如何在 fastAPI 中返回 xlsx 文件？

Question

How do I return a xlsx file in fastAPI?

Using the python module fastAPI , I can't figure out how to return a xlsx file.
Alse I use openpyxl library to edit my xlsx.

@router.get('/omission-download')
async def omissionFileDownload():
    file_like = open('./upload/test2.xlsx', mode="r") 
    return StreamingResponse(file_like, media_type='application/vnd.openxmlformats-officedocument.spreadsheetml.sheet')

And here is my error

  File "/usr/lib/python3.6/codecs.py", line 321, in decode
    (result, consumed) = self._buffer_decode(data, self.errors, final)
UnicodeDecodeError: 'utf-8' codec can't decode byte 0xc7 in position 12: invalid continuation byte

If anyone knows what to do, I would appreciate it if you could tell me how.

Answer 1

Like Charlie Clark said, you'll want to use FileResponse . Try this:

@router.get('/omission-download')
async def omissionFileDownload():
    return FileResponse('./upload/test2.xlsx')

---

However, to directly answer your question, the error stems from the fact that you're trying to parse a file in UTF-8 mode. Change the mode to bytes and it'll work.

@router.get('/omission-download')
async def omissionFileDownload():
    file_like = open('./upload/test2.xlsx', mode="rb") 
    return StreamingResponse(file_like, media_type='application/vnd.openxmlformats-officedocument.spreadsheetml.sheet')