I am working with a legacy project and we need to implement a Django Admin that helps download a csv report that was stored as a BinaryField.
The model is something like this:
class MyModel(models.Model):
csv_report = models.BinaryField(blank=True,null=True)
Everything seems to being stored as expected but I have no clue how to decode the field back to a csv file for later use.
I am using something like these (as an admin action on MyModelAdmin class)
class MyModelAdmin(admin.ModelAdmin):
...
...
actions = ["download_file",]
def download_file(self, request,queryset):
# just getting one for testing
contents = queryset[0].csv_report
encoded_data = base64.b64encode(contents).decode()
with open("report.csv", "wb") as binary_file:
# Write bytes to file
decoded_image_data = base64.decodebytes(encoded_data)
binary_file.write(decoded_image_data)
response = HttpResponse(encoded_data)
response['Content-Disposition'] = 'attachment; filename=report.csv'
return response
download_file.short_description = "file"
But all I download is a scrambled csv file. I don't seem to understand if it is a problem of the format I am using to decode (.decode('utf-8') does nothing either )
PD:
I know it is a bad practice to use BinaryField for this. But requirements are requirements. Nothing to do about it.
EDIT:
As #TimRoberts pointed out, encoding and then decoding is REALLY silly :$. I've changed the method like so:
def download_file(self, request,queryset):
# print(self,request)
contents = queryset[0].csv_report
# print(type(contents))
encoded_data = base64.b64decode(contents)
with open("my_file.csv", "wb") as binary_file:
binary_file.write(encoded_data)
response = HttpResponse(encoded_data)
response['Content-Disposition'] = 'attachment; filename=blob.csv'
return response
download_file.short_description = "file"
Still I am getting a csv file with something like this:
A big fat case of the old RTFM: I was getting carried away by the all base64.. Obviously I didn't have any idea of what I was doing.
After tampering with the shell and reading the docs, I just changed my method to:
def download_file(self, request,queryset):
**contents = bytes(queryset[0].csv_report)**
response = HttpResponse(contents)
response['ContentDisposition']='attachment;filename=report.csv'
return response
Note that I was scrambling the data on purpose by doing the encoded_data = base64.b64decode(contents) stuff. I just needed to apply bytes on my BinaryField and voilá
Related
Currently I am exporting an excel file, however, before it gets exported, I create an xls file on the host machine. To create the excel file, I use tablib. My export view looks like this:
#login_required
def export_xls(request):
# some irrelevant code
data = convert_json_to_dataset(json_data)
table = data.export('xls')
with open('/tmp/students.xls', 'wb') as f:
f.write(table)
response = FileResponse(open('/tmp/students.xls', 'rb'), as_attachment=True, filename="test.xls")
return response
What I am trying to achieve is to avoid writing always to /tmp/students.xls. I tried using BytesIO, however that did not work out.
#login_required
def export_xls(request):
# some irrelevant code
data = convert_json_to_dataset(json_data)
table = data.export('xls')
buffer = BytesIO()
buffer.write(table)
response = FileResponse(buffer.read(), as_attachment=True, filename="test.xls")
return response
Currently I am always overwriting the file, however, I will change the naming of the file and that will cause multiple files to be created, which I would like to avoid.
I want to render plot (matplotlib) in one place of code and then in ViewSet serve to it user without saving on disk. I tried to use io library to keep file in memory, but it seems that always something is wrong.
My code where I save plot on disk:
def some_func(self):
...generating plot...
filename = self.generate_path() # generate random name for file
plt.savefig(filename, transparent=True)
return filename
Code of ViewSet:
class SomeViewsSet(ViewSet):
def create(self, request):
... some code ...
path = some_func()
name = path.split('/')[-1]
with open(path, 'rb') as file:
response = HttpResponse(file, content_type=guess_type(path)[0])
response['Content-Length'] = len(response.content)
response['Content-Disposition'] = f'attachment; filename={name}'
return response
Make sure that you pass matplotlib a BytesIO object, and not a StringIO. Then get the bytes using getvalue(), and pass them to HttpResponse. If that's what you've already tried, please post your code and the error message you're seeing.
I'm using Python 2.7 and Django 1.7.
I have a method in my admin interface that generates some kind of a csv file.
def generate_csv(args):
...
#some code that generates a dictionary to be written as csv
....
# this creates a directory and returns its filepath
dirname = create_csv_dir('stock')
csvpath = os.path.join(dirname, 'mycsv_file.csv')
fieldnames = [#some field names]
# this function creates the csv file in the directory shown by the csvpath
newcsv(data, csvheader, csvpath, fieldnames)
# this automatically starts a download from that directory
return HttpResponseRedirect('/media/csv/stock/%s' % csvfile)
All in all I create a csv file, save it somewhere on the disk, and then pass its URL to the user for download.
I was thinking if all this can be done without writing to disc. I googled around a bit and maybe content disposition attachment might help me, but I got lost in documentation a bit.
Anyway if there's an easier way of doing this I'd love to know.
Thanks to #Ragora, you pointed me towards the right direction.
I rewrote the newcsv method:
from io import StringIO
import csv
def newcsv(data, csvheader, fieldnames):
"""
Create a new csv file that represents generated data.
"""
new_csvfile = StringIO.StringIO()
wr = csv.writer(new_csvfile, quoting=csv.QUOTE_ALL)
wr.writerow(csvheader)
wr = csv.DictWriter(new_csvfile, fieldnames = fieldnames)
for key in data.keys():
wr.writerow(data[key])
return new_csvfile
and in the admin:
csvfile = newcsv(data, csvheader, fieldnames)
response = HttpResponse(csvfile.getvalue(), content_type='text/csv')
response['Content-Disposition'] = 'attachment; filename=stock.csv'
return response
If it annoys you that you are saving a file to disk, just add the application/octet-stream content-type to the Content-Disposition header then delete the file from disk.
If this header (Content-Disposition) is used in a response with the application/octet- stream content-type, the implied suggestion is that the user agent should not display the response, but directly enter a `save response as...' dialog.
I'm trying to let the user download an xml file that i have generated.
This is my code:
tree.write('output.xml', encoding="utf-16")
# Pathout is the path to the output.xml
xmlFile = open(pathout, 'r')
myfile = FileWrapper(xmlFile.read())
response = HttpResponse(myfile, content_type='application/xml')
response['Content-Disposition'] = 'attachment; filename='+filename
return response
When I try to create my response I get this exception:
'\\'str\\' object has no attribute \\'read\\''
Can't figure out what I'm doing wrong. Any ideas?
Edit:
When I use this code I get no errors but instead the downloaded file is empty
tree.write('output.xml', encoding="utf-16")
xmlFile = open(pathout, 'r')
myfile = FileWrapper(xmlFile)
response = HttpResponse(myfile, content_type='application/xml')
response['Content-Disposition'] = 'attachment; filename='+filename
return response
You are calling xmlFile.read() - which yields a string - and passing the result to FileWrapper() which expects a readable file-like object. You should either just pass xmlFile to FileWrapper, or not use FileWrapper at all and pass the result of xmlFile.read() as your HttpResponse body.
Note that if you are creating the xml dynamically (which seems to be the case according to your snippet's first line), writing it to disk only to read it back a couple lines later is both a waste of time and resources and a potential cause of race conditions. You perhaps want to have a look at https://docs.python.org/2/library/xml.etree.elementtree.html#xml.etree.ElementTree.tostring
You're reading the file and passing the resulting string to FileWrapper, instead of passing the actual file object.
myfile = FileWrapper(xmlFile)
Alternatively from the other answers, I would recommend to completely go around the problem by utilizing the Django template system:
from django.http import HttpResponse
from django.template import Context, loader
def my_view(request):
# View code here...
t = loader.get_template('myapp/myfile.xml')
c = Context({'foo': 'bar'})
response = HttpResponse(t.render(c), content_type="application/xml")
response['Content-Disposition'] = 'attachment; filename=...'
return response
In this way create a myfile.xml template which is used to render the proper xml response without having to deal with writing any files to the filesystem. This is cleaner and faster, given that there is no other need to indeed create the xml and store it permanently.
I'm trying create and serve excel files using Django. I have a jar file which gets parameters and produces an excel file according to parameters and it works with no problem. But when i'm trying to get the produced file and serve it to the user for download the file comes out broken. It has 0kb size. This is the code piece I'm using for excel generation and serving.
def generateExcel(request,id):
if os.path.exists('./%s_Report.xlsx' % id):
excel = open("%s_Report.xlsx" % id, "r")
output = StringIO.StringIO(excel.read())
out_content = output.getvalue()
output.close()
response = HttpResponse(out_content,content_type='application/vnd.openxmlformats-officedocument.spreadsheetml.sheet')
response['Content-Disposition'] = 'attachment; filename=%s_Report.xlsx' % id
return response
else:
args = ['ServerExcel.jar', id]
result = jarWrapper(*args) # this creates the excel file with no problem
if result:
excel = open("%s_Report.xlsx" % id, "r")
output = StringIO.StringIO(excel.read())
out_content = output.getvalue()
output.close()
response = HttpResponse(out_content,content_type='application/vnd.openxmlformats-officedocument.spreadsheetml.sheet')
response['Content-Disposition'] = 'attachment; filename=%s_Report.xlsx' % id
return response
else:
return HttpResponse(json.dumps({"no":"excel","no one": "cries"}))
I have searched for possible solutions and tried to use File Wrapper also but the result did not changed. I assume i have problem with reading the xlsx file into StringIO object. But dont have any idea about how to fix it
Why on earth are you passing your file's content to a StringIO just to assign StringIO.get_value() to a local variable ? What's wrong with assigning file.read() to your variable directly ?
def generateExcel(request,id):
path = './%s_Report.xlsx' % id # this should live elsewhere, definitely
if os.path.exists(path):
with open(path, "r") as excel:
data = excel.read()
response = HttpResponse(data,content_type='application/vnd.openxmlformats-officedocument.spreadsheetml.sheet')
response['Content-Disposition'] = 'attachment; filename=%s_Report.xlsx' % id
return response
else:
# quite some duplication to fix down there
Now you may want to check weither you actually had any content in your file - the fact that the file exists doesn't mean it has anything in it. Remember that you're in a concurrent context, you can have one thread or process trying to read the file while another (=>another request) is trying to write it.
In addition to what Bruno says, you probably need to open the file in binary mode:
excel = open("%s_Report.xlsx" % id, "rb")
You can use this library to create excel sheets on the fly.
http://xlsxwriter.readthedocs.io/
For more information see this page. Thanks to #alexcxe
XlsxWriter object save as http response to create download in Django
my answer is:
def generateExcel(request,id):
if os.path.exists('./%s_Report.xlsx' % id):
with open('./%s_Report.xlsx' % id, "rb") as file:
response = HttpResponse(file.read(),content_type='application/vnd.openxmlformats-officedocument.spreadsheetml.sheet')
response['Content-Disposition'] = 'attachment; filename=%s_Report.xlsx' % id
return response
else:
# quite some duplication to fix down there
why using "rb"? because HttpResponse class init parameters is (self, content=b'', *args, **kwargs), so we should using "rb" and using .read() to get the bytes.