2

I want to read in backend this xlsx file. When I use the swagger I get it, but when I test in frontend I receive an Error 422 - devtools/Network detail'': ( ( loc'':(body file msg'':field required'', type'': value_error missing))).

router = APIRouter()

@router.post('/')
async def upload_file(file: Uploadfile = File(...)):

  try:
    arquivo = await file.read()
    df_cambio = pd.read_excel(arquivo)
    cambio_dict = df_cambio.to_dict('index')
    
    print(cambio_dict)
    return{"file_name": file.filename}
  
  except Exception as e:
    exception_handler(e)

react ->

export defaut function Dropzone() {
const [create, {isLoading}] = usePost();

const handleSubmit = (res) => {
    create('cambios', { data:res.file }})
};
if (isLoading) { return <LoadingPageSkeleton />;}

return (

<BasicForm
  initialValues={{}}
  onSubmit={handleSubmit}
>
    <FormInput id="file />
    <FormSubmitButton/>
</Basicform>
mviniciusserra
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  • `UploadFile` requires that you send a form with `enctype="multipart/form-data` and an `input type="file"` field. Since you haven't included what `create` does, I'm guessing that it submits the content as JSON and not as a multipart form post. If you want to submit a JSON body (as is common from a Javascript application such as react), you can instead use a pydantic model that matches your JSON strucuture and encode the file data as base64 data, which you deserialize on the server side. – MatsLindh Aug 11 '22 at 20:47
  • You would need to send the file data as `FormData`, using the key defined in your endpoint, i.e., `file`, and the file object obtained from an [``](https://developer.mozilla.org/en-US/docs/Web/HTML/Element/input/file) element. For instance, `var formData = new FormData(); formData.append("file", fileInput.files[0]);`. Please have a look at the answers [here](https://stackoverflow.com/a/70824288/17865804), as well as [here](https://stackoverflow.com/a/72029319/17865804) and [here](https://stackoverflow.com/a/70640522/17865804), for more details. – Chris Aug 12 '22 at 05:28

1 Answers1

1

I faced similar issue by trying to create python pop request for FastAPI demo file upload. Working from docs where it gave me ready code in form of curl line:

curl -X 'POST' \
  'http://127.0.0.1:8000/uploadfile/' \
  -H 'accept: application/json' \
  -H 'Content-Type: multipart/form-data' \
  -F 'file=@test_file'

(FastAPI auto generated docs page). Which i used as hint and from which I got that header for file name should be in the form of "type" : "multipart/form-data" , files = { "file" : test_file_descriptor } And it worked

More exact: key for the file descriptor should be file. My case (Python): test_response = requests.post(test_url, data={'filename': test_file , "msg":"hello" ,"type" : "multipart/form-data"}, files = { "file" : test_file } ) Where test_file is file descriptor. I know that my case might be different, but this is the only meaningful google link for this error, and someone with my case could also use it.

miksolo
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