I found some python docs relating to docxtpl at this link:
https://docxtpl.readthedocs.io/en/latest/
I followed the instruction and entered the code found at this site into a view and created the associated URL. When I go to the URL I would like for a doc to be generated - but I get an error that no HTTP response is being returned. I understand I am not defining one, but I am a bit confused about what HTTP response I need to define (I am still very new to this). The MS word template that I have saved is titled 'template.docx'.
Any help would be greatly appreciated!
VIEWS.PY
def doc_test(request):
doc = DocxTemplate("template.docx")
context = { 'ultimate_consignee' : "World company" }
doc.render(context)
doc.save("generated_doc.docx")
I would like accessing this view to generate the doc, where the variables are filled with what is defined in the context above.
Gist: Read the contents of the file and return the data in an HTTP response.
First of all, you'll have to save the file in memory so that it's easier to read. Instead of saving to a file name like doc.save("generated_doc.docx"), you'll need to save it to a file-like object.
Then read the contents of this file-like object and return it in an HTTP response.
import io
from django.http import HttpResponse
def doc_test(request):
doc = DocxTemplate("template.docx")
# ... your other code ...
doc_io = io.BytesIO() # create a file-like object
doc.save(doc_io) # save data to file-like object
doc_io.seek(0) # go to the beginning of the file-like object
response = HttpResponse(doc_io.read())
# Content-Disposition header makes a file downloadable
response["Content-Disposition"] = "attachment; filename=generated_doc.docx"
# Set the appropriate Content-Type for docx file
response["Content-Type"] = "application/vnd.openxmlformats-officedocument.wordprocessingml.document"
return response
Note: This code may or may not work because I haven't tested it. But the general principle remains the same i.e. read the contents of the file and return it in an HTTP response with appropriate headers.
So if this code doesn't work, maybe because the package you're using doesn't support writing to file-like objects or for some other reason, then it would be a good idea to ask the creator of the package or file an issue on their Github about how to read the contents of the file.
Here is a more concise solution:
import os
from io import BytesIO
from django.http import FileResponse
from docxtpl import DocxTemplate
def downloadWord(request, pk):
context = {'first_name' : 'xxx', 'sur_name': 'yyy'}
byte_io = BytesIO()
tpl = DocxTemplate(os.path.join(BASE_PATH, 'template.docx'))
tpl.render(context)
tpl.save(byte_io)
byte_io.seek(0)
return FileResponse(byte_io, as_attachment=True, filename=f'generated_{pk}.docx')
I am following the instruction from this page. I am building a slack slash command handling server and I can't rebuild the signature to validate slash request authenticity.
here is the code snippet from my django application (the view uses the django rest-framework APIView):
#property
def x_slack_req_ts(self):
if self.xsrts is not None:
return self.xsrts
self.xsrts = str(self.request.META['HTTP_X_SLACK_REQUEST_TIMESTAMP'])
return self.xsrts
#property
def x_slack_signature(self):
if self.xss is not None:
return self.xss
self.xss = self.request.META['HTTP_X_SLACK_SIGNATURE']
return self.xss
#property
def base_message(self):
if self.bs is not None:
return self.bs
self.bs = ':'.join(["v0", self.x_slack_req_ts, self.raw.decode('utf-8')])
return self.bs
#property
def encoded_secret(self):
return self.app.signing_secret.encode('utf-8')
#property
def signed(self):
if self.non_base is not None:
return self.non_base
hashed = hmac.new(self.encoded_secret, self.base_message.encode('utf-8'), hashlib.sha256)
self.non_base = "v0=" + hashed.hexdigest()
return self.non_base
This is within a class where self.raw = request.body the django request and self.app.signing_secret is a string with the appropriate slack secret string. It doesn't work as the self.non_base yield an innaccurate value.
Now if I open an interactive python repl and do the following:
>>> import hmac
>>> import hashlib
>>> secret = "8f742231b10e8888abcd99yyyzzz85a5"
>>> ts = "1531420618"
>>> msg = "token=xyzz0WbapA4vBCDEFasx0q6G&team_id=T1DC2JH3J&team_domain=testteamnow&channel_id=G8PSS9T3V&channel_name=foobar&user_id=U2CERLKJA&user_name=roadrunner&command=%2Fwebhook-collect&text=&response_url=https%3A%2F%2Fhooks.slack.com%2Fcommands%2FT1DC2JH3J%2F397700885554%2F96rGlfmibIGlgcZRskXaIFfN&trigger_id=398738663015.47445629121.803a0bc887a14d10d2c447fce8b6703c"
>>> ref_signature = "v0=a2114d57b48eac39b9ad189dd8316235a7b4a8d21a10bd27519666489c69b503"
>>> base = ":".join(["v0", ts, msg])
>>> hashed = hmac.new(secret.encode(), base.encode(), hashlib.sha256)
>>> hashed.hexdigest()
>>> 'a2114d57b48eac39b9ad189dd8316235a7b4a8d21a10bd27519666489c69b503'
You will recognise the referenced link example. If I use the values from my django app with one of MY examples, it works within the repl but doesn't within the django app.
MY QUESTION: I believe this is caused by the self.raw.decode() encoding not being consistent with the printout I extracted to copy/paste in the repl. Has anyone encountered that issue and what is the fix? I tried a few random things with the urllib.parse library... How can I make sure that the request.body encoding is consistent with the example from flask with get_data() (as suggested by the doc in the link)?
UPDATE: I defined a custom parser:
class SlashParser(BaseParser):
"""
Parser for form data.
"""
media_type = 'application/x-www-form-urlencoded'
def parse(self, stream, media_type=None, parser_context=None):
"""
Parses the incoming bytestream as a URL encoded form,
and returns the resulting QueryDict.
"""
parser_context = parser_context or {}
request = parser_context.get('request')
raw_data = stream.read()
data = QueryDict(raw_data, encoding='utf-8')
setattr(data, 'raw_body', raw_data) # setting a 'body' alike custom attr with raw POST content
return data
To test based on this question and the raw_body in the custom parser generates the exact same hashed signature as the normal "body" but again, copy pasting in the repl to test outside the DRF works. Pretty sure it's an encoding problem but completely at loss...
I found the problem which is very frustrating.
It turns out that the signing secret was stored in too short a str array and were missing trailing characters which obviously, resulted in bad hashing of the message.
I'm trying to collect the weather data for year 2000 from this site.
My code for the spider is:
import scrapy
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors import LinkExtractor
class weather(CrawlSpider):
name = 'data'
start_urls = [
'https://www1.ncdc.noaa.gov/pub/data/uscrn/products/daily01/'
]
custom_settings = {
'DEPTH_LIMIT': '2',
}
rules = (
Rule(LinkExtractor(restrict_xpaths=
('//table/tr[2]/td[1]/a',)),callback='parse_item',follow=True),
)
def parse_item(self, response):
for b in response.xpath('//table')
yield scrapy.request('/tr[4]/td[2]/a/#href').extract()
yield scrapy.request('/tr[5]/td[2]/a/#href').extract()
The paths with 'yield' are the links to two text file and i want to scrape the data from these text files and store it separately in two different files but I don't know how to continue.
I don't usually use CrawlSpider so I'm unfamiliar with it, but it seems like you should be able to create another xpath (preferably something more specific than "/tr[4]/td[2]/a/#href") and supply a callback function.
However, in a "typical" scrapy project using Spider instead of CrawlSpider, you would simply yield a Request with another callback function to handle extracting and storing to the database. For example:
def parse_item(self, response):
for b in response.xpath('//table')
url = b.xpath('/tr[4]/td[2]/a/#href').extract_first()
yield scrapy.Request(url=url, callback=extract_and_store)
url = b.xpath('/tr[5]/td[2]/a/#href').extract_first()
yield scrapy.Request(url=url, callback=extract_and_store)
def extract_and_store(self, response):
"""
Scrape data and store it separately
"""
pyGTrends does not seem to work. Giving errors in Python.
pyGoogleTrendsCsvDownloader seems to work, logs in, but after getting 1-3 requests (per day!) complains about exhausted quota, even though manual download with the same login/IP works flawlessly.
Bottom line: neither work. Searching through stackoverflow: many questions from people trying to pull csv's from Google, but no workable solution I could find...
Thank you in advance: whoever will be able to help. How should the code be changed? Do you know of another solution that works?
Here's the code of pyGoogleTrendsCsvDownloader.py
import httplib
import urllib
import urllib2
import re
import csv
import lxml.etree as etree
import lxml.html as html
import traceback
import gzip
import random
import time
import sys
from cookielib import Cookie, CookieJar
from StringIO import StringIO
class pyGoogleTrendsCsvDownloader(object):
'''
Google Trends Downloader
Recommended usage:
from pyGoogleTrendsCsvDownloader import pyGoogleTrendsCsvDownloader
r = pyGoogleTrendsCsvDownloader(username, password)
r.get_csv(cat='0-958', geo='US-ME-500')
'''
def __init__(self, username, password):
'''
Provide login and password to be used to connect to Google Trends
All immutable system variables are also defined here
'''
# The amount of time (in secs) that the script should wait before making a request.
# This can be used to throttle the downloading speed to avoid hitting servers too hard.
# It is further randomized.
self.download_delay = 0.25
self.service = "trendspro"
self.url_service = "http://www.google.com/trends/"
self.url_download = self.url_service + "trendsReport?"
self.login_params = {}
# These headers are necessary, otherwise Google will flag the request at your account level
self.headers = [('User-Agent', 'Mozilla/5.0 (Windows NT 6.1; WOW64; rv:12.0) Gecko/20100101 Firefox/12.0'),
("Accept", "text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8"),
("Accept-Language", "en-gb,en;q=0.5"),
("Accept-Encoding", "gzip, deflate"),
("Connection", "keep-alive")]
self.url_login = 'https://accounts.google.com/ServiceLogin?service='+self.service+'&passive=1209600&continue='+self.url_service+'&followup='+self.url_service
self.url_authenticate = 'https://accounts.google.com/accounts/ServiceLoginAuth'
self.header_dictionary = {}
self._authenticate(username, password)
def _authenticate(self, username, password):
'''
Authenticate to Google:
1 - make a GET request to the Login webpage so we can get the login form
2 - make a POST request with email, password and login form input values
'''
# Make sure we get CSV results in English
ck = Cookie(version=0, name='I4SUserLocale', value='en_US', port=None, port_specified=False, domain='www.google.com', domain_specified=False,domain_initial_dot=False, path='/trends', path_specified=True, secure=False, expires=None, discard=False, comment=None, comment_url=None, rest=None)
self.cj = CookieJar()
self.cj.set_cookie(ck)
self.opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(self.cj))
self.opener.addheaders = self.headers
# Get all of the login form input values
find_inputs = etree.XPath("//form[#id='gaia_loginform']//input")
try:
#
resp = self.opener.open(self.url_login)
if resp.info().get('Content-Encoding') == 'gzip':
buf = StringIO( resp.read())
f = gzip.GzipFile(fileobj=buf)
data = f.read()
else:
data = resp.read()
xmlTree = etree.fromstring(data, parser=html.HTMLParser(recover=True, remove_comments=True))
for input in find_inputs(xmlTree):
name = input.get('name')
if name:
name = name.encode('utf8')
value = input.get('value', '').encode('utf8')
self.login_params[name] = value
except:
print("Exception while parsing: %s\n" % traceback.format_exc())
self.login_params["Email"] = username
self.login_params["Passwd"] = password
params = urllib.urlencode(self.login_params)
self.opener.open(self.url_authenticate, params)
def get_csv(self, throttle=False, **kwargs):
'''
Download CSV reports
'''
# Randomized download delay
if throttle:
r = random.uniform(0.5 * self.download_delay, 1.5 * self.download_delay)
time.sleep(r)
params = {
'export': 1
}
params.update(kwargs)
params = urllib.urlencode(params)
r = self.opener.open(self.url_download + params)
# Make sure everything is working ;)
if not r.info().has_key('Content-Disposition'):
print "You've exceeded your quota. Continue tomorrow..."
sys.exit(0)
if r.info().get('Content-Encoding') == 'gzip':
buf = StringIO( r.read())
f = gzip.GzipFile(fileobj=buf)
data = f.read()
else:
data = r.read()
myFile = open('trends_%s.csv' % '_'.join(['%s-%s' % (key, value) for (key, value) in kwargs.items()]), 'w')
myFile.write(data)
myFile.close()
Although I don't know python, I may have a solution. I am currently doing the same thing in C# and though I didn't get the .csv file, I got created a custom URL through code and then downloaded that HTML and saved to a text file (also through code). In this HTML (at line 12) is all the information needed to create the graph that is used on Google Trends. However, this has alot of unnecessary text within it that needs to be cut down. But either way, you end up with the same result. The Google Trends data. I posted a more detailed answer to my question here:
Downloading .csv file from Google Trends
There is an alternative module named pytrends - https://pypi.org/project/pytrends/ It is really cool. I would recommend this.
Example usage:
import numpy as np
import pandas as pd
from pytrends.request import TrendReq
pytrend = TrendReq()
#It is the term that you want to search
pytrend.build_payload(kw_list=["Eminem is the Rap God"])
# Find which region has searched the term
df = pytrend.interest_by_region()
df.to_csv("path\Eminem_InterestbyRegion.csv")
Potentially if you have a list of terms to search you could make use of "for loop" to automate the insights as per your wish.
In my django app, I have a view which accomplishes file upload.The core snippet is like this
...
if (request.method == 'POST'):
if request.FILES.has_key('file'):
file = request.FILES['file']
with open(settings.destfolder+'/%s' % file.name, 'wb+') as dest:
for chunk in file.chunks():
dest.write(chunk)
I would like to unit test the view.I am planning to test the happy path as well as the fail path..ie,the case where the request.FILES has no key 'file' , case where request.FILES['file'] has None..
How do I set up the post data for the happy path?Can somebody tell me?
I used to do the same with open('some_file.txt') as fp: but then I needed images, videos and other real files in the repo and also I was testing a part of a Django core component that is well tested, so currently this is what I have been doing:
from django.core.files.uploadedfile import SimpleUploadedFile
def test_upload_video(self):
video = SimpleUploadedFile("file.mp4", "file_content", content_type="video/mp4")
self.client.post(reverse('app:some_view'), {'video': video})
# some important assertions ...
In Python 3.5+ you need to use bytes object instead of str. Change "file_content" to b"file_content"
It's been working fine, SimpleUploadedFile creates an InMemoryFile that behaves like a regular upload and you can pick the name, content and content type.
From Django docs on Client.post:
Submitting files is a special case. To POST a file, you need only
provide the file field name as a key, and a file handle to the file
you wish to upload as a value. For example:
c = Client()
with open('wishlist.doc') as fp:
c.post('/customers/wishes/', {'name': 'fred', 'attachment': fp})
I recommend you to take a look at Django RequestFactory. It's the best way to mock data provided in the request.
Said that, I found several flaws in your code.
"unit" testing means to test just one "unit" of functionality. So,
if you want to test that view you'd be testing the view, and the file
system, ergo, not really unit test. To make this point more clear. If
you run that test, and the view works fine, but you don't have
permissions to save that file, your test would fail because of that.
Other important thing is test speed. If you're doing something like
TDD the speed of execution of your tests is really important.
Accessing any I/O is not a good idea.
So, I recommend you to refactor your view to use a function like:
def upload_file_to_location(request, location=None): # Can use the default configured
And do some mocking on that. You can use Python Mock.
PS: You could also use Django Test Client But that would mean that you're adding another thing more to test, because that client make use of Sessions, middlewares, etc. Nothing similar to Unit Testing.
I do something like this for my own event related application but you should have more than enough code to get on with your own use case
import tempfile, csv, os
class UploadPaperTest(TestCase):
def generate_file(self):
try:
myfile = open('test.csv', 'wb')
wr = csv.writer(myfile)
wr.writerow(('Paper ID','Paper Title', 'Authors'))
wr.writerow(('1','Title1', 'Author1'))
wr.writerow(('2','Title2', 'Author2'))
wr.writerow(('3','Title3', 'Author3'))
finally:
myfile.close()
return myfile
def setUp(self):
self.user = create_fuser()
self.profile = ProfileFactory(user=self.user)
self.event = EventFactory()
self.client = Client()
self.module = ModuleFactory()
self.event_module = EventModule.objects.get_or_create(event=self.event,
module=self.module)[0]
add_to_admin(self.event, self.user)
def test_paper_upload(self):
response = self.client.login(username=self.user.email, password='foz')
self.assertTrue(response)
myfile = self.generate_file()
file_path = myfile.name
f = open(file_path, "r")
url = reverse('registration_upload_papers', args=[self.event.slug])
# post wrong data type
post_data = {'uploaded_file': i}
response = self.client.post(url, post_data)
self.assertContains(response, 'File type is not supported.')
post_data['uploaded_file'] = f
response = self.client.post(url, post_data)
import_file = SubmissionImportFile.objects.all()[0]
self.assertEqual(SubmissionImportFile.objects.all().count(), 1)
#self.assertEqual(import_file.uploaded_file.name, 'files/registration/{0}'.format(file_path))
os.remove(myfile.name)
file_path = import_file.uploaded_file.path
os.remove(file_path)
I did something like that :
from django.core.files.uploadedfile import SimpleUploadedFile
from django.test import TestCase
from django.core.urlresolvers import reverse
from django.core.files import File
from django.utils.six import BytesIO
from .forms import UploadImageForm
from PIL import Image
from io import StringIO
def create_image(storage, filename, size=(100, 100), image_mode='RGB', image_format='PNG'):
"""
Generate a test image, returning the filename that it was saved as.
If ``storage`` is ``None``, the BytesIO containing the image data
will be passed instead.
"""
data = BytesIO()
Image.new(image_mode, size).save(data, image_format)
data.seek(0)
if not storage:
return data
image_file = ContentFile(data.read())
return storage.save(filename, image_file)
class UploadImageTests(TestCase):
def setUp(self):
super(UploadImageTests, self).setUp()
def test_valid_form(self):
'''
valid post data should redirect
The expected behavior is to show the image
'''
url = reverse('image')
avatar = create_image(None, 'avatar.png')
avatar_file = SimpleUploadedFile('front.png', avatar.getvalue())
data = {'image': avatar_file}
response = self.client.post(url, data, follow=True)
image_src = response.context.get('image_src')
self.assertEquals(response.status_code, 200)
self.assertTrue(image_src)
self.assertTemplateUsed('content_upload/result_image.html')
create_image function will create image so you don't need to give static path of image.
Note : You can update code as per you code.
This code for Python 3.6.
from rest_framework.test import force_authenticate
from rest_framework.test import APIRequestFactory
factory = APIRequestFactory()
user = User.objects.get(username='#####')
view = <your_view_name>.as_view()
with open('<file_name>.pdf', 'rb') as fp:
request=factory.post('<url_path>',{'file_name':fp})
force_authenticate(request, user)
response = view(request)
As mentioned in Django's official documentation:
Submitting files is a special case. To POST a file, you need only provide the file field name as a key, and a file handle to the file you wish to upload as a value. For example:
c = Client()
with open('wishlist.doc') as fp:
c.post('/customers/wishes/', {'name': 'fred', 'attachment': fp})
More Information: How to check if the file is passed as an argument to some function?
While testing, sometimes we want to make sure that the file is passed as an argument to some function.
e.g.
...
class AnyView(CreateView):
...
def post(self, request, *args, **kwargs):
attachment = request.FILES['attachment']
# pass the file as an argument
my_function(attachment)
...
In tests, use Python's mock something like this:
# Mock 'my_function' and then check the following:
response = do_a_post_request()
self.assertEqual(mock_my_function.call_count, 1)
self.assertEqual(
mock_my_function.call_args,
call(response.wsgi_request.FILES['attachment']),
)
if you want to add other data with file upload then follow the below method
file = open('path/to/file.txt', 'r', encoding='utf-8')
data = {
'file_name_to_receive_on_backend': file,
'param1': 1,
'param2': 2,
.
.
}
response = self.client.post("/url/to/view", data, format='multipart')`
The only file_name_to_receive_on_backend will be received as a file other params received normally as post paramas.
In Django 1.7 there's an issue with the TestCase wich can be resolved by using open(filepath, 'rb') but when using the test client we have no control over it. I think it's probably best to ensure file.read() returns always bytes.
source: https://code.djangoproject.com/ticket/23912, by KevinEtienne
Without rb option, a TypeError is raised:
TypeError: sequence item 4: expected bytes, bytearray, or an object with the buffer interface, str found
from django.test import Client
from requests import Response
client = Client()
with open(template_path, 'rb') as f:
file = SimpleUploadedFile('Name of the django file', f.read())
response: Response = client.post(url, format='multipart', data={'file': file})
Hope this helps.
Very handy solution with mock
from django.test import TestCase, override_settings
#use your own client request factory
from my_framework.test import APIClient
from django.core.files import File
import tempfile
from pathlib import Path
import mock
image_mock = mock.MagicMock(spec=File)
image_mock.name = 'image.png' # or smt else
class MyTest(TestCase):
# I assume we want to put this file in storage
# so to avoid putting garbage in our MEDIA_ROOT
# we're using temporary storage for test purposes
#override_settings(MEDIA_ROOT=Path(tempfile.gettempdir()))
def test_send_file(self):
client = APIClient()
client.post(
'/endpoint/'
{'file':image_mock},
format="multipart"
)
I am using Python==3.8.2 , Django==3.0.4, djangorestframework==3.11.0
I tried self.client.post but got a Resolver404 exception.
Following worked for me:
import requests
upload_url='www.some.com/oaisjdoasjd' # your url to upload
with open('/home/xyz/video1.webm', 'rb') as video_file:
# if it was a text file we would perhaps do
# file = video_file.read()
response_upload = requests.put(
upload_url,
data=video_file,
headers={'content-type': 'video/webm'}
)
I am using django rest framework and I had to test the upload of multiple files.
I finally get it by using format="multipart" in my APIClient.post request.
from rest_framework.test import APIClient
...
self.client = APIClient()
with open('./photo.jpg', 'rb') as fp:
resp = self.client.post('/upload/',
{'images': [fp]},
format="multipart")
I am using GraphQL, upload for test:
with open('test.jpg', 'rb') as fp:
response = self.client.execute(query, variables, data={'image': [fp]})
code in class mutation
#classmethod
def mutate(cls, root, info, **kwargs):
if image := info.context.FILES.get("image", None):
kwargs["image"] = image
TestingMainModel.objects.get_or_create(
id=kwargs["id"],
defaults=kwargs
)