch (Exception e) { 
        return "You failed to upload " + fileName + ": " + e.getMessage() +"<br/>"; 
       } 
      } 
      return msg; 
     } else { 
      return "Unable to upload. File is empty."; 
     } 
    } 
} 

Answer 3

ファイルの中で：