CODEWITHSUNDEEP
javaopenapiswagger-codegen
fhueser
fhueser

Reputation: 649

Get stream from file upload in openapi

I have a backend application (Java 8) and an Angular frontend. Now, I added a file upload using multipart/formdata. We use openapi to generate the api code. Here's the part of the code used to generate it:

"post": {
   ...
   "consumes" : [
      "multipart/form-data"
   ],
   "produces": [
      "application/json"
   ],
   "parameters": [
      {
         "name": "updateFile",
         "in": "formData",
         "type": "file",
         "required": true,
      }
   ],
   ...
}

And the frontend prepares the file as such:

  const formData = new FormData();
  formData.append('updateFile', updateFile);

This actually works quite well and provides us a Part-Object. Unfortunately the machine this code will run on in production is quite weak and the uploaded file will be too large to be handled in the memory of it (~130mb). So, I need a workaround for this. I think accessing the incoming stream would be an elegant way to write the file to some temp directory "on-the-fly" and I've come across code samples doing exactly that. I can't figure out how to configure my endpoint to provide a stream instead of the "finished" Part-Object.

Unfortunately this project uses a quite old openapi version 2.

Upvotes: 7

Views: 1447

Answers (1)

fhueser
fhueser

Reputation: 649

The workaround I've found now is taking the ServletInpuStream directly from the HttpServletRequest and leaving the body in the API description empty to prevent the stream from being consumed before I actively do so. This of course forces me to parse the content myself. I found MultipartStream of apache.commons to be very helpful for this.

This solution feels a bit hacky but is at least viable until we updated to a more recent version that directly supports something like that.

Upvotes: 1

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