WebEditor's note: This answer uses pre_save, which no longer exists in Django REST framework 3.0.; In a sufficiently new version of Django REST framework, MultiPartParser should be available by default, which allows uploading file with no special handling. See an answer below for an example.; I'm using the same stack and was also looking for an … WebDec 3, 2024 · UploadedFile.read () returns the file data in bytes, not a file path or file-like object. In order to use pandas read_csv () function, you'll need to turn those bytes into a stream. Since your file is a csv, the most straightforward way would be to use bytes.decode () with io.StringIO (), like:
Python 在Django中上载JSON文件而不保存_Python_Django_File Upload_Django …
WebFeb 7, 2011 · from django import forms class FileUploadForm (forms.Form): file = forms.FileField () Closing Update: The most important difference between the helping answer and my situation is that I had to decode my input. See the following line as mine csv_file in handle_csv_data: StringIO (content.read ().decode ('utf-8-sig')) python django … WebJan 31, 2024 · 1 I have a question about Django InMemoryUploadedFile processing. I tried to upload a simple csv file whose content is as below username,name user1,user1 I use the snippet below to get the file data and save it locally (Django REST framework) norfolk county building department
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WebOct 21, 2024 · Uploading Files with Django Creating the Model Let's start off by defining a model of a Beast, which directly matches to a database table. A form can then be created to represent a blank slate of this model, allowing the user to fill in the details. WebJun 6, 2015 · def load_files (request): return render (request, 'forms.html', {}) def analysis (request): my_data = [] for name, uploaded_file in request.FILES.items (): # do something with uploaded_file and output 'data' my_data.append (data) return render (request, 'analysis.html', {'my_data': my_data}) forms.html WebJan 26, 2013 · 5 Answers. If you open the file first and then assign request.FILES to the open file object you can access your file. request = self.factory.post ('/') with open (file, 'r') as f: request.FILES ['file'] = f request.FILES ['file'].read () Now you can access request.FILES like you normally would. Remember that when you leave the open block ... norfolk county building dept