How to render data in the texfield? - django

Question

I'm new to Django, using Django 2.7.1 and python 3.7.2, trying to make a thing, which will take data from field forms on my site, then handle this data, and finally render the result in another field form on the same page.
I've tried to make it by the model's methods, but didn't understood how.
Now i'm trying to do it by views. So i understood what i should do, but still don't recognize how to render the result.
this is my models.py
from django.db import models
# Create your models here.
class CombinatorCols(models.Model):
first_column = models.TextField("Column 1", null=True, blank=True)
second_column = models.TextField("Column 2", null=True, blank=True)
third_column = models.TextField("Column 3", null=True, blank=True)
fourth_column = models.TextField("Column 4", null=True, blank=True)
result = models.TextField("Result", help_text='Halo', blank=True,)
def __str__(self):
return self.first_column
this is my forms.py
from django import forms
from .models import CombinatorCols
class CombinatorForm(forms.ModelForm):
class Meta:
model = CombinatorCols
fields = ('first_column', 'second_column', 'third_column','fourth_column', 'result',)
this is my views.py
from django.shortcuts import render, redirect
from django.http import HttpResponse
from django.template import loader
from .models import Combinate,CombinatorCols
from .forms import CombinatorForm,ResultForm
# Create your views here.
def get_col(request):
#column = get_object_or_404(CombinatorCols)
if request.method == "POST":
form = CombinatorForm(request.POST)
if form.is_valid():
column = form.save()
column.sender = request.user
first_handle = [i for i in column.first_column.split('\n') if i]
second_handle = column.second_column
third_handle = column.third_column
fourth_handle = column.fourth_column
column.result = first_handle
column.save()
print(first_handle)
print('Naiiiceee')
else:
form = CombinatorForm()
return render(request, 'index.html', {'form': form})
and the last, my index.html
<form action="" method="post">
{% csrf_token %}
{{ form.as_table }}
<input type="submit" value="Submit">
</form>
I expect the updating of 'result' form on my site, but i see only the source page. All that has been written in the form fields is staying on their places and sends to the DB. It's all the expected behavior, except the result field isn't updating. Print statements shows correctly.

Answer 1

When you get the POST call, after you have processed the form, in order to update the form on the same view, you need to reinitialise it with the new data that you want to show.
After saving result, reinitialise the form:
column.result = first_handle
column.save()
data = request.POST.dict().copy() # add these
data["result"] = first_handle # three lines
form = CombinatorForm(data) # to reinitialise the form
print(first_handle)
print('Naiiiceee')
This will solve you problem.
By the way, try using class based generic views for CRUD operation. You can learn about them here. They will make your life much more easier.