It is used to upload external files such as doc files, excel files, video files, image files, audio files, JSON files, CSV files, or dataset files in the Django application in the project then we use the file handling concept.
Django supports all HTML elements to design form hence we can use <input type="file" /> for file upload UI.
We will create <form > with enctype="multipart/form-data" and the method will be posted.
Django provides FileSystemStorage class under django.core.files.storage and save() to save the external file under the Django media directory.
note: don't forget to write the setting.py code and urls.py code under the project.
Step for file uploading on Django:-
1) Create a design file using .html under the templates folder.
<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body>
<form method="POST" enctype="multipart/form-data">
{% csrf_token %}
<input type="file" name="myfile" />
<button type="submit">Save</button>
</form>
{% if uploaded_file_url %}
<p>File uploaded at: <a href="{{ uploaded_file_url }}">{{ uploaded_file_url }}</a></p>
{% endif %}
</body>
</html>
Code file
from django.conf import settings
from django.core.files.storage import FileSystemStorage
def fileupload(request):
if request.method == 'POST' and request.FILES['myfile']:
myfile = request.FILES['myfile']
fs = FileSystemStorage()
filename = fs.save(myfile.name, myfile)
uploaded_file_url = fs.url(filename)
return render(request, 'scsapp/fileupload.html', {
'uploaded_file_url': uploaded_file_url
})
return render(request, 'scsapp/fileupload.html')
setting.py
MEDIA_URL = '/media/'
MEDIA_ROOT = os.path.join(BASE_DIR, 'media')
urls.py
from django.contrib import admin
from django.urls import path,include
from django.conf import settings
from django.conf.urls.static import static
urlpatterns = [
path('scs/', include('scsapp.urls')),
path('admin/', admin.site.urls),
]
if settings.DEBUG:
urlpatterns += static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)
Database Operation in Django?
1) Create Model Class
class Fupload(models.Model):
filepath = models.CharField(max_length=100)
2) Create urls.py
path('viewfile',views.viewfile,name='viewfile')
3)
def uploadfile(request):
if request.method == 'POST' and request.FILES['myfile']:
myfile = request.FILES['myfile']
fs = FileSystemStorage()
filename = fs.save(myfile.name, myfile)
data= Fupload(filepath=filename)
data.save()
res = fs.url(filename)
return render(request, 'dbapp/fileupload.html',{'key':res})
return render(request, 'dbapp/fileupload.html'
)
def viewfile(request):
s = Fupload.objects.all()
return render(request,"dbapp/viewfile.html",{'res':s})
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Document</title>
</head>
<body>
{% for p in res %}
<img src="/media/{{p.filepath}}" height="100" width="100" />
{% endfor %}
</body>
</html>
Hi Sir,
ReplyDeleteHow to upload the multiple files.
for multiple file you should use almost same code but in template file you should use following code.
ReplyDeleteyou should use multiple attribute into input type="file"
and in views.py you can use loop statement.
def handle_uploaded_file(f):
with open('some/file/name.txt', 'wb+') as destination:
for chunk in f.chunks():
destination.write(chunk)