Converting MultipartFile to java.io.File without copying to local machine

Shiju K Babu picture Shiju K Babu · Jan 13, 2014 · Viewed 33.4k times · Source

I have a Java Spring MVC web application. From client, through AngularJS, I am uploading a file and posting it to Controller as webservice.

In my Controller, I am gettinfg it as MultipartFile and I can copy it to local machine.

But I want to upload the file to Amazone S3 bucket. So I have to convert it to java.io.File. Right now what I am doing is, I am copying it to local machine and then uploading to S3 using jets3t.

Here is my way of converting in controller

MultipartHttpServletRequest mRequest=(MultipartHttpServletRequest)request;

Iterator<String> itr=mRequest.getFileNames();
        while(itr.hasNext()){
            MultipartFile mFile=mRequest.getFile(itr.next());
            String fileName=mFile.getOriginalFilename();

            fileLoc="/home/mydocs/my-uploads/"+date+"_"+fileName; //date is String form of current date.

Then I am using FIleCopyUtils of SpringFramework

File newFile = new File(fileLoc);

                  // if the directory does not exist, create it
                  if (!newFile.getParentFile().exists()) {
                    newFile.getParentFile().mkdirs();  
                  }
                FileCopyUtils.copy(mFile.getBytes(), newFile);

So it will create a new file in the local machine. That file I am uplaoding in S3

S3Object fileObject = new S3Object(newFile);

s3Service.putObject("myBucket", fileObject);

It creates file in my local system. I don't want to create.

Without creating a file in local system, how to convert a MultipartFIle to java.io.File?

Answer

Boris picture Boris · Jan 13, 2014

MultipartFile, by default, is already saved on your server as a file when user uploaded it. From that point - you can do anything you want with this file. There is a method that moves that temp file to any destination you want. http://docs.spring.io/spring/docs/3.0.x/api/org/springframework/web/multipart/MultipartFile.html#transferTo(java.io.File)

But MultipartFile is just API, you can implement any other MultipartResolver http://docs.spring.io/spring/docs/3.0.x/api/org/springframework/web/multipart/MultipartResolver.html

This API accepts input stream and you can do anything you want with it. Default implementation (usually commons-multipart) saves it to temp dir as a file.

But other problem stays here - if S3 API accepts a file as a parameter - you cannot do anything with this - you need a real file. If you want to avoid creating files at all - create you own S3 API.