I found this handy code in the Django Docs:
from reportlab.pdfgen import canvas
from django.http import HttpResponse
def some_view(request):
# Create the HttpResponse object with the appropriate PDF headers.
response = HttpResponse(content_type='application/pdf')
response['Content-Disposition'] = 'attachment; filename="somefilename.pdf"'
# Create the PDF object, using the response object as its "file."
p = canvas.Canvas(response)
# Draw things on the PDF. Here's where the PDF generation happens.
# See the ReportLab documentation for the full list of functionality.
p.drawString(100, 100, "Hello world.")
# Close the PDF object cleanly, and we're done.
p.showPage()
p.save()
return response
I would like to pass a variable to the function as variable name, but cannot figure it out (though I am sure it is super easy).
Anyone have an idea?
Thanks so much!
filename="somefilename.pdf" there is place where you can determine your filename.
You can use it as:
filename = 'somefilename.pdf'
response['Content-Disposition'] = 'attachment; filename="{}"'.format(filename)
Hope it might helps you:
Add one more argument named "name" to the view "some_view" and replace the filename with the argument value.
def hello_pdf(request, name):
# Create the HttpResponse object with the appropriate PDF headers.
response = HttpResponse(mimetype='application/pdf')
response['Content-Disposition'] = 'attachment; filename=%s.pdf'% name
temp = StringIO()
# Create the PDF object, using the StringIO object as its "file."
p = canvas.Canvas(temp)
# Draw things on the PDF. Here's where the PDF generation happens.
# See the ReportLab documentation for the full list of functionality.
p.drawString(100, 100, "Hello world.")
# Close the PDF object cleanly.
p.showPage()
p.save()
# Get the value of the StringIO buffer and write it to the response.
response.write(temp.getvalue())
return response
Check the name argument replacing the filename in the 4th line of code. Also don't forget to update the urls.py like below:
url(r'^(?P<name>\w+)/$', 'app.some_view')
After all pass the filename you wish to replace with url like below:
http://yourwebsite.com/name-of-file/
If you want to force viewing the PDF in browser with given filename but without automatic download, you can use "inline" instead of "attachment":
response["Content-Disposition"] = f'inline; filename="{file_name}"'
Related
I use Django 2.1.4 and use Django sample for create pdf report
def some_view(request):
# Create a file-like buffer to receive PDF data.
buffer = io.BytesIO()
# Create the PDF object, using the buffer as its "file."
p = canvas.Canvas(buffer)
# Draw things on the PDF. Here's where the PDF generation happens.
# See the ReportLab documentation for the full list of functionality.
p.drawString(100, 100, "Hello world.")
# Close the PDF object cleanly, and we're done.
p.showPage()
p.save()
# FileResponse sets the Content-Disposition header so that browsers
# present the option to save the file.
return FileResponse(buffer, as_attachment=True, filename='hello.pdf')
It started dowloaded but never finished. What is wrong?
Old version works properly
def some_view(request):
# Create the HttpResponse object with the appropriate PDF headers.
response = HttpResponse(content_type='application/pdf')
response['Content-Disposition'] = 'attachment; filename="somefilename.pdf"'
# Create the PDF object, using the response object as its "file."
p = canvas.Canvas(response)
# Draw things on the PDF. Here's where the PDF generation happens.
# See the ReportLab documentation for the full list of functionality.
p.drawString(100, 100, "Hello world.")
# Close the PDF object cleanly, and we're done.
p.showPage()
p.save()
return response
I would look at this example (except for the queryset),, see if that works, similar worked for me earlier today, How can i display my data in database and export it to pdf -Django
Get the value of the BytesIO buffer and write it to the response.
response = HttpResponse(content_type='application/pdf')
response['Content-Disposition'] = 'attachment; filename="WishList.pdf"'
from reportlab.pdfgen import canvas
buffer = BytesIO()
p = canvas.Canvas(buffer)
p.drawString(100, 100, "Hello world.")
p.showPage()
p.save()
pdf = buffer.getvalue()
buffer.close()
response.write(pdf)
return response
Believe it or not, the old version is the way I recommend that you do it. To get this working with a FileResponse, try the below:
# FileResponse sets the Content-Disposition header so that browsers
# present the option to save the file.
response = FileResponse(buffer, as_attachment=True, filename='hello.pdf', content_type='application/pdf')
# set the content length to let the browser know how many bytes to expect
response['Content-Length'] = bugger.getbuffer().nbytes
return response
my x variable is getting all the data in my database, i guess? someone help me how can i display all data and export it to pdf file.
response = HttpResponse(content_type='application/pdf')
response['Content-Disposition'] = 'attachment; filename="WishList.pdf"'
buffer = BytesIO()
# Create the PDF object, using the BytesIO object as its "file."
p = canvas.Canvas(buffer)
x = Item.objects.all()
p.drawString(100, 100,x)
p.drawString(200,300,"sad")
# Close the PDF object cleanly.
p.showPage()
p.save()
# Get the value of the BytesIO buffer and write it to the response.
pdf = buffer.getvalue()
buffer.close()
response.write(pdf)
return response
Your variable x won't return anything you can print into the PDF, because its a queryset and not a couple strings attached to each other. I just started working with Django and getting the values works something like this:
x = Item.objects.all[0].name
This code snippet will assign the vale of the row name of the first entry in your Item table to the variable x.
For more Information I can only recommend reading the Tutorial about writing queries on the django website.
I'm trying to fill pre-made pdf form with database data and flatten it. For example, if the user inputs a field called "name", it should be placed in the name field of the pdf form.
You can use reportlab library in combination with Django views :
from io import BytesIO
from reportlab.pdfgen import canvas
from django.http import HttpResponse
def some_view(request):
# Create the HttpResponse object with the appropriate PDF headers.
response = HttpResponse(content_type='application/pdf')
response['Content-Disposition'] = 'attachment; filename="somefilename.pdf"'
buffer = BytesIO()
# Create the PDF object, using the BytesIO object as its "file."
p = canvas.Canvas(buffer)
# Draw things on the PDF. Here's where the PDF generation happens.
# See the ReportLab documentation for the full list of functionality.
p.drawString(100, 100, "Hello world.")
# Close the PDF object cleanly.
p.showPage()
p.save()
# Get the value of the BytesIO buffer and write it to the response.
pdf = buffer.getvalue()
buffer.close()
response.write(pdf)
return response
Django documentation : https://docs.djangoproject.com/en/1.9/howto/outputting-pdf/
Reportlab documentation : http://www.reportlab.com/documentation/
I am displaying the list of objects in the html table.
i have the download link in front of every row which i want them to download the linked file.
I have made this function
def make_downloadable_link(path):
#Prepare the form for downloading
wrapper = FileWrapper(open(mypath))
response = HttpResponse(wrapper,'application/pdf')
response['Content-Length'] = os.path.getsize(mypath)
fname = mypath.split('/')[-1]
response['Content-Disposition'] = 'attachment; filename= fname'
return response
This is working fine if i use it for hard coded path in view for single file. But i want to make a generic view so that it works on all the files in the table
I hav the path of the file avaiable in object.path variable but i am confused how can i pass the path object to the downlaod file view. because i want to hide that actual path from the user.
I don't know what to write in the URLs.py file fo that download file view
What would you like to do is get actual file path from object. And as you have said the file path is stored in object.path that makes it easy.
For example:
urls.py
url(r'^download/(?P<object_id>\d+)/$', "yourapp.views.make_downloadable_link", name="downloadable")
In views.py:
def make_downloadable_link(object_id):
# get object from object_id
object = ObjectModel.objects.get(id=object_id)
mypath = object.path
#prepare to serve the file
wrapper = FileWrapper(open(mypath))
response = HttpResponse(wrapper,'application/pdf')
response['Content-Length'] = os.path.getsize(mypath)
fname = mypath.split('/')[-1]
response['Content-Disposition'] = 'attachment; filename= fname'
return response
I have a such problem - I am using Python 2.6 / Django 1.3 and I need to accept as POST variable with key 'f', which contains a binary data. After that, I need to save data in a file.
POST
T$topX$objectsX$versionY$archiverО©ҐR$0О©ҐО©ҐО©Ґull_=<---------------------- content of file -------------------->О©ҐО©Ґ_NSKeyedArchive(258:=CО©ҐО©Ґ
Code
from django.core.files.storage import default_storage
from django.core.files.base import ContentFile
def save(request):
upload_file = request.POST['f']
save_path = default_storage.save('%s%s' % (save_dir, filename),
ContentFile(upload_file))
When I am trying to do
nano /tmp/myfile.zip
It returns data like
T^#^#^#$^#^#^#t^#^#^#o^#^#^#p^#^#^#X^#^#^#$^#^#^#o^#^#^#b^#^#^#j^#^#^#e^#^#^#c^#^#^#t^#^#^#s^#^#^#X^#^#^#$^#^#^#v^#^#^#e^#^#^#r^#^#^#s^#^#^#i^#^#$
When its done, I am going to read saved file
def read(request):
user_file = default_storage.open(file_path).read()
file_name = get_filename(file_path)
response = HttpResponse(user_file, content_type = 'text/plain',
mimetype = 'application/force-download')
response['Content-Disposition'] = 'attachment; filename=%s' % file_name
response['Content-Length'] = default_storage.size(file_path)
return response
In case, when I am writing
print user_file
It returns a correct data, but when I am returning a HttpResponse it has a different data from a source
It would probably be easier, and more memory efficient if you just save the data into a file, and like #keckse said, let a browser stream it. Django is very inefficient in streaming data. It will all depend on the size of the data. If you want to stream it with django anyways, it can be done like this:
from django.http import HttpResponse
import os.path
import mimetypes
def stream(request, document, type=None):
doc = Document.objects.get(pk=document)
fsock = open(doc.file.path,"r")
file_name = os.path.basename(doc.file.path)
mime_type_guess = mimetypes.guess_type(file_name)
if mime_type_guess is not None:
response = HttpResponse(fsock, mimetype=mime_type_guess[0])
response['Content-Disposition'] = 'attachment; filename=' + file_name
return response
In your case you might want to set the mime type manually, you can try out application/octet-stream too. The mainpassing iterators difference is that you pass the "string" from file.read(), instead of the handle to the file directly. Please note: if you use read(), you will be loading the whole file into memory.
More on passing iterators to HttpResonse. And I might be wrong, but I think you can drop the content-type.