#django

self.object seems to work

but you're never going to be able to do a redirect based on self.object unless get_absolute_url() is defined

or you create the URL in get_success_url()

https://docs.djangoproject.com/en/dev/topics/ht... - if you pass redirect() a model, then it will attempt to call get_absolute_url(), otherwise you need to pass a view name or a URL

thx, yeah i saw that, i'm just not sure how get_absolute_url() works

get_absolute_url is defined on the model?

https://docs.djangoproject.com/en/dev/ref/model... - try looking at the examples and see? it's basically a way to define a URL for a model

yep you define it as part of your model

but the url needed is simply for the createview

then you can still define it in get_success_url() on the CreateView...you just can't use redirect(object) and expect it to direct to your object...you'd need to have get_success_url() return the actual URL you want

mmikeym: thanks for your help. i'm just not sure i follow.

What can I define it in get_success_url() on the CreateView?

joelones: instead of returningg redirect(object) you'd need to return the url path you want CreateView to go to on success...so '/path/to/where/I/want/to/go'

right, that i'm doing, just dont know how to return the object

to get the previously added object you'd probably need to pass a kwarg with the object when you're doing the redirect...not too sure how you'd do that to be honest

k, appreciate the help

no prob

Hi guys. I'm trying to loop through a result set, and when a certain field is true for the first time, I want to output something once and continue with the loop. I can't set a variable to use as a flag in the template. Any ideas how I could go about this?

I have a category and subcategory drop down. I would like to change the subcategory dropdown list based on the category selected. What is the best way to do this?

trewq, I think it's not really Django-specific question. However I'd say you should add JS onChange handler to your category dropdownbox and send AJAX request to your backend, then return subcategories in JSON or rendered HTML and replace your subcategory innerHTML with that.

Is there a way to have urls be dynamic? For example if I had a url like: asite.com/chicken/soup and asite.com/beef/soup return a list of soups I would like asite.com/soup to return the list of soups for both chicken and beef and split_pea soups and any other soups.

You can use a middleware class to see if a part of the URL matches some list and short-circuit the urlconf.

I'm sorry, I probably missed yesterday's answers, but can anyone advice a user registration app esp. if I target a Django 1.7 / Python 3.4 project ?

I have field on model as first_name = models.CharField(max_length=100). Db backend is sqlite3. If a db record is created via django shell, first_name is set to empty string. Is this due to sqlite3?

What were you expecting to happen?

Throw an error which says field is required

not form field, I mean

For historical reasons, model validation isn't enforced, you have to call it yourself.

"first_name" varchar(100) NOT NULL,

obj.save() doesn't call clean ?

What would be the best way to get all the ids of a manyToMany field?

i have a post that has a gallery. I pass down the landing page of a website.

I could loop trough all the posts in the view and make a array with the ids, but that doesn't seem verry efficient

Getting them from the template also doesn't make much sense to me

gingitsune: sounds fine to do it within the view

gingitsune: also, have a look at values_list for your use case

Is there a way to query an object by it's foreign key and only pull the objects for which all the foreign keys exist?

URE