I am new to django.I want to upload two files in single page.I have created form for uploading one file.But with same code I have tried upload two files by making some changes.But I can't get it .
Please help me to submitting two files in one submit button in a single page
views.py
from __future__ import unicode_literals
from django.shortcuts import render
from django.conf import settings
from django.core.files.storage import FileSystemStorage
from django.conf.urls import url
#import csv
def simple_upload(request):
if request.method == 'POST' and request.FILES['myfile']:
#request.FILES['myfile'] and request.FILES["myfile1"]:
myfile = request.FILES['myfile']
#myfile1=request.FILES["myfile1"]
fs = FileSystemStorage()
filename = fs.save(myfile.name, myfile)
#filename1=fs.save(myfile1.name, myfile1)
uploaded_file_url = fs.url(filename)
#uploaded_file_url1 = fs.url(filename1)
#data = [row for row in csv.reader(myfile.read().splitlines())]
return render(request, 'myapp/simple_upload.html', {
'uploaded_file_url': uploaded_file_url,
})
#upload_file = request.FILES['upload_file']
#data = [row for row in csv.reader(upload_file.read().splitlines())]
return render(request, 'myapp/simple_upload.html')
def home(request):
return render(request,'myapp/home.html')
html
<!doctype>
<html>
{% block content %}
<body>
<div class="col-md-12">
<form action="{% url "home" %}" method="post" enctype="multipart/form-data">
{% csrf_token %}
<div class="row">
<label for="fileupload" class="btn btn-primary col-md-2 col-sm-4 uploadBtn" >Upload GSTR 2A</label><br>
<input type="file" class="fileupload" id="fileupload" name="myfile" required="True">
</div>
<div class="col-md-1 col-sm-2" style="text-align:center">
<i class="fa fa-check-circle checkIcon"></i>
</div>
<div class="col-md-1 col-sm-2">
<p class="cancel">X</p>
</div>
</div>
<div class="col-md-12">
<form action="{% url "home" %}" method="post" enctype="multipart/form-data">
{% csrf_token %}
<div class="row">
<label for="fileupload" class="btn btn-primary col-md-2 uploadBtn" >Upload Purchase Account</label><br>
<input type="file" class="fileupload" id="fileupload" name="myfile" required="True" >
<div class="col-md-1">
<a"<p class="cancel">X</p><a>
</div>
</div>
<div class="buttonGroup">
<button type="submit" class="btn btn-primary recBtn">Reconcile</button> <span class="backBtn"> Back</span>
</div>
</form>
</div>
{% endblock %}
</body>
</html>
You only need to have two type file inputs inside the form, and get it in the view exactly like the first you created, like this:
<form method="post" enctype="multipart/form-data" required="True">
{% csrf_token %}
<input type="file" name="myfile">
<input type="file" name="myfile2">
<button type="submit">Upload</button>
</form>
and in your views, you get it and save it like the first one:
myfile = request.FILES['myfile']
myfile2 = request.FILES['myfile2']
In case you need to add the input file outsite the form, you need to add the "form" attribute to the input file. like this:
<input type="file" name="myfile2" form="FORM_ID">
just replace FORM_ID with the id of your form.
Related
Created a website with products. I need to make a window for editing them on the site in order to change the manufacturer and other characteristics. This must be done in a pop-up window. I have data displayed, I change it, but nothing changes when I save it. How can this problem be solved.
My vievs:
def parts(request):
added = ''
error = ''
PartAllView = Part.objects.order_by('-id')
if request.method == 'POST' and 'parts_add' in request.POST:
form = PartForm(request.POST, request.FILES)
if form.is_valid():
form.save()
added = 'Добавлено'
else:
error = 'Данная запчасть уже добавлена'
if request.method == 'POST' and 'parts_edit' in request.POST:
PartPost = int(request.POST['parts_edit'])
PartID = Part.objects.get(id=PartPost)
if PartID:
PartID.save()
added = 'Запчасть успешно отредактирована'
else:
error = 'Ошибка редактирования'
form = PartForm()
data = {
'added': added,
'error': error,
'form': form,
'PartAllView': PartAllView,
}
return render(request, 'kross/parts.html', data)
My HTML:
{% if PartAllView %}
{% for el in PartAllView %}
<form method="post" enctype="multipart/form-data">
{% csrf_token %}
<div class="modal fade" id="partEdit{{ el.id }}">
<div class="modal-dialog modal-dialog-centered text-center" role="document">
<div class="modal-content modal-content-demo">
<div class="modal-header">
<h6 class="modal-title">Добавление запчасти</h6><button aria-label="Close" class="btn-close"
data-bs-dismiss="modal"><span aria-hidden="true">×</span></button>
</div>
<div class="modal-body">
<div class="row row-sm">
<div class="col-lg-6">
<div class="form-group">
<input type="text" class="form-control" name="brand" value="{{ el.brand }}">
</div>
</div>
<div class="col-lg-6">
<div class="form-group">
<input type="text" class="form-control" value="{{ el.number }}">
</div>
</div>
<div class="col-lg-12">
<div class="form-group">
<input type="text" class="form-control" value="{{ el.name }}"><br>
<input type="textarea" class="form-control" rows="2" value="{{ el.description }}">
</div>
</div>
</div>
{{ el.analog }}
...
You can use updateView to edit an existing data in your website by simply:
from django.views.generic.edit import UpdateView
From MyApp models import #Model
class editview(UpdateView):
model = #Your Model You want to edit
fields = [#Add the fields you want to edit]
template_name = 'edit.html'
success_url = ('Home')
In your edit Template add:
<form method="post">
{% csrf_token %}
{{ form.as_p }}
<input type="submit" value="Update">
I hope it help.
HTML form on Django.
I want to perform or create web page in which the button performs the action, and in backend I am handling the Django function.
The code of HTML file is:
<form name="bookappointment" class="form" method="POST">
{% csrf_token %}
<br>
<input type="hidden", name="selecteddoctornumber" value="{{i.doctornumber}}">
<div class="row">
<div class="col-md-1"></div>
<div class="col-md-4">
<button class="btn approve" name="approvebtn">Approved</button>
</div>
<div class="col-md-1"></div>
<div class="col-md-4">
<button class="btn notapprove" name="notapprovebtn">Not Approved</button>
</div>
<br>
</div>
<a class="btn cancel" href="requests">Cancel</a>
<br>
</form>
and the other side the Django function is:
if request.method == 'POST':
if request.POST.get('approvebtn'):
for i in doctor:
if pendingDoctorNumber == i.contactNumber:
i.status = 'Approved'
return redirect('request')
if request.POST.get('notapprovebtn'):
for i in doctor:
if pendingDoctorNumber == i.contactNumber:
i.status = 'Not Approved'
return redirect('request')
but its not working any action just get me back to same page
<form action="{% url 'bookappointment' %}" method="POST">
you have to define bookappointment in your urls.py which redirect to views.py where your function lies with name bookappointment.
I have form, user must fill it and submit but I have no reaction from this form
First I thought there is a problem with action directive of the form, so used redirect method in the views but no help
def organization_info(request):
organization_form = OrganizationInformationForm()
context = {
'organization_form': organization_form
}
if request.method == "POST":
print("POST")
organization_form = OrganizationInformationForm(request.POST, request.FILES)
if organization_form.is_valid():
print("VALID")
new_org = OrganizationInformation.objects.create(**organization_form.cleaned_data)
print("FILLED")
return redirect(organization_list)
return render(request, 'organization_form.html', context)
<form method="POST" enctype="multipart/form-data" class="form-horizontal">
{% csrf_token %}
<div class="form-group">
<label for="name" class="col-sm-4 control-label">Organization Name:</label>
<div class="col-sm-4">
{{ organization_form.name }}
</div>
.
.
<div class="form-group">
<div class="col-sm-4 col-sm-offset-4">
<button type="submit" class="btn btn-pink">Submit</button>
</div>
</div>
</form>
I only have the "POST" printed on the log no any errors
add form handler path in action and try again
<form method="POST" enctype="multipart/form-data" class="form-horizontal" action={'your form handler path'}>
{% csrf_token %}
<div class="form-group">
<label for="name" class="col-sm-4 control-label">Organization Name:</label>
<div class="form-group">
<div class="col-sm-4 col-sm-offset-4">
<button type="submit" class="btn btn-pink">Submit</button>
</div>
</div>
</form>
You need to add an action attribute to your form.
<form method="POST" enctype="multipart/form-data" class="form-horizontal" **action="url-of-handler"**>
</form>
More on this topic: https://www.w3schools.com/tags/att_form_action.asp
I want to select several input files in my flask application, when I want to have an array list of my selected input files, the array is empty. How can I fix that ?
HTML
<form method="POST">
<div class="form-group">
<h6>Select files:</h6> <input type="file" name="inputfiles[]" multiple=""><br><br>
</div>
<div class=" form-group">
<button type="submit" class="btn btn-light text-primary" >GO</button>
</div>
</form>
Python
#app.route('/gp_bagging_several_apps', methods=['POST','GET'])
def gp_bagging_several_apps():
if request.method == 'POST':
print("HELLO")
f = request.files.getlist("inputfiles[]")
print(f)
I also tried with "inputfiles" instead of "insteadfiles[]". I had the same problem.
Thank you
I think in your input tag, you should have multiple. In your form tag, have action and enctype.
You can try something like below:
<form action="{% url 'function' %}" method="post" enctype="multipart/form-data">
{% csrf_token %}
<p><input type="file" name="files" required multiple/></p>
<p><input type="submit" value="Upload" class="btn btn-primary btn-large"/></p>
</form>
And in the function:
if request.method == 'POST':
files = request.FILES.getlist('files')
for file in files:
# process your file
Let me know if it works.
I have used django-adaptops for this purpose. i have implemented the mechanism shown in docs. I need to know how can achieve this using browser , for example - i have a template where user has the option to select the csv file, once i click the import button i get the call to my view , but here i cannot read the file to upload to my model. how i can perform this.
this is my template code.
<form action="/manage/dashboard/importProspects" method="post" class="form-horizontal"
enctype="multipart/form-data"> {% csrf_token %}
<div class="row-fluid" style="margin-top: 5%">
<div class="span6">
<div class="control-group">
<p>{{ csvform.fname.label_tag }} {{ csvform.fname }} {{ csvform.fname.url }}</p>
</div>
</div>
<button type="submit" class="btn btn-small btn-success " style="margin-left:10px; ">
Import from CSV
</button>
</div>
</form>
csvForm is form object passed when this template is loaded.
the form code is below:
class ImportCsvForm(forms.Form):
fname = forms.FileField(label='CSV')
and /manage/dashboard/importProspects url will call my view importcsv which right now does not perform anything.
how can i handle this so that i can read the csv file, i cannot get the csvfile path. or am i missing some thing here ? please help.
Follow this,
in template:
<form action="/manage/dashboard/importProspects" method="post" class="form-horizontal"
enctype="multipart/form-data"> {% csrf_token %}
<div class="row-fluid" style="margin-top: 5%">
<div class="span6">
<div class="control-group">
<p>
<input type="file" name="csvfile" />
</p>
</div>
</div>
<input type="submit" class="btn btn-small btn-success " style="margin-left:10px; " />
</div>
</form>
in urls.py:
urlpatterns = patterns("mine.views",
url(r'^', 'csvupload'),)
in views.py file:
def csvupload(request):
if request.method == "POST":
f = request.FILES['csvfile']
with open("/tmp/csvfile.csv", 'wb+') as destination:
for chunk in f.chunks():
destination.write(chunk)
then use csvfile whatever you want.....Hope this helps to you.