Im working through Alex Pale's minimal django file upload example for django 1.8 -
https://github.com/axelpale/minimal-django-file-upload-example/tree/master/src/for_django_1-8/myproject/myproject/myapp
I know how to get the file extension in the form, but how can I get this in the view. I'm aware I can access the file thru -
docfile=request.FILES['docfile']
View -
from django.shortcuts import render_to_response
from django.template import RequestContext
from django.http import HttpResponseRedirect
from django.core.urlresolvers import reverse
from myproject.myapp.models import Document
from myproject.myapp.forms import DocumentForm
def list(request):
# Handle file upload
if request.method == 'POST':
form = DocumentForm(request.POST, request.FILES)
if form.is_valid():
docfile=request.FILES['docfile']
# Redirect to the document list after POST
return HttpResponseRedirect(reverse('myproject.myapp.views.list'))
else:
form = DocumentForm() # A empty, unbound form
# Load documents for the list page
documents = Document.objects.all()
# Render list page with the documents and the form
return render_to_response(
'list.html',
{'documents': documents, 'form': form},
context_instance=RequestContext(request)
)
Form -
from django import forms
class DocumentForm(forms.Form):
docfile = forms.FileField(
label='Select a file'
)
If you are asking for the extension try this:
>>> f = "this.that.ext"
>>> e = f.split(".")[-1]
>>> f
'this.that.ext'
>>> e
'ext'
>>>
You know how to get the complete filename. What you want is the last string after the "."
Related
I'm attempting to process an uploaded CSV file using Django. The main logic of how I go about doing this is expressed in both the models.py and views.py scripts. Once I've uploaded the file, I'm unable to process any of the content (in my views.py). Here are the two scripts, but if there's any more information I can provide, I'd be happy to.
In my models.py file, I have two classes, one for the document itself, and the other class for the fields in the file.
models.py:
from django.db import models
import os
class Document(models.Model):
docfile = models.FileField(upload_to='documents')
class DocumentEntry(models.Model):
document = models.ForeignKey(Document, on_delete=models.CASCADE)
field = models.CharField(max_length=250, default="TEST")
Next, in my views.py I get the file that was uploaded via the request.FILES['docfile'] and pass it to the handle_files() function. However, when I try to loop through the reader, I'm unable to access any of the elements in the file that was uploaded.
views.py:
from django.shortcuts import render
from django.conf import settings
from django.http import HttpResponseRedirect
from django.core.urlresolvers import reverse
import csv
from .models import Document, DocumentEntry
from .forms import UploadFileForm
def process_file(request):
# Handle file upload
if request.method == 'POST':
form = UploadFileForm(request.POST, request.FILES)
if form.is_valid():
handle_files(request.FILES['docfile'])
# Redirect to the document list after POST
return HttpResponseRedirect(reverse('process_files'))
else:
form = UploadFileForm() # A empty, unbound form
# Load documents for the list page
documents = Document.objects.all()
# Render list page with the documents and the form
return render(
request,
'upload/process_files.html',
{'documents': documents, 'form': form}
)
def handle_files(csv_file):
newdoc = Document(docfile=csv_file)
newdoc.save()
reader = csv.DictReader(open(csv_file))
for row in reader:
field = row['field']
entry = DocumentEntry(document=newdoc, field=field)
entry.save()
Updated
Here is full example of handler function:
def handle_files(csv_file):
newdoc = Document(docfile=csv_file)
newdoc.save()
with open(newdoc.docfile.name) as f:
reader = csv.DictReader(f)
for row in reader:
field = row['field']
entry = DocumentEntry(document=newdoc, field=field)
entry.save()
open() expects the path to the file, not the actual file data, which is contained in request.FILES['docfile'].
Replace:
reader = csv.DictReader(open(csv_file))
with:
import io
io_file = io.TextIOWrapper(csv_file.file)
reader = csv.DictReader(io_file)
I am trying to have a simple form that once filled, will direct to a different webpage or remain on the same page if invalid. The page should have a text box and submit form and once a user enters anything it should direct you to a separate page.
My directory structure is as follows:
appName/
app/
forms.py
urls.py
views.py
templates/
app/
goodbye.html
name.html
library.html
thanks.html
appName/
settings.py
urls.py
My app/urls.py is as follows:
from django.conf.urls import url
from . import views
app_name = 'app'
urlpatterns = [
url(r'^$', views.index2, name = 'index'),
url(r'^hello/$', views.hello, name = 'hello'),
url(r'^goodbye/$', views.goodbye, name = 'goodbye'),
#url(r'^library$', views.library, name = 'library'),
url(r'^library/$', views.library, name = 'library'),
url(r'^library/(?P<book_id>[0-9]+)/$', views.book, name = 'book'),
url(r'^getname/$', views.get_name, name = 'get_name'),
url(r'^your-name/$',views.get_name, name='get_name'),
url(r'^thanks/$',views.say_thanks,name='thanks'),
#url(r'^thanks/(?P<name_id>[a-zA-Z]+)/$', views.say_thanks,name='thanks'),
]
My forms.py is :
from django import forms
class NameForm(forms.Form):
your_name = forms.CharField(label = 'Your name', max_length=100)
My app/views.py is:
from django.http import HttpResponse
from django.template import loader
from django.shortcuts import render
from django.http import HttpResponseRedirect
#forms
from .forms import NameForm
# Create your views here.
def index2(request):
return HttpResponse("hello world")
def hello(request):
text = """<h1>Welcome to my app! </h1>"""
return HttpResponse(text)
def goodbye(request):
template = loader.get_template("app/goodbye.html")
context = {
'output' : 'This is output from goodby views.py request handler'
}
return HttpResponse(template.render(context,request))
def library(request):
template = loader.get_template("app/library.html")
context = {
'output' : 'Welcome to the libary!!'
}
return HttpResponse(template.render(context, request))
def book(request, book_id):
return HttpResponse("You're looking at book %s. " % book_id)
def get_name(request):
# if this is a POST request we need to process the form data
if request.method == 'POST':
# create a form instance and populate it with data from the request:
form = NameForm(request.POST)
# check whether it's valid:
if form.is_valid():
#process the data in form.cleaned_data as required
locationGo = "/thanks/"
template = loader.get_template("app/thanks.html")
return HttpResponse(template.render({'name':'name'},request))
else:
form = NameForm()
template = loader.get_template("app/name.html")
context = {'form': form}
return HttpResponse(template.render(context, request))
def say_thanks(request):
template = loader.get_template("app/thanks.html")
return HttpResponse(template.render({'name': 'name'},request))
My templates include:
name.html :
<form action = "/getname/" method = "post">
{% csrf_token %}
{{ form }}
<input type = "submit" value = "Submit" />
</form>
goodbye.html
<h1>Goodbye to Template Romance</h1>
Go Back
thanks.html
Thanks {{name}}!
What I would like is for:
A user to visit to : website.com/getname/ to show the name.html file (which it does)
If a user hits submit to stay on the same page (website.com/getname/) (which it doesn't - it gives: ValueError at /getname/ ->The view app.views.get_name didn't return an HttpResponse object. It returned None instead.
If a user enters in the submit field, to be redirected to website.com/thanks/ (which it sort of does. It currently loads the thanks.html template, but the URL stays on website.com/getname/)
Inside the get_name(request): function, the POST and GET if...else doesn't seem to be firing based on the Submit button, and it doesn't seem to be loading the correct page, OR change the current URL address once it gets processed. I have tried using HttpRedirect() which works, however, I would also like to pass the forms data (which is another issue).
Any suggestions would be a big help!
Your first problem is that you are not returning a response when the request method is post and the form is invalid. You can fix that by changing the indentation of your view, so that you always return a response at the end of your view.
def get_name(request):
# if this is a POST request we need to process the form data
if request.method == 'POST':
...
else:
form = NameForm()
template = loader.get_template("app/name.html")
context = {'form': form}
return HttpResponse(template.render(context, request))
If you want to redirect to the /thanks/ view, then you can use the redirect shortcut.
if form.is_valid():
return redirect('thanks')
Note that it isn't possible to redirect and pass the form data (see this question for an explanation). You should do any processing you want with the data before redirecting. You could use the messages framework to create a message 'Thanks <name>' before redirecting.
This works because you have name='thanks' in your url pattern.
You can simplify your views by using the render shortcut. Instead of
template = loader.get_template("app/name.html")
context = {'form': form}
return HttpResponse(template.render(context, request))
you can simply do:
return render(request, "app/name.html", context)
Remember to add the imports for the shortcuts:
from django.shortcuts import redirect, render
I want to make a application when the user upload the file, so get it to system, and execute the linux command like "xxd".
cSite---cSite---settings.py
| |__etc...
|
--myapp---
Here is my file upload codes.
views.py :
from django.shortcuts import render_to_response
from django.template import RequestContext
from django.http import HttpResponseRedirect
from django.core.urlresolvers import reverse
from myapp.models import Document
from myapp.forms import DocumentForm
def list(request):
# Handle file upload
if request.method == 'POST':
form = DocumentForm(request.POST, request.FILES)
if form.is_valid():
newdoc = Document(docfile = request.FILES['docfile'])
newdoc.save()
# Redirect to the document list after POST
return HttpResponseRedirect(reverse('myapp.views.list'))
else:
form = DocumentForm() # A empty, unbound form
# Load documents for the list page
documents = Document.objects.all()
# Render list page with the documents and the form
return render_to_response(
'list.html',
{'documents': documents, 'form': form},
context_instance=RequestContext(request)
)
models.py :
from django.db import models
class Document(models.Model):
docfile = models.FileField(upload_to='documents/%Y/%m/%d')
As you know, this code is widespreaded simple file upload form.
Now, here is my trouble.
Want to use the salt when I save the file
(Can I use the "Keeping Original Filename for FileField in Django"? and if I can, how to apply this?
I don't know what is representing the file path variable in FileField on ./views.py/ It always contain the error. (The command must be executed at the right after same time of file uploaded in user view.)
i am doing a copy cat with Need a minimal Django file upload example
i changed the view.py with my code to dump the csv file into sqlite database.
i have already created the table in default sqlite database.
import sqlite3
import csv
import sys
from django.shortcuts import render_to_response
from django.template import RequestContext
from django.http import HttpResponseRedirect
from django.core.urlresolvers import reverse
from myproject.myapp.models import Document
from myproject.myapp.forms import DocumentForm
def list(request):
# Handle file upload
if request.method == 'POST':
form = DocumentForm(request.POST, request.FILES)
if form.is_valid():
newdoc = Document(docfile = request.FILES['docfile'])
newdoc.save()
gfile= csv.reader(open(newdoc))
gon = sqlite3.connect("database.sqlite")
gon.text_factory = str
gon.execute("DELETE FROM abc where rowID > 0 ")
gon.executemany("insert into abc values (?, ?, ?, ?, ?)", gfile)
gon.commit()
gon.close()*
return HttpResponseRedirect(reverse('myproject.myapp.views.list'))
else:
form = DocumentForm() # A empty, unbound form
# Load documents for the list page
documents = Document.objects.all()
# Render list page with the documents and the form
return render_to_response(
'myapp/list.html',
{'documents': documents, 'form': form},
context_instance=RequestContext(request)
)
my code is starting from first introduction of gfile
Error # line 20 : Coercing to unicode : need string or buffer, Document Found
please help
You are passing Document instance to open. Instead you should pass the file, that was uploaded directly to csv.reader:
gfile = csv.reader(request.FILES['docfile'])
I am trying to pass the id through reverse. But it's not working. I'm getting this error
Reverse for 'reg.views.thanks' with arguments '(20,)' and keyword arguments '{}' not found.
Here is my views.py:
from django.http import HttpResponse, Http404, HttpResponseRedirect
from django.core.urlresolvers import reverse
from reg.models import registration, registrationform
from django.shortcuts import render_to_response, get_object_or_404
from django.template import RequestContext
def registration(request):
if request.method == 'POST':
form = registrationform(request.POST)
if form.is_valid():
data = form.save()
id = data.id
return thanks(request,id)
else:
form = registrationform()
return render_to_response('registration.html', {'form' : form}, context_instance=RequestContext(request))
def thanks(request, id):
p = get_object_or_404(registration, pk=id)
return render_to_response('thanks.html', {'reg' : p})
Here is my urls.py:
from django.conf.urls import patterns, include, url
url(r'^registration/$', 'reg.views.registration'),
url(r'^thanks/$', 'reg.views.thanks'),
url(r'^$','django.views.generic.simple.direct_to_template', {'template' : 'index.html'}),
)
Here is thanks.html:
<html>
<body>
<p>Thank you for registration mr.{{reg.username}}</p>
</body>
</html>
and I'm also showing my models.py:
from django.db import models
from django.forms import ModelForm
class registration(models.Model):
username = models.CharField(max_length=100)
password = models.CharField(max_length=100)
def __unicode__(self):
return self.name
class registrationform(ModelForm):
class Meta:
model = registration
Thanks.
from this links (django tutorial):
https://docs.djangoproject.com/en/dev/topics/http/urls/#django.core.urlresolvers.reverse
example:
def myview(request):
return HttpResponseRedirect(reverse('arch-summary', args=[1945]))
so your code goes to:
in urls.py:
url(r'^thanks/(?P<id>\d+)$', 'reg.views.thanks', name='my_thanks_url')
in your function:
return HttpResponseRedirect(reverse('my_thanks_url', args=[id]))
This line
return HttpResponseRedirect(reverse('reg.views.thanks', args=(id,)))
Is trying to construct a url to your view reg.views.thanks, with the id variable used as a parameter.
This line in urls.py
url(r'^thanks/$', 'reg.views.thanks'),
Does not have anywhere for that parameter to go.
The first thing that you need to figure out is whether you actually want to send an HTTP redirect to the browser to tell it to go to the 'thanks' page. If you really do, then you need a way to send that id in the URL. You can do it as part of the URL path itself, as #moguzalp suggests, or you can put it in the query string, like
/thanks/?id=12345
Or you can do other things, like stashing the id in the user's session, and pulling it out when they request the thanks page. That's a bit more complicated, though.
If you don't actually need to issue an HTTP redirect, then there's nothing stopping you from just calling the thanks() function from inside your view function, like this:
def registration(request):
if request.method == 'POST':
form = registrationform(request.POST)
if form.is_valid():
data = form.save()
id = data.id
return thanks(request, id)
else:
form = registrationform()
return render_to_response('registration.html', {'form' : form}, context_instance=RequestContext(request))
The URL won't change in the browser, but the correct ID will be used, and doesn't need to appear anywhere else, in the URL, the query parameters, or the session