I have a django model as follows:
class ExportFile(BaseExportFile):
created_timestamp = models.DateTimeField(auto_now=True, editable=False)
data = models.FileField(upload_to='exports')
and a view function that renders a template to create a csv file:
def create_csv(request):
context = Context({'data': MyModel.objects.all()})
rendered = render_to_string('mytemplate.html', context)
# create tradefile and save
cf = ContentFile(rendered)
tf = ExportFile()
tf.data.save('myfile.csv', cf)
tf.save()
response = HttpResponse(mimetype='text/csv')
response['Content-Disposition'] = 'attachment; filename=%s' % 'myfile.csv'
response.write(rendered)
return response
The view not only saves the csv data to a FileField but it also returns it to the browser. The problem I have is the browser file works perfectly, but the file saved on the model is twice the size and when I use a diff program I can see extra hidden characters. I think it must be to do with the mime type vs django auto saving utf8 but I just can't figure it out!
Solved the problem!
The ContentFile is a subclass of cStringIO.StringIO - which deals with ASCII encoded files. The string therefore needs to be encoded as ASCII as everything in django is UTF8 by default
cf = ContentFile(rendered.encode('ascii'))
Related
Hi i'm making my own webserver using django.
i just want to upload local file to django server.
i google every method but i can't get answer.
every method using form or html but i don't want to using form and html
example : from www.localfolder/example.txt to /media/examplefolder.
i don't know how to do.. any help?
this is my code.
#csrf_exempt
def download_file(request, file):
fl_path = 'media/'
filename = str(file)
fl = open(fl_path, 'r')
mime_type, _ = mimetypes.guess_type(fl_path)
response = HttpResponse(fl, content_type=mime_type)
response['Content-Disposition'] = "attachment; filename = %s" % filename
return response
What did you search for when you googled? These were the top 2 results for Django files
https://docs.djangoproject.com/en/3.0/topics/http/file-uploads/
https://docs.djangoproject.com/en/3.0/topics/files/
Seems to have everything you are looking for.
I want to download a csv file with custom filename in a django project, but somehow the downloaded filename just display as "download.csv" instead of using the value of filename in Content-Disposition. I also tried to print csv_response['Content-Disposition'] out, but I'm getting a very strange string =?utf-8?b?YXR0YWNobWVudDsgZmlsZW5hbWU9Iuivvueoi+aKpeWQjeaDheWGtV8yMDE5MTEyODA3NDI0Ny5jc3Yi?=
the code snippet is :
#action(detail=False, methods=['GET'])
def download(self, request):
registrations = self.filter_queryset(self.get_queryset())
csv_response = HttpResponse(content_type='text/csv')
csv_response['Content-Disposition'] = 'attachment; filename="some_custom_name_{time}.csv"'.format(
time=time.strftime("%Y%m%d%H%M%S", time.localtime())
)
writer = csv.writer(csv_response)
writer.writerow([
some content,
])
for registration in registrations:
term_title = '{order} th'.format(order=registration.term.order)
course_title = registration.course.title
writer.writerow([
registration.user.email,
course_title,
term_title,
str(registration.confirmation_code),
str(registration.payment_due),
str(registration.payment_paid),
str(registration.source),
str(registration.created_at),
str(registration.updated_at),
str(registration.payment_source),
])
return csv_response
the django I am using is 2.2
any ideas why this is happening? I am a newb.
Thx in advance
The response header in chrome Dev tools:
I resolved the problem, by following the answer in the below post:
HttpResponse Django does not change file name
I guess that it is that because the string of Content-Disposition needs to be encoded, and if no, then somehow cannot operate on that, by using urlquote, it is solved.
Explanation about urlquote is here
UPDATE:
Also, a simpler way to resolve this without importing urlquote is to add encode(), like below:
csv_response['Content-Disposition'] = 'attachment; filename="some_custom_name_{time}.csv"'.format(
time=time.strftime("%Y%m%d%H%M%S", time.localtime())
).encode()
Change to this:
csv_response['Content-Disposition'] = 'attachment; filename="some_custom_name_{}.csv"'.format(
time.strftime("%Y%m%d%H%M%S", time.localtime())
)
I have a django form in which the user can upload an image that will then be displayed on a page. After the form request is submitted, I'm trying to return an httpresponse to the user of the uploaded image using the following code:
image_data = open("/path/to/my/image.png", "rb").read()
return HttpResponse(image_data, content_type="image/png")
The issue is that I can't get the absolute path from image submitted in the form request. By doing the following, I can get the name of the image, but not the local path to it:
name = ""
for filename, file in request.FILES.iteritems():
name = request.FILES[filename]
imageFileName = name
I've tried using the function file_to_string() based on an SO post, but it looks like the function is deprecated now. How can I get the absolute file path so I can pass it to the open function to return the image?
models.py
class PhotoUploader(models.Model):
title = models.CharField(max_length =120)
image = models.ImageField()
Here the solutions:
once you save the image,then you may got the path like that:
instance = PhotoUploader.objects.get(id=instance_id);
image_full_path = instance.image.path
image_data = open(image_full_path, "rb").read()
return HttpResponse(image_data, content_type="image/png")
"image_full_path" this should be your uploaded image full path.
In django 1.8 I have a couple of functions that read pdf files and return them, and that generate a pdf with reportlab and return it.
In some cases the file is served correctly, but sometimes the PDF is opened by the browser as if it were html and what is even more strange, pdf source is displayed in my django base template.
In this case, if reloading the page after the error, the pdf is served.
This is the code of a view:
fpdf = open (path, 'rb')
return HttpResponse (FileWrapper (fpdf), content_type = 'application/pdf')
and this is the code of the other:
pdf = pisa.CreatePDF (StringIO.StringIO (html.encode ("UTF-8")), result)
if not pdf.err:
response = HttpResponse (result.getvalue (), content_type = 'application / pdf')
response ['Content-Disposition'] = 'attachment; filename =% S.pdf '% (doc.name.replace ("", "_"))
return response
#Return HttpResponse (result.getvalue (), content_type = 'application/pdf')
Returning the PDF as an attachment is a test that I made to see if solved, because the desired behavior would be directly open the file.
Unfortunately, the error still occurs even so.
Change this line
response = HttpResponse (result.getvalue (), content_type = 'application / pdf')
To this line
response = HttpResponse (result.getvalue (), content_type = 'application/octet-stream')
This will make the file to be treated as a binary, and downloaded to the user instead of opening it in the browser.
If you view it inside the browser, follow Igor Pomaranskiy advice, and remove the space inside your content_type variable by doing the following
Change this
content_type = 'application / pdf'
to this
content_type = 'application/pdf'
I have a web app that needs to do the following:
Present a form to request a client side file for CSV import.
Validate the data in the CSV file or ask for another filename.
At one point, I was doing the CSV data validation in the view, after the form.is_valid() call from getting the filename (i.e. I have the imported CSV file into memory in a dictionary using csv.DictReader). After running into problems trying to pass errors back to the original form, I'm now trying to validate the CONTENTS of the CSV file in the form's clean() method.
I'm currently stumped on how to access the in memory file from clean() as the request.FILES object isn't valid. Note that I have no problems presenting the form to the client browser and then manipulating the resulting CSV file. The real issue is how to validate the contents of the CSV file - if I assume the data format is correct I can import it to my models. I'll post my forms.py file to show where I currently am after moving the code from the view to the form:
forms.py
import csv
from django import forms
from io import TextIOWrapper
class CSVImportForm(forms.Form):
filename = forms.FileField(label='Select a CSV file to import:',)
def clean(self):
cleaned_data = super(CSVImportForm, self).clean()
f = TextIOWrapper(request.FILES['filename'].file, encoding='ASCII')
result_csvlist = csv.DictReader(f)
# first line (only) contains additional information about the event
# let's validate that against its form definition
event_info = next(result_csvlist)
f_eventinfo = ResultsForm(event_info)
if not f_eventinfo.is_valid():
raise forms.ValidationError("Error validating 1st line of data (after header) in CSV")
return cleaned_data
class ResultsForm(forms.Form):
RESULT_CHOICES = (('Won', 'Won'),
('Lost', 'Lost'),
('Tie', 'Tie'),
('WonByForfeit', 'WonByForfeit'),
('LostByForfeit', 'LostByForfeit'))
Team1 = forms.CharField(min_length=10, max_length=11)
Team2 = forms.CharField(min_length=10, max_length=11)
Result = forms.ChoiceField(choices=RESULT_CHOICES)
Score = forms.CharField()
Event = forms.CharField()
Venue = forms.CharField()
Date = forms.DateField()
Div = forms.CharField()
Website = forms.URLField(required=False)
TD = forms.CharField(required=False)
I'd love input on what's the "best" method to validate the contents of an uploaded CSV file and present that information back to the client browser!
I assume that when you want to access that file is in this line inside the clean method:
f = TextIOWrapper(request.FILES['filename'].file, encoding='ASCII')
You can't use that line because request doesn't exist but you can access your form's fields so you can try this instead:
f = TextIOWrapper(self.cleaned_data.get('filename'), encoding='ASCII')
Since you have done super.clean in the first line in your method, that should work. Then, if you want to add custom error message to you form you can do it like this:
from django.forms.util import ErrorList
errors = form._errors.setdefault("filename", ErrorList())
errors.append(u"CSV file incorrect")
Hope it helps.