I cannot find any example on how to attach files(pdf) that are within my root folder of the site in python (google app engine) send_mail function.
url_test = "https://mywebsite.com/pdf/test.pdf"
test_file = urlfetch.fetch(url_test)
if test_file.status_code == 200:
test_document = test_file.content
mail.send_mail(sender=EMAIL_SENDER,
to=['test#test.com'],
subject=subject,
body=theBody,
attachments=[("testing",test_document)])
Decided to try it with EmailMessage:
message = mail.EmailMessage( sender=EMAIL_SENDER,
subject=subject,body=theBody,to=['myemail#gmail.com'],attachments=
[(attachname, blob.archivoBlob)])
message.send()
The above blob attachment is successfully sending however attaching a file with relative path always says "invalid attachment"
new_file = open(os.path.dirname(__file__) +
'/../pages/pdf/test.PDF').read()
message = mail.EmailMessage( sender=EMAIL_SENDER,
subject=subject,body=theBody,to=['myemail#gmail.com'],attachments=
[('testing',new_file )])
message.send()
In debugging I have also tried to see if the file is being read by doing this:
logging.info(new_file)
It seems to be reading the file as it outputs some unicode characters
Please help why am I not able to attach a PDF while I can attach a blob
When calling the attachments, the File type has to be indicated on the file title, for example attachments= [('testing.pdf',new_file )]). View this link
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.
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'm using Tweepy, a tweeting python library, django-storages and boto. I have a custom manage.py command that works correctly locally, it gets an image from the filesystem and tweets that image. If I change the storage to Amazon S3, however, I can't access the file. It gives me this error:
raise TweepError('Unable to access file: %s' % e.strerror)
I tried making the images in the bucket "public". Didn't work. This is the code (it works without S3):
filename = model_object.image.file.url
media_ids = api.media_upload(filename=filename) # ERROR
params = {'status': tweet_text, 'media_ids': [media_ids.media_id_string]}
api.update_status(**params)
This line:
model_object.image.file.url
Gives me the complete url of the image I want to tweet, something like this:
https://criptolibertad.s3.amazonaws.com/OrillaLibertaria/195.jpg?Signature=xxxExpires=1467645897&AWSAccessKeyId=yyy
I also tried constructing the url manually, since it is a public image stored in my bucket, like this:
filename = "https://criptolibertad.s3.amazonaws.com/OrillaLibertaria/195.jpg"
But it doesn't work.
¿Why do I get the Unable to access file error?
The source code from tweepy looks like this:
def media_upload(self, filename, *args, **kwargs):
""" :reference: https://dev.twitter.com/rest/reference/post/media/upload
:allowed_param:
"""
f = kwargs.pop('file', None)
headers, post_data = API._pack_image(filename, 3072, form_field='media', f=f) # ERROR
kwargs.update({'headers': headers, 'post_data': post_data})
def _pack_image(filename, max_size, form_field="image", f=None):
"""Pack image from file into multipart-formdata post body"""
# image must be less than 700kb in size
if f is None:
try:
if os.path.getsize(filename) > (max_size * 1024):
raise TweepError('File is too big, must be less than %skb.' % max_size)
except os.error as e:
raise TweepError('Unable to access file: %s' % e.strerror)
Looks like Tweepy can't get the image from the Amazon S3 bucket, but how can I make it work? Any advice will help.
The issue occurs when tweepy attempts to get file size in _pack_image:
if os.path.getsize(filename) > (max_size * 1024):
The function os.path.getsize assumes it is given a file path on disk; however, in your case it is given a URL. Naturally, the file is not found on disk and os.error is raised. For example:
# The following raises OSError on my machine
os.path.getsize('https://criptolibertad.s3.amazonaws.com/OrillaLibertaria/195.jpg')
What you could do is to fetch the file content, temporarily save it locally and then tweet it:
import tempfile
with tempfile.NamedTemporaryFile(delete=True) as f:
name = model_object.image.file.name
f.write(model_object.image.read())
media_ids = api.media_upload(filename=name, f=f)
params = dict(status='test media', media_ids=[media_ids.media_id_string])
api.update_status(**params)
For your convenience, I published a fully working example here: https://github.com/izzysoftware/so38134984
I'm writing a web app for customer/order handling in python with asana integration.
For a registered incoming order, an invoice is created as .pdf. This file I want to send to asana as an email attachment using mandrill, because the asana python API doesn't provide attachments yet.
Because mandrill wants the content of the attachment as a base64-encoded string, I create a binary of the pdf using this function:
def binaryFile(self, pathToFile):
binary_obj = xmlrpclib.Binary( open(pathToFile).read() )
return binary_obj
Together with the path of the file, I throw this into mandrill like so:
'attachments': [{'content': binaryFile,
'name': pathOfFile,
'type': 'application/pdf'}]
When I try to send the whole thing, this is what I get:
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site- packages/mandrill.py", line 1215, in send
return self.master.call('messages/send', _params)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/mandrill.py", line 131, in call
params = json.dumps(params)
OverflowError: Overlong 3 byte UTF-8 sequence detected when encoding string
Who can hint at what I am doing wrong?
Thank you.
All right ok, I found the mistake myself, it's in the base64-encoding of course. I'm now doing it like so:
import base64
def filetobase64(self, inputfilename):
return base64.b64encode(open(inputfilename, 'rb').read())
Everything works fine now.
No harm meant!
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.