I'm working on a web. User can upload a file. This file is in docx format. After he uploads a file and choose which languages he wants to translate the file to, I want to redirect him to another page, where he can see prices for translations. The prices depends on particular language and number of characters in the docx file.
I can't figure out how to handle the file uploaded. I have a function which get's path to file and returns a number of characters. After uploading file and click on submit, I want to call this function so I can render new page with estimated prices.
I've read that I can call temporary_file_path on request.FILES['file'] but it raises
'InMemoryUploadedFile' object has no attribute 'temporary_file_path'
I want to find out how many characters uploaded file contains and send it in a request to another view - /order-estimation.
VIEW:
def create_order(request):
LanguageLevelFormSet = formset_factory(LanguageLevelForm, extra=5, max_num=5)
language_level_formset = LanguageLevelFormSet(request.POST or None)
job_creation_form = JobCreationForm(request.POST or None, request.FILES or None)
context = {'job_creation_form': job_creation_form,
'formset': language_level_formset}
if request.method == 'POST':
if job_creation_form.is_valid() and language_level_formset.is_valid():
cleaned_data_job_creation_form = job_creation_form.cleaned_data
cleaned_data_language_level_formset = language_level_formset.cleaned_data
for language_level_form in [d for d in cleaned_data_language_level_formset if d]:
language = language_level_form['language']
level = language_level_form['level']
Job.objects.create(
customer=request.user,
text_to_translate=cleaned_data_job_creation_form['text_to_translate'],
file=cleaned_data_job_creation_form['file'],
short_description=cleaned_data_job_creation_form['short_description'],
notes=cleaned_data_job_creation_form['notes'],
language_from=cleaned_data_job_creation_form['language_from'],
language_to=language,
level=level,
)
path = request.FILES['file'].temporary_file_path
utilities.docx_get_characters_number(path) # THIS NOT WORKS
return HttpResponseRedirect('/order-estimation')
else:
return render(request, 'auth/jobs/create-job.html', context=context)
return render(request, 'auth/jobs/create-job.html', context=context)
The InMemoryUploadedFile does not provide temporary_file_path. The content lives 'in memory' - as the class name implies.
By default Django uses InMemoryUploadedFile for files up to 2.5MB size, larger files use TemporaryFileUploadHandler. where the later provides the temporary_file_path method in question. Django Documentation
So an easy way would be to change your settings for FILE_UPLOAD_HANDLERS to always use TemporaryFileUploadHandler:
FILE_UPLOAD_HANDLERS = [
'django.core.files.uploadhandler.TemporaryFileUploadHandler',
]
Just keep in mind that this is not the most efficient way when you have a site with a lot of concurrent small upload requests.
Related
I am trying to compress a folder before saving it to database/file storage system using Django. For this task I am using ZipFile library. Here is the code of view.py:
class BasicUploadView(View):
def get(self, request):
file_list = file_information.objects.all()
return render(self.request, 'fileupload_app/basic_upload/index.html',{'files':file_list})
def post(self, request):
zipfile = ZipFile('test.zip','w')
if request.method == "POST":
for upload_file in request.FILES.getlist('file'): ## index.html name
zipfile.write(io.BytesIO(upload_file))
fs = FileSystemStorage()
content = fs.save(upload_file.name,upload_file)
data = {'name':fs.get_available_name(content), 'url':fs.url(content)}
zipfile.close()
return JsonResponse(data)
But I am getting the following error:
TypeError: a bytes-like object is required, not 'InMemoryUploadedFile'
Is there any solution for this problem? Since I may have to upload folder with large files, do I have to write a custom TemporaryFileUploadHandler for this purpose? I have recently started working with Django and it is quite new to me. Please help me with some advice.
InMemoryUploadedFile is an object that contains more than just file you should open file and read it content ( InMemoryUploadedFile.file is the file)
InMemoryUploadedFile.open()
You should open file with open() and then read() it's content, also you should check if you have uploaded files correctly also you could use with syntax for both zip and file
https://www.pythonforbeginners.com/files/with-statement-in-python
I have a Django app where users would upload Images and certain functions would take the image file path to process them and save the result in the same model.
all of this would happen in the file upload view the problem is My functions take the file path which isn't created/committed in the DB yet as I don't save before calling those functions.
I tried overriding the save method in the models.py and it didn't work so how can I call the functions after the upload in a convenient way ??
here's the function:
# The view for analysing images
def upload_view(request,pk=None):
patient = get_object_or_404(Patient,pk=pk)
if request.method == 'POST':
form = forms.ImageForm(request.POST,request.FILES)
if form.is_valid():
image = form.save(commit=False)
image.patient = patient
image.enhanced = main(image.original.path)
image.segmented = segment(image.enhanced.path)
image.enhanced.name = image.enhanced.path.split('media')[1]
image.segmented.name = image.enhanced.path.split('media')[1]
messages.success(request,"Image added successfully!")
image.save()
return HttpResponseRedirect(reverse_lazy('patients:patient_detail', kwargs={'pk' : image.patient.pk}))
else:
form = forms.ImageForm()
return render(request, 'patients/upload.html', {'form': form})
else:
form = forms.ImageForm()
return render(request, 'patients/upload.html', {'form': form})
image.original is the uploaded image
the problem is the file path isn't passed correctly and the functions return errors bec of that. (it worked when I made the processing in a different view where it was accessed after the upload)
Before you call save() on your model, the path to your image doesn't exist, or is temporary. You can fix this by first creating the model from the form, no need for commit=False, assuming main and segment are PIL based functions that return images, you can use the ImageField.save() method to set the name at the same time:
if form.is_valid():
image = form.save() # this creates the model, the upload paths are created here!
image.patient = patient
image.enhanced.save("enhanced_" + image.original.name, main(image.original.path)) # I made this up to avoid conflicts in your storage
image.segmented.save("segmented_" + image.original.name, segment(image.enhanced.path))
image.save() # this updates the model with the new images
messages.success(request,"Image added successfully!") # I moved this here, more accurate, in case save fails....
return HttpResponseRedirect(reverse_lazy('patients:patient_detail', kwargs={'pk' : image.patient.pk}))
As you can see, you need two hits to the database to save your images. This assumes that enhanced and segmented fields are not required in the model.
Also, because image transformation is an expensive task, I'd check how to get this out of the request cycle, by using something like Celery or Django-RQ, that way, your app can still service request while making the transformations on the background.
I am looking to attach a file to an email which includes all the content a user inputs from a contact form. I currently refer a PDF which records their inputs, and I attach that PDF from a file destination. However, I do not know how to attach additional files which the user provides on the contact form. In this case, this is represented by "msg.attach_file(upload_file)." My thoughts are:
Have the file be uploaded to a destination; however, it needs to renamed to a uniform name each time so I can refer to it during the attachment process (msg.attach_file).
Figure out a way to use request.FILES to attach it immediately without having to worry about its file name or upload destination (I am not sure if msg.attach_file is a valid command for this method).
Is there a right way to perform this action? I am attempting to perform method 2 with my views.py file which refers to my forms.py file, but it is giving me an error.
Views.py
def quote_req(request):
submitted = False
if request.method == 'POST':
form = QuoteForm(request.POST, request.FILES)
company = request.POST['company']
contact_person = request.POST['contact_person']
upload_file = request.FILES['upload_file']
description = 'You have received a sales contact form'
if form.is_valid():
data_dict = {
'company_': str(company),
'contact_person_': str(contact_person),
}
write_fillable_pdf(INVOICE_TEMPLATE_PATH, INVOICE_OUTPUT_PATH, data_dict)
form.save()
# assert false
msg = EmailMessage('Contact Form', description, settings.EMAIL_HOST_USER, ['sample#mail.com'])
msg.attach_file('/uploads/file.pdf')
msg.attach_file(upload_file)
msg.send(fail_silently=False)
return HttpResponseRedirect('/quote/?submitted=True')
else:
form = QuoteForm()
if 'submitted' in request.GET:
submitted = True
Error Log
TypeError at /quote/
expected str, bytes or os.PathLike object, not InMemoryUploadedFile
Request Method: POST
Request URL: http://www.mytestingwebsitesample.com/quote/
Django Version: 2.1.3
Exception Type: TypeError
Exception Value:
expected str, bytes or os.PathLike object, not InMemoryUploadedFile
Can you try the following? Since InMemoryUploadedFile doesn't work, might have to process it first
upload_file = request.FILES['upload_file']
content = upload_file.read()
attachment = (upload_file.name, content, 'application/pdf')
# . . .
msg.attach(attachment)
upload_file.read() will return bytes. You might want to try using attach instead of attach_file. attach_file requires the file to be saved to your filesystem, while attach can take data. However, I believe that with attach, you should be able to use request.FILES['upload_file'] directly.
https://docs.djangoproject.com/en/2.2/topics/email/#emailmessage-objects
I have resolved my issue by employing a storage.py file that overwrites files with the same name; in my case, I am uploading each file, renaming it to a uniform name, and then having the storage file overwrite it later on rather than Django adding an extension to a file name with the same title.
I'm experimenting with a site that will allow users to upload audio files. I've read every doc that I can get my hands on but can't find much about validating files.
Total newb here (never done any file validation of any kind before) and trying to figure this out. Can someone hold my hand and tell me what I need to know?
As always, thank you in advance.
You want to validate the file before it gets written to disk. When you upload a file, the form gets validated then the uploaded file gets passed to a handler/method that deals with the actual writing to the disk on your server. So in between these two operations, you want to perform some custom validation to make sure it's a valid audio file
You could:
check if the the file is less then a certain size (good practice)
then check if the submitted file has a certain content type (i.e. an audio file)
this is pretty useless as someone could easily spoof it
then check that the file ends in a certain extension (or extensions)
this is also pretty useless
try read the file and see if it's actually audio
(I haven't tested this code)
models.py
class UserSong(models.Model):
title = models.CharField(max_length=100)
audio_file = models.FileField()
forms.py
class UserSongForm(forms.ModelForm):
# Add some custom validation to our file field
def clean_audio_file(self):
file = self.cleaned_data.get('audio_file',False):
if file:
if file._size > 4*1024*1024:
raise ValidationError("Audio file too large ( > 4mb )")
if not file.content-type in ["audio/mpeg","audio/..."]:
raise ValidationError("Content-Type is not mpeg")
if not os.path.splitext(file.name)[1] in [".mp3",".wav" ...]:
raise ValidationError("Doesn't have proper extension")
# Here we need to now to read the file and see if it's actually
# a valid audio file. I don't know what the best library is to
# to do this
if not some_lib.is_audio(file.content):
raise ValidationError("Not a valid audio file")
return file
else:
raise ValidationError("Couldn't read uploaded file")
views.py
from utils import handle_uploaded_file
def upload_file(request):
if request.method == 'POST':
form = UserSongForm(request.POST, request.FILES)
if form.is_valid():
# If we are here, the above file validation has completed
# so we can now write the file to disk
handle_uploaded_file(request.FILES['file'])
return HttpResponseRedirect('/success/url/')
else:
form = UploadFileForm()
return render_to_response('upload.html', {'form': form})
utils.py
# from django's docs
def handle_uploaded_file(f):
ext = os.path.splitext(f.name)[1]
destination = open('some/file/name%s'%(ext), 'wb+')
for chunk in f.chunks():
destination.write(chunk)
destination.close()
https://docs.djangoproject.com/en/dev/topics/http/file-uploads/#file-uploads
https://docs.djangoproject.com/en/dev/ref/forms/fields/#filefield
https://docs.djangoproject.com/en/dev/ref/files/file/#django.core.files.File
Can someone illuminate me as to exactly when an uploaded file is actually written to the location returned by "upload_to" in the FileField, in particular with regards to the order of field, model, and form validation and cleaning?
Right now I have a "clean" method on my model which assumes the uploaded file is in place, so it can do some validation on it. It looks like the file isn't yet saved, and may just be held in a temporary location or in memory. If that is the case, how do I "open" it or find a path to it if I need to execute some external process/program to validate the file?
Thanks,
Ian
The form cleansing has nothing to do with actually saving the file, or with saving any other data for that matter. The file isn't saved until to you run the save() method of the model instance (note that if you use ModelName.objects.create() this save() method is called for you automatically).
The bound form will contain an open File object, so you should be able to do any validation on that object directly. For example:
form = MyForm(request.POST, request.FILES)
if form.is_valid():
file_object = form.cleaned_data['myFile']
#run any validation on the file_object, or define a clean_myFile() method
# that will be run automatically when you call form.is_valid()
model_inst = MyModel('my_file' = file_object,
#assign other attributes here....
)
model_inst.save() #file is saved to disk here
What do you need to do on it? If your validation will work without a temporary file, you can access the data by calling read() on what your file field returns.
def clean_field(self):
_file = self.cleaned_data.get('filefield')
contents = _file.read()
If you do need it on the disk, you know where to go from here :) write it to a temporary location and do some magic on it!
Or write it as a custom form field. This is the basic idea how I go about verification of an MP3 file using the 'mutagen' library.
Notes:
first check the file size then if correct size write to tmp location.
Will write the file to temporary location specified in SETTINGS check its MP3 and then delete it.
The code:
from django import forms
import os
from mutagen.mp3 import MP3, HeaderNotFoundError, InvalidMPEGHeader
from django.conf import settings
class MP3FileField(forms.FileField):
def clean(self, *args, **kwargs):
super(MP3FileField, self).clean(*args, **kwargs)
tmp_file = args[0]
if tmp_file.size > 6600000:
raise forms.ValidationError("File is too large.")
file_path = getattr(settings,'FILE_UPLOAD_TEMP_DIR')+'/'+tmp_file.name
destination = open(file_path, 'wb+')
for chunk in tmp_file.chunks():
destination.write(chunk)
destination.close()
try:
audio = MP3(file_path)
if audio.info.length > 300:
os.remove(file_path)
raise forms.ValidationError("MP3 is too long.")
except (HeaderNotFoundError, InvalidMPEGHeader):
os.remove(file_path)
raise forms.ValidationError("File is not valid MP3 CBR/VBR format.")
os.remove(file_path)
return args