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.)
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 would like to create a website-app to run a bash script located in a server. Basically I want this website for:
Upload a file
select some parameters
Run a bash script taking the input file and the parameters
Download the results
I know you can do this with php, javascript... but I have never program in these languages. However I can program in python. I have used pyQT library in python for similar purposes.
Can this be done with django? or should I start learning php & javascript?
I cannot find any tutorial for this specific task in Django.
This can be done in Python using the Django framework.
First create a form including a FileField and the fields for the other parameters:
from django import forms
class UploadFileForm(forms.Form):
my_parameter = forms.CharField(max_length=50)
file = forms.FileField()
Include the UploadFileForm in your view and call your function for handling the uploaded file:
from django.http import HttpResponseRedirect
from django.shortcuts import render
from .forms import UploadFileForm
# Imaginary function to handle an uploaded file.
from somewhere import handle_uploaded_file
def upload_file(request):
if request.method == 'POST':
form = UploadFileForm(request.POST, request.FILES)
if form.is_valid():
my_parameter = form.cleaned_data['my_parameter']
# Handle the uploaded file
results = handle_uploaded_file(request.FILES['file'], title)
# Clear the form and parse the results
form = UploadFileForm()
return render(request, 'upload.html', {'form': form, 'results': results})
else:
form = UploadFileForm()
return render(request, 'upload.html', {'form': form})
Create the function to handle the uploaded file and call your bash script:
import subprocess
import os
def handle_uploaded_file(f, my_parameter):
file_path = os.path.join('/path/to/destination/', f.name)
# Save the file
with open(file_path, 'wb+') as destination:
for chunk in f.chunks():
destination.write(chunk)
# Call your bash script with the
output = subprocess.check_output(['./my_script.sh',str(file_path),str(my_parameter)], shell=True)
return output
Check out https://docs.djangoproject.com/en/1.10/topics/http/file-uploads/ for more examples and instructions on how the handle file uploads in Django.
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 "."
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