django-admin action in 1.1

Question

I am writing a action in django.I want to now about the rows which are updated by the action or say id field of the row.I want to make a log of all the actions.

I am having a field status which has 3 values :'activate','pending','reject'.I have made action for changing the status to activate.when i perform the action i want to have the log of rows updated so i need some value which can be stored in log such as id coresponding to that row

Answer 1

As far as i can understand you want make an admin log-entry for the object you update using your custom action. I actually did something like that, purely as django does it. As its your custom action you can add this piece of code.

Edit: Call this function after your action finishes, or rather i should say, after you change the status and save the object.

def log_it(request, object, change_message):
    """
    Log this activity
    """
    from django.contrib.admin.models import LogEntry
    from django.contrib.contenttypes.models import ContentType

    LogEntry.objects.log_action(
        user_id         = request.user.id, 
        content_type_id = ContentType.objects.get_for_model(object).pk, 
        object_id       = object.pk, 
        object_repr     = change_message, # Message you want to show in admin action list
        change_message  = change_message, # I used same
        action_flag     = 4
    )

# call it after you save your object
log_it(request, status_obj, "Status %s activated" % status_obj.pk)

You can always get which object you updated by fetching LogEntry object

log_entry = LogEntry.objects.filter(action_flag=4)[:1]
log_entry[0].get_admin_url()

Hope this helps.

Answer 2

it is very easy

just make a loop of your queryset,then u can access each field of that row and store it where u want

for e in queryset:

if (e.status != "pending"): flag = False