Django 1.9: Add a custom button to run a python script on clicking in admin site of a app/model

Question

I have created an app in a django project. This app has four models. I can add/modify/delete from admin site for all four models. But for one of the four models (say - ModelXYZ), I need to add a custom button for each entry in this table. Currently, I can see "Save", "Delete", etc. buttons for each entry present in ModelXYZ table. I need to add another button "Run", clicking on it will execute a python script written in that app only. Currently, I am running that script like this -

python manage.py shell
>>> execfile("my_app/my_script_1.py")
>>> execfile("my_app/my_script_2.py")

The script name is also stored in ModelXYZ table for each entry.

Django docs say that the admin site is customizable, but I am not very sure how can I run a python script by clicking on a button.

Answer 1

I solved it like this and it solves my purpose :-

class Site(models.Model):
    name = models.CharField(null=False, blank=False, max_length=256, unique=True)
    script = models.CharField(null=False, blank=False, max_length=256)
    def __unicode__(self):
       return u"{0}".format(self.name)

Then in my_app/admin.py, I wrote:-

from django.contrib import admin

from .models import *

def run(self, request, queryset):
    id=request.POST.get('_selected_action')
    siteObj=self.model.objects.get(pk=id)
    self.message_user(request, "Running: " + siteObj.script)
    execfile(siteObj.script)
run.short_description = "Run the script"

class SiteAdmin(admin.ModelAdmin):
    actions = [run]
    model = Site

# Register your models here.
admin.site.register(Site, SiteAdmin)