How to get the name of uploaded file in Django - python

I have seen some question about it already but these couldn't solve my issue, that's why I'm asking a new question.So, don't mark this as duplicate, please!
Using Python(3.6) & Django(1.10)
I'm trying to get the name of uploaded file, but it returns
AttributeError: 'NoneType' object has no attribute 'name'
Here's what I have tried:
From models.py
sourceFile = models.FileField(upload_to='archives/', name='sourceFile', blank=True)
From HTML template:
<div class="form-group" hidden id="zipCode">
<label class="control-label" style="font-size: 1.5rem; color: black;">Select File</label>
<input id="sourceFile" name="sourceFile" type="file" class="file" multiple
data-allowed-file-extensions='["zip"]'>
<small id="fileHelp" class="form-text control-label" style="color:black; font-size: .9rem;">
Upload a Tar or Zip
archive which include Dockerfile, otherwise your deployment will fail.
</small>
</div>
From views.py:
if form.is_valid():
func_obj = form
func_obj.sourceFile = form.cleaned_data['sourceFile']
func_obj.save()
print(func_obj.sourceFile.name)
what's wrong here?
Help me, please!
Thanks in advance!

To get the filename, you simply use the request.FILES dictionary (I assume that there is only 1 file being uploaded)
Example:
try:
print(next(iter(request.FILES))) # this will print the name of the file
except StopIteration:
print("No file was uploaded!")
Note that this requires that the files were sent as a part of a form by the POST method.
To change their name to a random string, I recommend uuid.uuid4, as this generates a random string that is VERY unlikely to collide with anything already there. Also, you need to edit your upload_to= section of your sourceFile model by providing a function to generate the name:
# In models.py
def content_file_name(instance, filename):
filename = "{}.zip".format(str(uuid.uuid4().hex))
return os.path.join('archives', filename)
# later....
sourceFile = models.FileField(upload_to=content_file_name, name='sourceFile', blank=True)
Hope this helps!

Related

form.is_valid() always returns false perhaps issue with "This field is required" error

I am learning django. I am stuck with this problem.
The problem is that form.is_valid() always returns false.
I tried to debug and I think that the problem is because of "This field is required" error and I think that the field is file_name. I tried to resolve the issue but I am unable to do so.
Just to give a context of what I am trying to do -
I have created a form in which a user uploads a text file and selects a gender.
In the backend I want to save the name of the text file along with the gender in a model. The purpose of doing this is because when multiple users will use the application, I should know which user selected what gender so that I can produce the desired output.
Here is the link to my git repository -
git repository
As I already said I am new to django and some help will be appreciated.
Your fields in form are fields = ['file_name','gender'] and in the template you have
<label for="myfile"> <b> Select a file: </b> </label> <br/>
<input type="file" name="document"> <br/><br/>
<br/><br/>
{{ form.gender }}
... fields document and gender, no file_name. Add {{ form.file_name }} to the template or, if you want to stick to manually writing code for input, file input with a name file_name. Input names need to match field names in form.
Also, read this: Working with forms.
I think you are not saving the form. In your views line 24 you have
if request.method == 'POST':
form = TranscriptForm(request.POST)
return HttpResponse(str(form.errors))
If you are posting you will always return there,therefore there is no check for is_valid().
Same as line 60:
context = {'form': TranscriptForm()}
return render(request, 'tutorial/upload.html', context)
you havent accessed/saved the form yet for it to be valid

Python/Django - using a view method on an upload file

I have this view:
def index(request):
file = open("SK ✌😜✌.txt", encoding="UTF-8")
data = file.read()
file.close()
lines = data.split("\n")
...More code...
In this view i open a file from the very first moment the app starts and i do some work on the file, is a story, and when i start the server and go to http://127.0.0.1:8000/(Name Of The App), i see all the work that i have done on that file.
What i want to do is to do that same work, starting with the reading of the file, BUT i want to do that with the file that the user uploads in that moment. I have this that i took from bootstrap:
<div class="form-group">
<label for="exampleInputFile">File input</label>
<input type="file" id="exampleInputFile">
</div>
I guess i have to use in some way the id of the input but i`m not really sure how to pass this file that the user uploads in the ui to the method that i have in my views.py
Any help will be really appreciated
You need to have a name attribute in your <input> template code.
<input type="file" id="exampleInputFile" name="some_file">
Then to access the file in your view, you need to use request.FILES attribute.
As per the Django docs on HttpRequest.FILES attribute:
A dictionary-like object containing all uploaded files. Each key in
FILES is the name from the <input type="file" name="" />. Each
value in FILES is an UploadedFile.
Your code should be something like:
def index(request):
if request.method=="POST":
uploaded_file = request.FILES['some_file'] # get the uploaded file
# do something with the file
Note: request.FILES will only contain data if the request method was POST and the <form> that posted to the request had enctype="multipart/form-data. Otherwise, FILES will be a blank dictionary-like object.

Getting file location from an html form

EDIT: Added the solution to my code, as suggested by Martijn Pieters.
I'm writing a web application that should have an image upload feature. Actually it is uploading directly to imgur.com through their api. So far I can upload an image to imgur through the terminal by running my python script. But I want the user to be able to select an image through an html form. All I actually need (I think) is the path for the file, then I have a script that converts the image to base64. As I write this I realize that the script might only work when it's run locally, and not from a browser. I have been googling for two days. I get a lot of results about cgi.FieldStorage() but can't seem to get that to work as I get a KeyError.
This is the file input part of my form:
<div class="container">
<form role="form" class="form-horizontal form-inline" method="post" enctype="multipart/form-data">
<div class="col-md-12">
<div class="form-group">
<label for="image" class="control-label">Product image</label>
<input type="file" id="image" name="imageurl" value=""/>
</div>
</div>
And then I need to be able to handle the input in my python script:
class MainPage(BlogHandler):
def get(self):
self.render("front.html")
def post(self):
image = self.request.POST.get("imageurl")
logging.info(getUrl(image))
This is the code for sending the image to the imgur api:
def getUrl(image):
API_URL = "https://api.imgur.com/3/upload.json"
image_read = image.file.read()
b64 = base64.b64encode(image_read)
data = {
'image': b64,
'type': base64,
'title': 'testupload'
}
response = requests.post(API_URL, data, headers={'Authorization': 'Client-ID my-cient-id'})
url = response.json()['data']['link']
return url
I developing in Google App Engine and using Jinja2 as the templating engine.
I'm a beginner programmer so I hope to get some pointers on my problem. I feel like I'm stuck on searching for the answer. I've read that the browser does not know anything about the file system of the computer for security reasons, but websites exist where you can choose a file from your computer to upload, so I figure it must be possible to achieve. :)
EDIT: Added the solution to my code, as suggested by Martijn Pieters.
Uploading an image requires that you change the form encoding to multipart/form-data:
<form encoding="multipart/form-data" role="form" class="form-horizontal form-inline" method="post">
See application/x-www-form-urlencoded or multipart/form-data?
Because this is a POST request, you'll find the fields in request.POST:
def post(self):
imageurl = self.request.POST.get("imageurl")
logging.info(imageurl)
although request.get() will search both query parameters and POST form data.
A form upload is returned as a cgi.FieldStorage() object; this has a file attribute, which gives you a file-like interface. You could just read the data into memory:
image = self.request.POST.get("imageurl")
image_read = image.file.read()
b64 = base64.b64encode(image_read)

uploading file in django and python?

<input type="file" name="doc-file" multiple/>
file_path = request.FILES.get('doc-file')
My return value of file_path is None
But when I do file_path = request.POST.get('doc-file') it returns filename . What is the best way to upload a file in python and django?
It sounds like you might be forgetting to include enctype="multipart/form-data" in your form tag. See Django's docs on file uploads for details.
Edit: formatting

Upload files in Google App Engine

I am planning to create a web app that allows users to downgrade their visual studio project files. However, It seems Google App Engine accepts files uploading and flat file storing on the Google Server through db.TextProperty and db.BlobProperty.
I'll be glad anyone can provide code sample (both the client and the server side) on how this can be done.
In fact, this question is answered in the App Egnine documentation. See an example on Uploading User Images.
HTML code, inside <form></form>:
<input type="file" name="img"/>
Python code:
class Guestbook(webapp.RequestHandler):
def post(self):
greeting = Greeting()
if users.get_current_user():
greeting.author = users.get_current_user()
greeting.content = self.request.get("content")
avatar = self.request.get("img")
greeting.avatar = db.Blob(avatar)
greeting.put()
self.redirect('/')
Here is a complete, working file. I pulled the original from the Google site and modified it to make it slightly more real world.
A few things to notice:
This code uses the BlobStore API
The purpose of this line in the
ServeHandler class is to "fix" the
key so that it gets rid of any name
mangling that may have occurred in
the browser (I didn't observe any in
Chrome)
blob_key = str(urllib.unquote(blob_key))
The "save_as" clause at the end of this is important. It will make sure that the file name does not get mangled when it is sent to your browser. Get rid of it to observe what happens.
self.send_blob(blobstore.BlobInfo.get(blob_key), save_as=True)
Good Luck!
import os
import urllib
from google.appengine.ext import blobstore
from google.appengine.ext import webapp
from google.appengine.ext.webapp import blobstore_handlers
from google.appengine.ext.webapp import template
from google.appengine.ext.webapp.util import run_wsgi_app
class MainHandler(webapp.RequestHandler):
def get(self):
upload_url = blobstore.create_upload_url('/upload')
self.response.out.write('<html><body>')
self.response.out.write('<form action="%s" method="POST" enctype="multipart/form-data">' % upload_url)
self.response.out.write("""Upload File: <input type="file" name="file"><br> <input type="submit" name="submit" value="Submit"> </form></body></html>""")
for b in blobstore.BlobInfo.all():
self.response.out.write('<li>' + str(b.filename) + '')
class UploadHandler(blobstore_handlers.BlobstoreUploadHandler):
def post(self):
upload_files = self.get_uploads('file')
blob_info = upload_files[0]
self.redirect('/')
class ServeHandler(blobstore_handlers.BlobstoreDownloadHandler):
def get(self, blob_key):
blob_key = str(urllib.unquote(blob_key))
if not blobstore.get(blob_key):
self.error(404)
else:
self.send_blob(blobstore.BlobInfo.get(blob_key), save_as=True)
def main():
application = webapp.WSGIApplication(
[('/', MainHandler),
('/upload', UploadHandler),
('/serve/([^/]+)?', ServeHandler),
], debug=True)
run_wsgi_app(application)
if __name__ == '__main__':
main()
There is a thread in Google Groups about it:
Uploading Files
With a lot of useful code, that discussion helped me very much in uploading files.
Google has released a service for storing large files. Have a look at blobstore API documentation. If your files are > 1MB, you should use it.
I try it today, It works as following:
my sdk version is 1.3.x
html page:
<form enctype="multipart/form-data" action="/upload" method="post" >
<input type="file" name="myfile" />
<input type="submit" />
</form>
Server Code:
file_contents = self.request.POST.get('myfile').file.read()
If your still having a problem, check you are using enctype in the form tag
No:
<form encoding="multipart/form-data" action="/upload">
Yes:
<form enctype="multipart/form-data" action="/upload">
You can not store files as there is not a traditional file system. You can only store them in their own DataStore (in a field defined as a BlobProperty)
There is an example in the previous link:
class MyModel(db.Model):
blob = db.BlobProperty()
obj = MyModel()
obj.blob = db.Blob( file_contents )
Personally I found the tutorial described here useful when using the Java run time with GAE. For some reason, when I tried to upload a file using
<form action="/testservelet" method="get" enctype="multipart/form-data">
<div>
Myfile:<input type="file" name="file" size="50"/>
</div>
<div>
<input type="submit" value="Upload file">
</div>
</form>
I found that my HttpServlet class for some reason wouldn't accept the form with the 'enctype' attribute. Removing it works, however, this means I can't upload any files.
There's no flat file storing in Google App Engine. Everything has to go in to the Datastore which is a bit like a relational database but not quite.
You could store the files as TextProperty or BlobProperty attributes.
There is a 1MB limit on DataStore entries which may or may not be a problem.
I have observed some strange behavior when uploading files on App Engine. When you submit the following form:
<form method="post" action="/upload" enctype="multipart/form-data">
<input type="file" name="img" />
...
</form>
And then you extract the img from the request like this:
img_contents = self.request.get('img')
The img_contents variable is a str() in Google Chrome, but it's unicode in Firefox. And as you now, the db.Blob() constructor takes a string and will throw an error if you pass in a unicode string.
Does anyone know how this can be fixed?
Also, what I find absolutely strange is that when I copy and paste the Guestbook application (with avatars), it works perfectly. I do everything exactly the same way in my code, but it just won't work. I'm very close to pulling my hair out.
There is a way of using flat file system( Atleast in usage perspective)
There is this Google App Engine Virtual FileSystem project. that is implemented with the help of datastore and memcache APIs to emulate an ordinary filesystem. Using this library you can use in you project a similar filesystem access(read and write).

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