java Multipart/form-data Restful 请求

Multipart/form-data Restful request

提问by phalco

I am working with multipart/form-data POST request which is divided by boundaries.

我正在处理按边界划分的 multipart/form-data POST 请求。

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

How can I handle this multipart request? I also want to validate the data before making POST request by using Spring4 and REST.

我该如何处理这个多部分请求？我还想在使用 Spring4 和 REST 发出 POST 请求之前验证数据。

采纳答案by user1849477

you should follow the below code :

您应该遵循以下代码：

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.";
        }
    }
} 

回答by phalco

This is how I handled the Multipart Request using Spring 4 (This is the the case when you don't know about filenames or total no. of files in Multipart Request)

这就是我使用 Spring 4 处理多部分请求的方式（当您不知道多部分请求中的文件名或文件总数时就是这种情况）

 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);
        }
    }
}

In Jersey2:

在泽西2：

 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
}

回答by Furquan Ahmed

@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\"}";
    }

You will get the file being uploaded in file argument and the String parameter name in your name argument.

您将在 file 参数中获取正在上传的文件，并在 name 参数中获取 String 参数名称。