django-modal-actions 0.1.5

A Django app for adding modal actions to the admin interface

Django Modal Actions

Django Modal Actions is a reusable Django app that provides a convenient way to add modal-based actions to your Django admin interface. It allows you to create custom actions that open in a modal dialog, enhancing the user experience and functionality of your Django admin.

Features

• Easy integration with Django admin
• Support for both list-view and object-view actions
• Customizable modal forms
• AJAX-based form submission
• Skip confirmation dialog for immediate action execution

Requirements

• Python (>= 3.7)
• Django (>= 3.2)

Installation

1. Install the package using pip:

   pip install django-modal-actions
   

2. Add 'django_modal_actions' to your INSTALLED_APPS setting:

   INSTALLED_APPS = [
       ...
       'django_modal_actions',
       ...
   ]
   

Usage

1. In your admin.py, import the necessary components:

   from django.contrib import admin
   from django_modal_actions import ModalActionMixin, modal_action
   

2. Create a custom admin class that inherits from ModalActionMixin and your base admin class:

   @admin.register(YourModel)
   class YourModelAdmin(ModalActionMixin, admin.ModelAdmin):
       list_display = ['name', 'status']
       modal_actions = ["approve"]
       list_modal_actions = ["bulk_approve"]
   
       @modal_action(
           modal_header="Approve Item",
           modal_description="Are you sure you want to approve this item?"
       )
       def approve(self, request, obj, form_data=None):
           obj.status = 'approved'
           obj.save()
           return f"{obj} has been approved."
   
       @modal_action(
           modal_header="Bulk Approve",
           modal_description="Are you sure you want to approve the selected items?"
       )
       def bulk_approve(self, request, queryset, form_data=None):
           count = queryset.update(status='approved')
           return f"{count} items have been approved."
   

3. If you need a custom form for your action, create a form class:

   from django import forms
   
   class ApprovalForm(forms.Form):
       reason = forms.CharField(label="Approval Reason", required=True, widget=forms.Textarea)
   
       def clean_reason(self):
           reason = self.cleaned_data["reason"]
           if len(reason) < 10:
               raise forms.ValidationError("Reason must be at least 10 characters long")
           return reason
   

   Then, use it in your action:

   @modal_action(
       modal_header="Approve with Reason",
       modal_description="Please provide a reason for approval",
       form_class=ApprovalForm
   )
   def approve_with_reason(self, request, obj, form_data=None):
       obj.status = 'approved'
       obj.approval_reason = form_data['reason']
       obj.save()
       return f"{obj} has been approved with reason: {form_data['reason']}"
   

Conditional Fields Example

You can create forms with fields that are shown or hidden based on the value of another field. This is useful for creating dynamic forms that adapt to user input. Here's an example:

from django import forms
from django_modal_actions import conditional_field

class NotificationForm(forms.Form):
    notification_type = forms.ChoiceField(
        label="Notification Type",
        choices=[
            ('email', 'Email'),
            ('sms', 'SMS'),
            ('none', 'No notification')
        ],
        initial='none'
    )

    # This field will only be shown when notification_type is 'email'
    email_address = conditional_field(
        dependent_field='notification_type',
        values=['email']
    )(
        forms.EmailField(
            label="Email Address",
            required=False
        )
    )

    # This field will only be shown when notification_type is 'sms'
    phone_number = conditional_field(
        dependent_field='notification_type',
        values=['sms']
    )(
        forms.CharField(
            label="Phone Number",
            required=False
        )
    )

@admin.register(YourModel)
class YourModelAdmin(ModalActionMixin, admin.ModelAdmin):
    modal_actions = ['send_notification']

    @modal_action(
        modal_header="Send Notification",
        modal_description="Send a notification to the user",
        form_class=NotificationForm
    )
    def send_notification(self, request, obj, form_data=None):
        if form_data['notification_type'] == 'email':
            # Send email notification
            return f"Email will be sent to {form_data['email_address']}"
        elif form_data['notification_type'] == 'sms':
            # Send SMS notification
            return f"SMS will be sent to {form_data['phone_number']}"
        else:
            return "No notification will be sent"

Permissions Example

You can add custom permission checks to your modal actions using the permissions parameter of the modal_action decorator. Here's an example:

from django.contrib import admin
from django_modal_actions import ModalActionMixin, modal_action
from .models import YourModel

def can_approve(request, obj=None):
    return request.user.has_perm('yourapp.can_approve_items')

@admin.register(YourModel)
class YourModelAdmin(ModalActionMixin, admin.ModelAdmin):
    list_display = ['name', 'status']
    modal_actions = ['approve']

    @modal_action(
        modal_header="Approve Item",
        modal_description="Are you sure you want to approve this item?",
        permissions=can_approve
    )
    def approve(self, request, obj, form_data=None):
        obj.status = 'approved'
        obj.save()
        return f"{obj} has been approved."

In this example, the can_approve function checks if the user has the can_approve_items permission. The approve action will only be available to users who have this permission.

You can also use multiple permission checks by passing a list of functions:

def is_staff(request, obj=None):
    return request.user.is_staff

@modal_action(
    modal_header="Approve Item",
    modal_description="Are you sure you want to approve this item?",
    permissions=[can_approve, is_staff]
)
def approve(self, request, obj, form_data=None):
    obj.status = 'approved'
    obj.save()
    return f"{obj} has been approved."

In this case, the user must both have the can_approve_items permission and be a staff member to see and use the approve action.

Skip Confirmation Dialog

Sometimes you may want to execute an action immediately without showing a confirmation dialog. You can use the skip_confirmation parameter to achieve this:

from django.contrib import admin
from django_modal_actions import ModalActionMixin, modal_action

@admin.register(YourModel)
class YourModelAdmin(ModalActionMixin, admin.ModelAdmin):
    list_display = ['name', 'status']
    modal_actions = ['toggle_status']
    list_modal_actions = ['bulk_toggle_status']

    @modal_action(
        modal_header="Toggle Status",
        skip_confirmation=True
    )
    def toggle_status(self, request, obj, form_data=None):
        if obj.status == 'active':
            obj.status = 'inactive'
        else:
            obj.status = 'active'
        obj.save()
        return f"{obj} status toggled to {obj.status}"

    @modal_action(
        modal_header="Bulk Toggle Status",
        skip_confirmation=True
    )
    def bulk_toggle_status(self, request, queryset, form_data=None):
        for obj in queryset:
            obj.status = 'inactive' if obj.status == 'active' else 'active'
            obj.save()
        return f"Toggled status for {queryset.count()} items"

In this example, both actions will execute immediately when clicked, without showing a confirmation modal. The page will reload automatically after the action completes.

Important Note: You cannot use skip_confirmation=True together with form_class. The skip confirmation feature is designed for actions that don't require any user input. If you need to collect form data, the modal must be shown to display the form.

Custom Admin Templates

If you need to customize the admin templates while still using the modal actions, you can override the change_form_template and change_list_template in your ModelAdmin class. Here's how to do it:

1. In your admin.py, add the change_form_template or change_list_template attribute to your ModelAdmin class:

   @admin.register(YourModel)
   class YourModelAdmin(ModalActionMixin, admin.ModelAdmin):
       change_form_template = 'admin/yourapp/yourmodel/change_form.html'
       change_list_template = 'admin/yourapp/yourmodel/change_list.html'
       # ... rest of your ModelAdmin code
   

2. Create the custom template files in your app's template directory. For example:

   yourapp/
   └── templates/
       └── admin/
           └── yourapp/
               └── yourmodel/
                   ├── change_form.html
                   └── change_list.html
   

3. In your custom templates, extend the default admin templates and add the modal action buttons. Here's an example for change_form.html:

   {% extends "admin/change_form.html" %}
   {% load i18n admin_urls %}
   
   {% block object-tools %}
       <ul class="object-tools">
           {% block object-tools-items %}
               {{ block.super }}
               {% if modal_action_buttons %}
                   <li>{{ modal_action_buttons }}</li>
               {% endif %}
           {% endblock %}
       </ul>
   {% endblock %}
   

   And for change_list.html:

   {% extends "admin/change_list.html" %}
   {% load i18n admin_urls %}
   
   {% block object-tools %}
       <ul class="object-tools">
           {% block object-tools-items %}
               {{ block.super }}
               {% if list_modal_action_buttons %}
                   <li>{{ list_modal_action_buttons }}</li>
               {% endif %}
           {% endblock %}
       </ul>
   {% endblock %}
   

These custom templates will include the modal action buttons while allowing you to make other customizations to your admin interface.

Testing

To run the tests, execute:

python -m unittest discover django_modal_actions/tests

Contributing

Contributions are welcome! Please feel free to submit a Pull Request.

License

This project is licensed under the MIT License.