I am having trouble uploading excel file to my django application. It is a very simple application that should allow a user to upload an excel file with 3 columns. The application will read the contents of this file and process it into bunch of calculations
here is my forms.py:
class InputForm(forms.Form):
FileLocation = forms.FileField(label='Import Data',required=True,widget=forms.FileInput(attrs={'accept': ".xlsx"}))
settings.py:
FILE_UPLOAD_HANDLERS = ["django_excel.ExcelMemoryFileUploadHandler",
"django_excel.TemporaryExcelFileUploadHandler"]
views.py:
import xlrd
from django.shortcuts import render_to_response, render
from django.conf.urls.static import static
from django.contrib import admin
from django.contrib.staticfiles.urls import staticfiles_urlpatterns
from django.template.context_processors import csrf
from io import TextIOWrapper
from WebApp.forms import *
from django.core.mail import send_mail
from django.utils.safestring import mark_safe
from django.db import connection
import os
import csv
def analyze(request):
if request.method == 'POST':
form = InputForm(request.POST,request.FILES['FileLocation'])
if form.is_valid():
book = xlrd.open_workbook(request.FILES('FileLocation'))
for sheet in book.sheets():
number_of_rows = sheet.nrows
number_of_columns = sheet.ncols
print(number_of_rows)
I upload the file in the form and it gives me an error:
AttributeError at /app/analyze/
'ExcelInMemoryUploadedFile' object has no attribute 'get'
Request Method: POST
Request URL: http://127.0.0.1:8000/data/analyze/
Django Version: 1.11
Exception Type: AttributeError
Exception Value:
Exception Location: C:\Python36\lib\site-packages\django\forms\widgets.py in value_from_datadict, line 367
Python Executable: C:\Python36\python.exe
Python Version: 3.6.4
I am also able to upload a .csv file successfully using the following views.py code:
def analyze(request):
c={}
context = RequestContext(request)
c.update(csrf(request))
abc=['a','b','c']
if request.method == 'POST':
form = InputForm(request.POST,request.FILES)
dataType = request.POST.get("DataType")
print(dataType)
if form.is_valid():
cd = form.cleaned_data #print (cd)
a = TextIOWrapper(request.FILES['FileLocation'].file,encoding='ascii',errors='replace')
#print (request.FILES.keys())
data = csv.reader(a)
row1csv = next(data)
region = row1csv[0]
metric = row1csv[2]
I have tried django-excel with same error.
You're correctly initialising your form for the .CSV case but not in your Excel case:
form = InputForm(request.POST, request.FILES)
Don't initialise using request.FILES['FileLocation'] as that's passing the wrong type to the form. It's expecting a MultiValueDict of uploaded files, not a single uploaded file. That's why it fails when calling get on it.
Next, you can't pass an ExcelInMemoryUploadedFile to xlrd.get_workbook(). You need to save the file to disk first, then pass it's path to the get_workbook() method. The documentation of django-excel gives some easier methods:
book = request.FILES['FileLocation'].get_book() # note the square brackets!
or to directly access a sheet:
sheet = request.FILES['FileLocation'].get_sheet('sheet1')
Related
I have a view that create two csv and my goal is to zip them and add to a model.FileField
zip = zipfile.ZipFile('myzip.zip','w')
zip.writestr('file1.csv', file1.getvalue())
zip.writestr('file2.csv', file2.getvalue())
I have tried this, the zip is upload but when I download it I have the error 'the archive is damaged or unknown format'
Mymodel.objects.create(zip = File(open('myzip.zip','rb'))
This example just worked for me,
from django.core.files import File
from django.http import HttpResponse
from .models import ZipFile
import zipfile
from django.views import View
class ZipWriteView(View):
def get(self, request, *args, **kwargs):
with zipfile.ZipFile("myzip.zip", "w") as zip_obj:
zip_obj.write("polls/views.py") # add file 1
zip_obj.write("polls/admin.py") # add file 2
with open(zip_obj.filename, "rb") as f:
ZipFile.objects.create(file=File(f))
return HttpResponse("Done")
I am integrating razorpay payment gateway in django project but i am getting error while importing razorpay as :- Import razorpay could not be resolved
from django.shortcuts import render
import razorpay # Here i am getting error
from .models import coffee
This is my full code
from django.shortcuts import render
import razorpay
from .models import coffee
# Create your views here.
def index(request):
if request.method=='POST':
Name = request.POST.get("Name")
Amount = int(request.POST.get("Amount")) * 100
client = razorpay.Client(auth= ("rzp_test_YhfEhfejrkkjdkfju","t5MRPkjfijdh23845kejkej"))
payment = client.order.create({'Amount':Amount, 'currency':'INR','payment_capture':'1'})
print(payment)
Coffee = coffee(Name=Name, Amount=Amount , payment_id = payment['id'] )
return render(request,'index.html',{'payment':payment})
return render(request,'index.html')
def success(request):
if request.method == 'POST':
a = request.POST
print(a)
return render(request,"success.html")
This is my terminal
File "D:\Project 3\payment\paymentapp\urls.py", line 18, in <module>
from .import views
File "D:\Project 3\payment\paymentapp\views.py", line 3, in <module>
import razorpay
ModuleNotFoundError: No module named 'razorpay'
There are few basic thing you have to know before using razorpay gateway in you project first is your amount is considered in paisa so you have to * 100 to converte it in to rupee as I can see you are multiplying * 10 to amount next if you want to use razorpay you must use
pip install razorpay
And I will also recommend you to read the full documentation for using becuase seems that you are missing lot of thing like you have to write JavaScript code handle etc.
https://razorpay.com/docs/payment-gateway/web-integration/standard/
I am trying to read csv file in django but I don't know how to achieve it. I am fresher recently joined organization and working singly, please any one help me
from django.shortcuts import render
from django.http import HttpResponse
import csv
from csv import reader
def fun(request):
with open(r"C:\Users\Sagar\Downloads\ULB_Sp_ThreePhaseUpdate.csv") as file:
reader = csv.reader(file)
ip = request.GET["id"]
flag = False
for rec in reader:
if rec[1]=="id":
return HttpResponse("MID: ",rec[2])
return render(request, "index.html")
I am trying to show/read pdfs from server , but getting erros. Below I have attached my view.py . Please help me to solve it
views.py
from django.shortcuts import render
from django.http import HttpResponse
from .models import PDF
def pdf_view(request):
a = PDF.objects.get(id=id)
with open('a.pdf', 'rb') as pdf:
response = HttpResponse(pdf.read(), contenttype='application/pdf')
response['Content-Disposition'] = 'filename=a.pdf'
return response
pdf.closed
you can use use Django templates to show/read pdf on sever. create 'templates' folder inside your django project. inside it create a html file which contain link of you pdf.
I have developed an API (Python 3.5, Django 1.10, DRF 3.4.2) that uploads a video file to my media path when I request it from my UI. That part is working fine. I try to write a test for this feature but cannot get it to run successfully.
#views.py
import os
from rest_framework import views, parsers, response
from django.conf import settings
class FileUploadView(views.APIView):
parser_classes = (parsers.FileUploadParser,)
def put(self, request, filename):
file = request.data['file']
handle_uploaded_file(file, filename)
return response.Response(status=204)
def handle_uploaded_file(file, filename):
dir_name = settings.MEDIA_ROOT + '/scene/' + filename + '/cam1'
new_filename = 'orig.mp4'
if not os.path.exists(dir_name):
os.makedirs(dir_name)
file_path = os.path.join(dir_name, new_filename)
with open(file_path, 'wb+') as destination:
for chunk in file.chunks():
destination.write(chunk)
and
#test.py
import tempfile
import os
from django.test import TestCase
from django.conf import settings
from django.core.files import File
from django.core.files.uploadedfile import SimpleUploadedFile
from rest_framework.test import APIRequestFactory
from myapp.views import FileUploadView
class UploadVideoTestCase(TestCase):
def setUp(self):
settings.MEDIA_ROOT = tempfile.mkdtemp(suffix=None, prefix=None, dir=None)
def test_video_uploaded(self):
"""Video uploaded"""
filename = 'vid'
file = File(open('media/testfiles/vid.mp4', 'rb'))
uploaded_file = SimpleUploadedFile(filename, file.read(), 'video')
factory = APIRequestFactory()
request = factory.put('file_upload/'+filename,
{'file': uploaded_file}, format='multipart')
view = FileUploadView.as_view()
response = view(request, filename)
print(response)
dir_name = settings.MEDIA_ROOT + '/scene/' + filename + '/cam1'
new_filename = 'orig.mp4'
file_path = os.path.join(dir_name, new_filename)
self.assertTrue(os.path.exists(file_path))
In this test, I need to use an existing video file ('media/testfiles/vid.mp4') and upload it since I need to test some processings on the video data after: that's why I reset the MEDIA_ROOT using mkdtemp.
The test fails since the file is not uploaded. In the def put of my views.py, when I print request I get <rest_framework.request.Request object at 0x10f25f048> and when I print request.data I get nothing. But if I remove the FileUploadParser in my view and use request = factory.put('file_upload/' + filename, {'filename': filename}, format="multipart") in my test, I get <QueryDict: {'filename': ['vid']}> when I print request.data.
So my conclusion is that the request I generate with APIRequestFactory is incorrect. The FileUploadParseris not able to retrieve the raw file from it.
Hence my question: How to generate a file upload (test) request with Django REST Framework's APIRequestFactory?
Several people have asked questions close to this one on SO but I had no success with the proposed answers.
Any help on that matter will be much appreciated!
It's alright now! Switching from APIRequestFactory to APIClient, I managed to have my test running.
My new test.py:
import os
import tempfile
from django.conf import settings
from django.core.files import File
from django.core.files.uploadedfile import SimpleUploadedFile
from django.urls import reverse
from rest_framework.test import APITestCase, APIClient
from django.contrib.auth.models import User
class UploadVideoTestCase(APITestCase):
def setUp(self):
settings.MEDIA_ROOT = tempfile.mkdtemp()
User.objects.create_user('michel')
def test_video_uploaded(self):
"""Video uploaded"""
filename = 'vid'
file = File(open('media/testfiles/vid.mp4', 'rb'))
uploaded_file = SimpleUploadedFile(filename, file.read(),
content_type='multipart/form-data')
client = APIClient()
user = User.objects.get(username='michel')
client.force_authenticate(user=user)
url = reverse('file_upload:upload_view', kwargs={'filename': filename})
client.put(url, {'file': uploaded_file}, format='multipart')
dir_name = settings.MEDIA_ROOT + '/scene/' + filename + '/cam1'
new_filename = 'orig.mp4'
file_path = os.path.join(dir_name, new_filename)
self.assertTrue(os.path.exists(file_path))
Below, testing file upload using APIRequestFactory as requested (and ModelViewSet).
from rest_framework.test import APIRequestFactory, APITestCase
from my_project.api.views import MyViewSet
from io import BytesIO
class MyTestCase(APITestCase):
def setUp(self):
fd = BytesIO(b'Test File content') # in-memory file to upload
fd.seek(0) # not needed here, but to remember after writing to fd
reqfactory = APIRequestFactory() # initialize in setUp if used by more tests
view = MyViewSet({'post': 'create'}) # for ViewSet {action:method} needed, for View, not.
request = factory.post('/api/new_file/',
{
"title": 'test file',
"fits_file": self.fd,
},
format='multipart') # multipart is default, but for clarification that not json
response = view(request)
response.render()
self.assertEqual(response.status_code, 201)
Note that there is no authorization for clarity, as with: 'DEFAULT_PERMISSION_CLASSES': ['rest_framework.permissions.AllowAny'].