can any1 tell me the how to use celery in windows?? I got async issue many time

Can someone rectify the error, I am getting page not found error on running this url (http://127.0.0.1:8000/music/01)?

what version? Celery does not support Windows from v4 onwards IIRC

ya but I tried celery3.x but it still has some issue with other dependency pike kombu

Hello everyone I am uploading multiple files using single form field in dhango and successfully uploaded to target directory also but how can i assign media url for these pictures because only one media url will be assigned to one picture... Below is my code for uploading pictures def post(self, request, *args, **kwargs): context={} if request.method =='POST': form = AddRoomForm(request.POST) s = request.POST['service'] service = Service.objects.get(id=s) sKey = service.serviceKey if form.is_valid(): for count, x in enumerate(request.FILES.getlist("roomPictures")): def process(f): with open("/Users/JARVIS/Desktop/obap/media/images/rooms/"+sKey+"_"+ str(count), 'wb+') as destination: for chunk in f.chunks(): destination.write(chunk) process(x) form.save() return redirect('roomsdata')

I deleted the 0001.initial.py file from migrations

and dont forget to run makemigrations again

But still it shows the filenotfounderror of 0001_initial.py

Hello everyone I am uploading multiple files using single form field in dhango and successfully uploaded to target directory also but how can i assign media url for these pictures because only one media url will be assigned to one picture... Below is my code for uploading pictures def post(self, request, *args, **kwargs): context={} if request.method =='POST': form = AddRoomForm(request.POST) s = request.POST['service'] service = Service.objects.get(id=s) sKey = service.serviceKey if form.is_valid(): for count, x in enumerate(request.FILES.getlist("roomPictures")): def process(f): with open("/Users/JARVIS/Desktop/obap/media/images/rooms/"+sKey+"_"+ str(count), 'wb+') as destination: for chunk in f.chunks(): destination.write(chunk) process(x) form.save() return redirect('roomsdata')

I can't share photos to this channel

woah woah what are you trying to do??

woah woah what are you trying to do??

Actually i am uploading multiple images with single file input element So when i upload images they are storing in targeted directory Upto this everything is ok.. But when i view the files uploaded post it is showing no files are associated to the attribute pictures... I realized that i am only uploading multiple images but bot linking them with post object, as we do it using upload_to field in model

thats true, so you already search how to properly store FileField now?

you should share your model first

Sachin: models.py code class Rooms(models.Model): roomPictures = models.ImageField(upload_to='images/rooms/', null=True, blank=True) roomNo = models.textInput(max_length=255, null=True, blank=True)

so you are uploading 5 room data?

then why your model only have one room pic?

Because we can't put limit to user that how many images he can upload For example consider facebook post, each post can contain many images uploaded by user only on single input i am trying to apply same logic here

Because we can't put limit to user that how many images he can upload For example consider facebook post, each post can contain many images uploaded by user only on single input i am trying to apply same logic here

so, imagine the images is simply texts, if you have ["text1","text2", ... ] how do you want to store those texts in your database?

Ya in this case what i am thinking is 1. for all these images i am creating a directory which contains unique random number as name before uploading 2.and for storing each image in that directory i am using for loop 3. In third step i am trying to attach the first uploaded image path as url in the imagefiled for Rooms model

But in third step i am failing because the imagefield not storing the path i given

class Rooms(models.Model): room_pictures = models.TextField() room_no = models.PositiveIntegerField() given input of room pictures: ["text1","text2", ... ] can you show me how you solve this problem?

But why for text We don't nees multiple titles for single post right....?

I want to make a relational database where two tables named patient and metabolite table will have one to one relation . The webportal will provide form where user has to submit the two excel sheet and based on the data submitted they can browse the metabolites based on some filters. so is it possible if browsing retrieves some metabolite eg methane . is there any way when clicked on methane it should search the methane just like google does

But why for text We don't nees multiple titles for single post right....?

i m just telling the row1 of patient table correspond to row1 of metabolite one

titles?

I mean Why ["text1","text2", ... ] only one text field is enough per post right

i want to know is it possible to incorporate custom google search in my web app

you are the one who want to upload 5 images and not one, i just simplify it to texts so you able to find wheres mistake you made

No no inchidi I mean Then tell me If one user wanted to upload 2 images Another user wnated to upload 3 images Another user wanted to upload 7 images How to write model filefield for this case

you should make pictures many to one relationship with room. consider retake django tutorial tho https://t.me/pythonres/21

class Rooms(models.Model): room_pictures = models.TextField() room_no = models.PositiveIntegerField() given input of room pictures: ["text1","text2", ... ] can you show me how you solve this problem?

if you already know how to solve this problem, you can back to your main problem which is how to store image properly

can anyone tell me how i can display pdf file in html page from django model

you can use <iframe> for that

Hi! I am a beginner in Django, am following step-by-step the tutorial; I am in the 2nd step: https://docs.djangoproject.com/en/2.1/intro/tutorial02/; there is this link: http://127.0.0.1:8000/admin/, whenever I try to open it, I have this:

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