✨ Add support for declaring UploadFile parameters without explicit File() (#4469)

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Sebastián Ramírez authored and GitHub committed 2022-01-23 19:14:13 +00:00
1 parent 59b1f353b3
commit 1bf55200a9
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from fastapi import FastAPI, File, UploadFile
app = FastAPI()
@app.post("/files/")
async def create_file(file: bytes = File(..., description="A file read as bytes")):
return {"file_size": len(file)}
@app.post("/uploadfile/")
async def create_upload_file(
file: UploadFile = File(..., description="A file read as UploadFile")
):
return {"filename": file.filename}