Django: how to increase object value - django

Question

I'm trying to implement a simple 'like' button, however, the value of the object is not changing. I have only a week of experience with django, any help is appreciated!
model>
class Mainnews(models.Model):
author = models.ForeignKey(Author, on_delete=models.DO_NOTHING, default= True)
title = models.CharField(max_length=200)
description = models.TextField()
image = models.ImageField(upload_to = 'photos/%Y/%m/%d/')
is_published = models.BooleanField(default = True)
publish_date = models.DateTimeField(default = datetime.now, blank = True)
views_counter = models.IntegerField(default=0)
likes = models.IntegerField(default=0)
def __str__(self):
return self.title
html>
<form action = "{% url 'like' %}" method = "POST">
{% csrf_token %}
<button type ='submit'>Like</button>
</form>
view>
def like(request):
if request.method == 'POST':
thelike = Mainnews.likes + 1
thelike.save()
url> on like redirect to homepage
path('news/', like, name = 'like'),

Answer 1

the ideal way to do it is using ajax ,so your page will not refresh every time you click the button,what missing in your question the condition of model which you want to increase ,in my code i set it news id you can change it ,check this :
<input id='news_id' name='news_id' >
<input type='button' onclick='Increase()' name='increase' >
<srcipt>
function Increase(){
var new_id= document.getElementById("news_id").value;
$.ajax({
type:"POST",
data:JSON.stringify({'data':new_id}),
success:function(responsedata){
// process on data
}
})
}
But if you want to use ajax you have to add a csrf_exempt decorator on your view
from django.views.decorators.csrf import csrf_exempt
#then write this before your view
#csrf_exempt
you can use F() object to generate a SQL expression that describes the required operation without actually having to pull them out of the database into Python memory check here you can use this :
if request.is_ajax():
new_id = json.loads(request.body.decode('utf-8'))["data"]
_reponse = Mainnews.objects.filter(id=new_id).update(likes=F('likes') + 1)
return JsonResponse({'data_to_return':'data'})

Answer 2

As stated in the comments by both Mahdi and Abdul, you shouldn't need a form to increase the value of Likes for a particular Mainnews instance. This can be done just by creating a URL and view specific to this purpose.
First, you will need to create a view specifically for increasing likes against a Mainnews instance. That view should look something like this:
def add_likes(request, news_id):
news_instance = Mainnews.objects.get(id=news_id)
news_instance.likes = news_instance.likes += 1
news_instance.save()
# Assuming that you want to display the same news instance again, you would need to send back the same instance to the news instance view
context = {
'news_instance': news_instance,
}
return render(request, 'your_template_here', context=context)
You will also need to create a URL path for updating likes. That should look something like this:
path('/news_item/add_like/<int:news_id>', add_likes, name='your-url-name'),
Then finally you would need to update your news template something like this (assuming you are using CSS to drive your front end):
Add Like
This will result in the page being refreshed by the user when they click on the "Like" button. If that is not the behavior you want, the answer submitted by Belhadjer would probably help prevent the refresh.