I am rolling my own fileupload handler in django and would like to know the file name. I am supporting more than one file format and want to do different processing in the receive_data_chunk method depending on which file format the uploaded file has. I thought I would be pragmatic and just judge file format based on file ending but I can't figure out how to get hold of the file name. If I try to extract the file name with something like the following code (before that method is called):
if request.method == 'POST':
p = re.compile('^.*\.sdf$', re.IGNORECASE)
if ( p.search(request.FILES['filecontent'].name) ) :
self.sdf = True
else:
self.sdf = False
It seems I never reach the receive_data_chunk method. I presume the call to request.FILES trigger the loading somehow and then it's already done? How can I do different processing based on file ending in my receive_data_chunk method?
Have you tried using
data=request.POST.copy()
and then working on the copy? I have used this for other things but may work in this case as well.
Related
I'm writing a Django function that takes some user input, and generates a pdf for the user. However, the process for generating the pdf is quite intensive, and I'll get a lot of repeated requests so I'd like to store the generated pdfs on the server and check if they already exist before generating them.
The problem is that django-wkhtmltopdf (which I'm using for generation) is meant to return to the user directly, and I'm not sure how to store it on the file.
I have the following, which works for returning a pdf at /pdf:
urls.py
urlpatterns = [
url(r'^pdf$', views.createPDF.as_view(template_name='site/pdftemplate.html', filename='my_pdf.pdf'))
]
views.py
class createPDF(PDFTemplateView):
filename = 'my_pdf.pdf'
template_name = 'site/pdftemplate.html'
So that works fine to create a pdf. What I'd like is to call that view from another view and save the result. Here's what I've got so far:
#Create pdf
pdf = createPDF.as_view(template_name='site/pdftemplate.html', filename='my_pdf.pdf')
pdf = pdf(request).render()
pdfPath = os.path.join(settings.TEMP_DIR,'temp.pdf')
with open(pdfPath, 'w') as f:
f.write(pdf.content)
This creates temp.pdf and is about the size I'd expect but the file isn't valid (it renders as a single completely blank page).
Any suggestions?
Elaborating on the previous answer given: to generate a pdf file and save to disk do this anywhere in your view:
...
context = {...} # build your context
# generate response
response = PDFTemplateResponse(
request=self.request,
template=self.template_name,
filename='file.pdf',
context=context,
cmd_options={'load-error-handling': 'ignore'})
# write the rendered content to a file
with open("file.pdf", "wb") as f:
f.write(response.rendered_content)
...
I have used this code in a TemplateView class so request and template fields were set like that, you may have to set it to whatever is appropriate in your particular case.
Well, you need to take a look to the code of wkhtmltopdf, first you need to use the class PDFTemplateResponse in wkhtmltopdf.views to get access to the rendered_content property, this property get us access to the pdf file:
response = PDFTemplateResponse(
request=<your_view_request>,
template=<your_template_to_render>,
filename=<pdf_filename.pdf>,
context=<a_dcitionary_to_render>,
cmd_options={'load-error-handling': 'ignore'})
Now you could use the rendered_content property to get access to the pdf file:
mail.attach('pdf_filename.pdf', response.rendered_content, 'application/pdf')
In my case I'm using this pdf to attach to an email, you could store it.
I have an API endpoint with Django Rest Framework to upload an image.
class MyImageSerializer(serializers.ModelSerializer):
image = serializers.ImageField(source='image')
I can upload images but they are saved with the filename that is sent from the client which can result to collisions. I would like instead to upload the file to my CDN with a timestamp filename.
Generating the filename is not the problem, just saving the image with it.
Any one knows how to do that?
Thanks.
If your image is of type ImageField from django, then you don't really have to do anything, not even declare it in your serializer like you did. It's enough to add it in the fields attribute and django will handle collisions. This means django will add _index on each new file which might generate a collision, so if you upload a file named 'my_pic.jpg' 5 times, you will actually have files 'my_pic.jpg', 'my_pic_1.jpg', 'my_pic_2.jpg', 'my_pic_3.jpg', 'my_pic_4.jpg' on your server.
Now, this is done using django's implementation for FileSystemStorage (see here), but if you want it to append a timestamp to your filename, all you have to do is write a storage class where you overwrite the get_available_name(name) method. Example:
class MyFileSystemStorage(FileSystemStorage):
def get_available_name(self, name):
''' name is the current file name '''
now = time.time()
stamp = datetime.datetime.fromtimestamp(now).strftime('%Y-%m-%d-%H-%M-%S')
return '{0}_{1}'.format(name, str(stamp))
And the image field in your model:
image = models.ImageField(upload_to='your upload dir', storage= MyFileSystemStorage)
Important update
As of August 20, 2014 this is no longer an issue, since Django found a vulnerability related to this behaviour (thanks #mlissner for pointing it out) . Important excerpt :
We’ve remedied the issue by changing the algorithm for generating file
names if a file with the uploaded name already exists.
Storage.get_available_name() now appends an underscore plus a random 7
character alphanumeric string (e.g. "_x3a1gho"), rather than iterating
through an underscore followed by a number (e.g. "_1", "_2", etc.).
I would like to read a file, update the website, read more lines, update the site, etc ...The logic is below but it's not working.
It only shows the first line from the logfile and stops. Is there a way to iterate over 'return render_to_response'?
#django view calling a remote python script that appends output to the logfile
proc = subprocess.Popen([program, branch, service, version, nodelist])
logfile = 'text.log'
fh = open(logfile, 'r')
while proc.poll() == None:
where = fh.tell()
line = fh.read()
if not line:
time.sleep(1)
fh.seek(where,os.SEEK_SET)
else:
output = cgi.escape(line)
output = line.replace('\n\r', '<br>')
return render_to_response('hostinfo/deployservices.html', {'response': output})
Thank you for your help.
You can actually do this, by making your function a generator - that is, using 'yield' to return each line.
However, you would need to create the response directly, rather than using render to response.
render_to_response will render the first batch to the website and stop. Then the website must call this view again somehow if you want to send the next batch. You will also have to maintain a record of where you were in the log file so that the second batch can be read from that point.
I assume that you have some logic in the templates so that the second post to render_to_response doesnt overwrite the first
If your data is not humongous, you should explore sending over the entire contents you want to show on the webpage each time you read some new lines.
Instead of re-inventing the wheel, use django_logtail
I wrote a cmd line routine to import a kml file into a geoDjango application, which works fine when you feed it a locally saved KML file path (using the datasource object).
Now I am writing a web file upload dialog, to achieve the same thing. This is the beginning of the code that I have, problem is, that the GDAL DataSource object does not seem to understand Djangos UploadedFile format. It is held in memory and not a file path as expected.
What would be the best strategy to convert the UploadedFile to a normal file, and access this through a path? I dont want to keep the file after processing.
def createFeatureSet(request):
if request.method == 'POST':
inMemoryFile = request.FILES['myfile']
name = inMemoryFile.name
POSTGIS_SRID = 900913
ds = DataSource(inMemoryFile) #This line doesnt work!!!
for layer in ds:
if layer.geom_type in (OGRGeomType('Point'), OGRGeomType('Point25D'), OGRGeomType('MultiPoint'), OGRGeomType('MultiPoint25D')):
layerGeomType = OGRGeomType('MultiPoint').django
elif layer.geom_type in (OGRGeomType('LineString'),OGRGeomType('LineString25D'), OGRGeomType('MultiLineString'), OGRGeomType('MultiLineString25D')):
layerGeomType = OGRGeomType('MultiLineString').django
elif layer.geom_type in (OGRGeomType('Polygon'), OGRGeomType('Polygon25D'), OGRGeomType('MultiPolygon'), OGRGeomType('MultiPolygon25D')):
layerGeomType = OGRGeomType('MultiPolygon').django
DataSource is a wrapper around GDAL's C API and needs an actual file. You'll need to write your upload somewhere on the disk, for insance using a tempfile. Then you can pass the file to DataSource.
Here is a suggested solution using a tempfile. I put the processing code in its own function which is now called.
f = request.FILES['myfile']
temp = tempfile.NamedTemporaryFile(delete=False)
temp.write(f.read())
temp.close()
createFeatureSet(temp.name, source_SRID= 900913)
I'd like to store uploaded files into a specific directory that depends on the URI of the POST request. Perhaps, I'd also like to rename the file to something fixed (the name of the file input for example) so I have an easy way to grep the file system, etc. and also to avoid possible security problems.
What's the preferred way to do this in Django?
Edit: I should clarify that I'd be interested in possibly doing this as a file upload handler to avoid writing a large file twice to the file system.
Edit2: I suppose one can just 'mv' the tmp file to a new location. That's a cheap operation if on the same file system.
Fixed olooney example. It is working now
#csrf_exempt
def upload_video_file(request):
folder = 'tmp_dir2/' #request.path.replace("/", "_")
uploaded_filename = request.FILES['file'].name
BASE_PATH = '/home/'
# create the folder if it doesn't exist.
try:
os.mkdir(os.path.join(BASE_PATH, folder))
except:
pass
# save the uploaded file inside that folder.
full_filename = os.path.join(BASE_PATH, folder, uploaded_filename)
fout = open(full_filename, 'wb+')
file_content = ContentFile( request.FILES['file'].read() )
try:
# Iterate through the chunks.
for chunk in file_content.chunks():
fout.write(chunk)
fout.close()
html = "<html><body>SAVED</body></html>"
return HttpResponse(html)
except:
html = "<html><body>NOT SAVED</body></html>"
return HttpResponse(html)
Django gives you total control over where (and if) you save files. See: http://docs.djangoproject.com/en/dev/topics/http/file-uploads/
The below example shows how to combine the URL and the name of the uploaded file and write the file out to disk:
def upload(request):
folder = request.path.replace("/", "_")
uploaded_filename = request.FILES['file'].name
# create the folder if it doesn't exist.
try:
os.mkdir(os.path.join(BASE_PATH, folder))
except:
pass
# save the uploaded file inside that folder.
full_filename = os.path.join(BASE_PATH, folder, uploaded_filename)
fout = open(full_filename, 'wb+')
# Iterate through the chunks.
for chunk in fout.chunks():
fout.write(chunk)
fout.close()
Edit: How to do this with a FileUploadHandler? It traced down through the code and it seems like you need to do four things to repurpose the TemporaryFileUploadHandler to save outside of FILE_UPLOAD_TEMP_DIR:
extend TemporaryUploadedFile and override init() to pass through a different directory to NamedTemporaryFile. It can use the try mkdir except for pass I showed above.
extend TemporaryFileUploadHandler and override new_file() to use the above class.
also extend init() to accept the directory where you want the folder to go.
Dynamically add the request handler, passing through a directory determined from the URL:
request.upload_handlers = [ProgressBarUploadHandler(request.path.replace('/', '_')]
While non-trivial, it's still easier than writing a handler from scratch: In particular, you won't have to write a single line of error-prone buffered reading. Steps 3 and 4 are necessary because FileUploadHandlers are not passed request information by default, I believe, so you'll have to tell it separately if you want to use the URL somehow.
I can't really recommend writing a custom FileUploadHandler for this. It's really mixing layers of responsibility. Relative to the speed of uploading a file over the internet, doing a local file copy is insignificant. And if the file's small, Django will just keep it in memory without writing it out to a temp file. I have a bad feeling that you'll get all this working and find you can't even measure the performance difference.