Right now I'm just checking the response of the link like so:
self.client = Client()
response = self.client.get(url)
self.assertEqual(response.status_code, 200)
Is there a Django-ic way to test a link to see if a file download event actually takes place? Can't seem to find much resource on this topic.
If the url is meant to produce a file rather than a "normal" http response, then its content-type and/or content-disposition will be different.
the response object is basically a dictionary, so you could so something like
self.assertEquals(
response.get('Content-Disposition'),
"attachment; filename=mypic.jpg"
)
more info:
https://docs.djangoproject.com/en/dev/ref/request-response/#telling-the-browser-to-treat-the-response-as-a-file-attachment
UPD:
If you want to read the actual contents of the attached file, you can use response.content. Example for a zip file:
try:
f = io.BytesIO(response.content)
zipped_file = zipfile.ZipFile(f, 'r')
self.assertIsNone(zipped_file.testzip())
self.assertIn('my_file.txt', zipped_file.namelist())
finally:
zipped_file.close()
f.close()
Related
I am receiving an http request from a desktop application with a screenshot. I cannot speak with the developer or see source code, so all I have is the http request I am getting.
The file isn't in request.FILES, it is in request.POST.
#csrf_exempt
def create_contract_event_handler(request, contract_id, event_type):
keyboard_events_count = request.POST.get('keyboard_events_count')
mouse_events_count = request.POST.get('mouse_events_count')
screenshot_file = request.POST.get('screenshot_file')
barr2 = bytes(screenshot_file.encode(encoding='utf8'))
with open('.test/output.jpeg', 'wb') as f:
f.write(barr2)
f.close()
The file is corrupted.
The binary starts like this, I don't know if that helps:
����JFIFHH��C
%# , #&')*)-0-(0%()(��C
(((((((((((((((((((((((((((((((((((((((((((((((((((�� `"��
Also, if I try to open the image with PIL, I get the following error:
from PIL import Image
im = Image.open('./test/output.jpg')
#OSError: cannot identify image file './test/output.jpg'
Finally, I managed to touch the code in the other hand, the 'filename' was missing in the header and for that reason I was getting the file in the POST instead of in the FILES dictionary.
Right now I'm just checking the response of the link like so:
self.client = Client()
response = self.client.get(url)
self.assertEqual(response.status_code, 200)
Is there a Django-ic way to test a link to see if a file download event actually takes place? Can't seem to find much resource on this topic.
If the url is meant to produce a file rather than a "normal" http response, then its content-type and/or content-disposition will be different.
the response object is basically a dictionary, so you could so something like
self.assertEquals(
response.get('Content-Disposition'),
"attachment; filename=mypic.jpg"
)
more info:
https://docs.djangoproject.com/en/dev/ref/request-response/#telling-the-browser-to-treat-the-response-as-a-file-attachment
UPD:
If you want to read the actual contents of the attached file, you can use response.content. Example for a zip file:
try:
f = io.BytesIO(response.content)
zipped_file = zipfile.ZipFile(f, 'r')
self.assertIsNone(zipped_file.testzip())
self.assertIn('my_file.txt', zipped_file.namelist())
finally:
zipped_file.close()
f.close()
I have taken a look at other questions related to multipart/form POST requests in Python but unfortunately, they don't seem to address my exact question. Basically, I normally use CURL in order to hit an API service that allows me to upload zip files in order to create HTML5 assets. The CURL command I use looks like this:
curl -X POST -H "Authorization: api: 222111" --form "type=html" --form "file=Folder1/Folder2/example.zip" "https://example.api.com/upload?ins_id=123"
I am trying to use a python script to iterate through a folder of zip files in order to upload all of these files and receive a "media ID" back. This is what my script looks like:
import os
import requests
import json
ins_id = raw_input("Please enter your member ID: ")
auth = raw_input("Please enter your API authorization token: ")
for filename in os.listdir("zips"):
if filename.endswith(".zip"):
file_path = os.path.abspath(filename)
url = "https://example.api.com/upload?
ins_id="+str(ins_id)
header = {"Authorization": auth}
response = requests.post(url, headers=header, files={"form_type":
(None, "html"), "form_file_upload": (None, str(file_path))})
api_response = response.json()
print api_response
This API service requires the file path to be included when submitting the POST. However, when I use this script, the response indicates that "file not provided". Am I including this information correctly in my script?
Thanks.
Update:
I think I am heading in the right direction now (thanks to the answer provided) but now, I receive an error message stating that there is "no such file or directory". My thinking is that I am not using os.path correctly but even if I change my code to use "relpath" I still get the same message. My script is in a folder and I have a completely different folder called "zips" (in the same directory) which is where all of my zip files are stored.
To upload files with the request library, you can include the file handler directly in the JSON as described in the documentation. This is the corresponding example that I have taken from there:
url = 'http://httpbin.org/post'
files = {'file': open('path_to_your_file', 'rb')}
r = requests.post(url, files=files)
If we integrate this in your script, it would look as follows (I also made it slightly more pythonic):
import os
import requests
import json
folder = 'zips'
ins_id = raw_input("Please enter your member ID: ")
auth = raw_input("Please enter your API authorization token: ")
url = "https://example.api.com/upload?"
header = {"Authorization": auth}
for filename in os.listdir(folder):
if not filename.endswith(".zip"):
continue
file_path = os.path.abspath(os.path.join(folder, filename))
ins_id="+str(ins_id)"
response = requests.post(
url, headers=header,
files={"form_type": (None, "html"),
"form_file_upload": open(file_path, 'rb')}
)
api_response = response.json()
print api_response
As I don't have the API end point, I can't actually test this code block - but it should be something along these lines.
I looked at the various questions similar to mine, but I could not find anything a fix for my problem.
In my code, I want to serve a freshly generated excel file residing in my app directory in a folder named files
excelFile = ExcelCreator.ExcelCreator("test")
excelFile.create()
response = HttpResponse(content_type='application/vnd.ms-excel')
response['Content-Disposition'] = 'attachment; filename="test.xls"'
return response
So when I click on the button that run this part of the code, it sends to the user an empty file. By looking at my code, I can understand that behavior because I don't point to that file within my response...
I saw some people use the file wrapper (which I don't quite understand the use). So I did like that:
response = HttpResponse(FileWrapper(excelFile.file),content_type='application/vnd.ms-excel')
But then, I receive the error message from server : A server error occurred. Please contact the administrator.
Thanks for helping me in my Django quest, I'm getting better with all of your precious advices!
First, you need to understand how this works, you are getting an empty file because that is what you are doing, actually:
response = HttpResponse(content_type='application/vnd.ms-excel')
response['Content-Disposition'] = 'attachment; filename="test.xls"'
HttpResponse receives as first arg the content of the response, take a look to its contructor:
def __init__(self, content='', mimetype=None, status=None, content_type=None):
so you need to create the response with the content that you wish, is this case, with the content of your .xls file.
You can use any method to do that, just be sure the content is there.
Here a sample:
import StringIO
output = StringIO.StringIO()
# read your content and put it in output var
out_content = output.getvalue()
output.close()
response = HttpResponse(out_content, mimetype='application/vnd.ms-excel')
response['Content-Disposition'] = 'attachment; filename="test.xls"'
I would recommend you use:
python manage.py runserver
to run your application from the command line. From here you will see the console output of your application and any exceptions that are thrown as it runs. This may provide a quick resolution to your problem.
I want to return some files in a HttpResponse and I'm using the following function. The file that is returned always has a filesize of 1kb and I do not know why. I can open the file, but it seems that it is not served correctly. Thus I wanted to know how one can return files with django/python over a HttpResponse.
#login_required
def serve_upload_files(request, file_url):
import os.path
import mimetypes
mimetypes.init()
try:
file_path = settings.UPLOAD_LOCATION + '/' + file_url
fsock = open(file_path,"r")
#file = fsock.read()
#fsock = open(file_path,"r").read()
file_name = os.path.basename(file_path)
file_size = os.path.getsize(file_path)
print "file size is: " + str(file_size)
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
except IOError:
response = HttpResponseNotFound()
return response
Edit:
The bug is actually not a bug ;-)
This solution is working in production on an apache server, thus the source is ok.
While writing this question I tested it local with the django development server and was wondering why it does not work. A friend of mine told me that this issue could arise if the mime types are not set in the server. But he was not sure if this is the problem. But one thing for sure.. it has something to do with the server.
Could it be that the file contains some non-ascii characters that render ok in production but not in development?
Try reading the file as binary:
fsock = open(file_path,"rb")
Try passing the fsock iterator as a parameter to HttpResponse(), rather than to its write() method which I think expects a string.
response = HttpResponse(fsock, mimetype=...)
See http://docs.djangoproject.com/en/dev/ref/request-response/#passing-iterators
Also, I'm not sure you want to call close on your file before returning response. Having played around with this in the shell (I've not tried this in an actual Django view), it seems that the response doesn't access the file until the response itself is read. Trying to read a HttpResponse created using a file that is now closed results in a ValueError: I/O operation on closed file.
So, you might want to leave fsock open, and let the garbage collector deal with it after the response is read.
Try disabling "django.middleware.gzip.GZipMiddleware" from your MIDDLEWARE_CLASSES in settings.py
I had the same problem, and after I looked around the middleware folder, this middleware seemed guilty to me and removing it did the trick for me.