CODEWITHSUNDEEP
pythondjangodjango-templates
comar3
comar3

Reputation: 1

django multiple submit buttons

I have many buttons in my django template. Each button has a different name {{transaction.id}}. How can I refer to (submit) a single button in my view?

base.html

<h3>Open transactions</h3>
    {% for transaction in transactions_open %}
    {{ transaction.id }} {{transaction.timestamp}} {{ transaction.currency }} {{ transaction.sell_or_buy }} {{ transaction.bid }} {{ transaction.date_add }}
    <form action="" method="POST">
    {% csrf_token %}
    <button type="submit" name={{ transaction.id }} value="close">Close</button>
    </form>
    </p>
    {% endfor %}

views.py

class MainView(View):

def get(self, request):

    transactions_open = DealModel.objects.filter(open_or_closed='OPEN')
    transactions_closed = DealModel.objects.filter(open_or_closed='CLOSED')
    content_dict = {'transactions_open': transactions_open,
                    'transactions_closed': transactions_closed}
    return TemplateResponse(request, 'base.html', content_dict)

def post(self,request):

    if request.method == 'POST':
        transactions_open = DealModel.objects.filter(open_or_closed='OPEN')
        transactions_closed = DealModel.objects.filter(open_or_closed='CLOSED')

        if request.POST.get('name???') == 'close':
        content_dict['answer'] = "Closed!!!"
        return TemplateResponse(request, 'base.html', content_dict)

this is my example

Upvotes: 0

Views: 1640

Answers (2)

comar3
comar3

Reputation: 1

Thank you for your answer. Here is my solution to the problem:

class MainView(View):

    def post(self,request):

    if 'close' in request.POST.values():
        for key, value in request.POST.items():
            if value == 'close':
                deal_id = key # deal_id is the value of {{transaction.id}} in template file.

    if request.POST.get(deal_id) == 'close':
        closing_deal = DealModel.objects.get(id=deal_id)
        closing_deal.open_or_closed = 'Closed'
        closing_deal.save()
        content_dict['answer'] = "Closed!!!"
        return TemplateResponse(request, 'base.html', content_dict)

Upvotes: 0

A.Raouf
A.Raouf

Reputation: 2320

First

def post(self,request): It will be called when the request method is POST so you don't need to check it

Second

This solution depends on what I got it from the question
You don't need a form to have a close button You can handle it using javascript if you want to hide it

<ul>{% for transaction in transactions_open %}
<li class="close_button" data-pk="{{ transaction.id }}">{{ transaction.id }} {{transaction.timestamp}} {{ transaction.currency }} {{ transaction.sell_or_buy }} {{ transaction.bid }} {{ transaction.date_add }}

<a type="submit"  value="close">Close</a>

</li>
{% endfor %}</ul>

at the javascript file using jquery

$('.close_button a').onclick(function(){
     $(this).parent(".close_button").hide(); // something like that
$.ajax{
            url: "{% url 'app_label:view_name' %}",
            data: {'data':$(this).data("pk")},
            method: "POST",
            success: function (result,status,xhr) {
                
                alert('closed Successfully.');

            },

})

Handle closing the transaction at the view:

class MyView(View):
    def post(self, request, *args, **kwargs):
        if request.is_ajax():
             id = request.post.get("pk")
             Transaction.objects.get(id=id).close() #develop close method at your model classes

Upvotes: 1

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