Re: Can't do any manipulation on a db object
Oh.
Boy do I feel stupid.
Thanks for all your help. I'll try to be less careless next time.
On Oct 22, 8:43 am, Malcolm Tredinnick <[EMAIL PROTECTED]>
wrote:
> On Tue, 2008-10-21 at 23:33 -0700, guruyaya wrote:
> > This is actually worse
>
> You mean top-posting? Yes, it's a terrible habit and you should stop
> doing it.
>
>
>
> > blog/models.py
> >
> > from django.shortcuts import render_to_response as RenderToResponse
> > from myblog.blog.models import Post
> > def listposts(request):
> > a = Post.objects.all()
> > a = list(a)
> > a[0].title = 'hihi'
> > a[0].content = 'haha'
> > return RenderToResponse('blog/showlist.html',{'postlist': a});
>
> > Just shows me a blank page (with Hello on top).
>
> > What am I missing?
>
> I'll bet you didn't change your template, so it's still trying to
> iterate over postlist.all, rather than just postlist now. Sorry, I cut a
> corner and didn't point out all the changes necessary, just pointing you
> in the right direction. You're passing a list now, not a queryset, so
> you'll need to adjust the template if you really want to make these
> changes in the view like this (although, as I indicated, it's a pretty
> uncommon pattern when you start to do things out of experimentation
> mode, so you might want to avoid the hassle altogether).
>
> Regards,
> Malcolm
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Re: Can't do any manipulation on a db object
On Tue, 2008-10-21 at 23:33 -0700, guruyaya wrote:
> This is actually worse
You mean top-posting? Yes, it's a terrible habit and you should stop
doing it.
>
> blog/models.py
>
> from django.shortcuts import render_to_response as RenderToResponse
> from myblog.blog.models import Post
> def listposts(request):
> a = Post.objects.all()
> a = list(a)
> a[0].title = 'hihi'
> a[0].content = 'haha'
> return RenderToResponse('blog/showlist.html',{'postlist': a});
>
> Just shows me a blank page (with Hello on top).
>
> What am I missing?
I'll bet you didn't change your template, so it's still trying to
iterate over postlist.all, rather than just postlist now. Sorry, I cut a
corner and didn't point out all the changes necessary, just pointing you
in the right direction. You're passing a list now, not a queryset, so
you'll need to adjust the template if you really want to make these
changes in the view like this (although, as I indicated, it's a pretty
uncommon pattern when you start to do things out of experimentation
mode, so you might want to avoid the hassle altogether).
Regards,
Malcolm
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Re: Can't do any manipulation on a db object
Sorry if it appears twice. My message seem to have vanished.
What I said was that this is even worse
blog/views.py
--
from django.shortcuts import render_to_response as RenderToResponse
from myblog.blog.models import Post
def listposts(request):
a = Post.objects.all()
a = list(a)
a[0].title = 'hihi'
a[0].content = 'haha'
return RenderToResponse('blog/showlist.html',{'postlist': a});
It just shows nothing. Not even the posts I didn't change.
Any ideas?
On Oct 22, 8:17 am, Malcolm Tredinnick <[EMAIL PROTECTED]>
wrote:
> On Tue, 2008-10-21 at 22:30 -0700, guruyaya wrote:
>
> [...]
>
> > blog/views.py
> > ---
> > from django.shortcuts import render_to_response as RenderToResponse
> > from myblog.blog.models import Post
> > def listposts(request):
> > a = Post.objects.all()
> > a[0].title = 'hihi'
> > a[0].content = 'haha'
> > print a[0].title
> > print a[0].content
> > return RenderToResponse('blog/showlist.html',{'postlist': a});
>
> [...]
>
> > As you may have guest, I expected the first post title to be 'hihi'
> > and it's content to be 'haha'. Well... it's not. I still get the same
> > post I have in my database. Yet the "print" lines give me the right
> > post content in the server debug messages.
>
> That's because a[0] is not updating 'a' in place. Instead, it's
> returning a new queryset, that contains a slice of the original one and
> then that clone is updated. You are still passing the original 'a' to
> the template, however.
>
> If you want to do updating in place, you'll need to convert 'a' to an
> object that supports those types of updates, namely a list:
>
> a = list(a)
> a[0].title = 'Now updated in place because a is a list!'
>
> Realise that this isn't the sort of thing that is normally going to be a
> problem because it's just not that normal a usage pattern. Typically,
> you will either be extracting a bunch of data in a queryset and passing
> that to a template, *or* you will be iterating through a queryset or
> getting one particular object, updating the object(s) and then saving
> them. Not mixing both at once.
>
> Regards,
> Malcolm
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Re: Can't do any manipulation on a db object
This is actually worse
blog/models.py
from django.shortcuts import render_to_response as RenderToResponse
from myblog.blog.models import Post
def listposts(request):
a = Post.objects.all()
a = list(a)
a[0].title = 'hihi'
a[0].content = 'haha'
return RenderToResponse('blog/showlist.html',{'postlist': a});
Just shows me a blank page (with Hello on top).
What am I missing?
On Oct 22, 8:17 am, Malcolm Tredinnick <[EMAIL PROTECTED]>
wrote:
> On Tue, 2008-10-21 at 22:30 -0700, guruyaya wrote:
>
> [...]
>
> > blog/views.py
> > ---
> > from django.shortcuts import render_to_response as RenderToResponse
> > from myblog.blog.models import Post
> > def listposts(request):
> > a = Post.objects.all()
> > a[0].title = 'hihi'
> > a[0].content = 'haha'
> > print a[0].title
> > print a[0].content
> > return RenderToResponse('blog/showlist.html',{'postlist': a});
>
> [...]
>
> > As you may have guest, I expected the first post title to be 'hihi'
> > and it's content to be 'haha'. Well... it's not. I still get the same
> > post I have in my database. Yet the "print" lines give me the right
> > post content in the server debug messages.
>
> That's because a[0] is not updating 'a' in place. Instead, it's
> returning a new queryset, that contains a slice of the original one and
> then that clone is updated. You are still passing the original 'a' to
> the template, however.
>
> If you want to do updating in place, you'll need to convert 'a' to an
> object that supports those types of updates, namely a list:
>
> a = list(a)
> a[0].title = 'Now updated in place because a is a list!'
>
> Realise that this isn't the sort of thing that is normally going to be a
> problem because it's just not that normal a usage pattern. Typically,
> you will either be extracting a bunch of data in a queryset and passing
> that to a template, *or* you will be iterating through a queryset or
> getting one particular object, updating the object(s) and then saving
> them. Not mixing both at once.
>
> Regards,
> Malcolm
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Re: Can't do any manipulation on a db object
On Tue, 2008-10-21 at 22:30 -0700, guruyaya wrote:
[...]
> blog/views.py
> ---
> from django.shortcuts import render_to_response as RenderToResponse
> from myblog.blog.models import Post
> def listposts(request):
> a = Post.objects.all()
> a[0].title = 'hihi'
> a[0].content = 'haha'
> print a[0].title
> print a[0].content
> return RenderToResponse('blog/showlist.html',{'postlist': a});
[...]
> As you may have guest, I expected the first post title to be 'hihi'
> and it's content to be 'haha'. Well... it's not. I still get the same
> post I have in my database. Yet the "print" lines give me the right
> post content in the server debug messages.
That's because a[0] is not updating 'a' in place. Instead, it's
returning a new queryset, that contains a slice of the original one and
then that clone is updated. You are still passing the original 'a' to
the template, however.
If you want to do updating in place, you'll need to convert 'a' to an
object that supports those types of updates, namely a list:
a = list(a)
a[0].title = 'Now updated in place because a is a list!'
Realise that this isn't the sort of thing that is normally going to be a
problem because it's just not that normal a usage pattern. Typically,
you will either be extracting a bunch of data in a queryset and passing
that to a template, *or* you will be iterating through a queryset or
getting one particular object, updating the object(s) and then saving
them. Not mixing both at once.
Regards,
Malcolm
--~--~-~--~~~---~--~~
You received this message because you are subscribed to the Google Groups
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To post to this group, send email to django-users@googlegroups.com
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Can't do any manipulation on a db object
I'm kinda new to django, and python, so I'm probably gonna make myself
look a bit stupid. Here goes anyway:
As a simple django (1.0) project, (using python 2.6) I've decided to
write a blog application. Here it goes:
blog/urls.py
--
from django.conf.urls.defaults import *
urlpatterns = patterns('myblog.blog',
(r'','views.listposts'),
)
blog/models.py
---
from django.conf.urls.defaults import *
urlpatterns = patterns('myblog.blog',
(r'','views.listposts'),
)
blog/models.py
from django.db import models
from django.http import HttpResponse
from tinymce.widgets import TinyMCE
import re
class Post(models.Model):
title = models.CharField(max_length=200)
content = models.TextField()
pub_date = models.DateTimeField('date
written',auto_now_add=True, null=True)
change_date = models.DateTimeField('date
changed',auto_now=True, null=True)
def __unicode__(self):
return self.title
def save(self, force_insert=False, force_update=False):
... Here come some long lines of code I think is irrelevant, but ask
me if you think they are
blog/views.py
---
from django.shortcuts import render_to_response as RenderToResponse
from myblog.blog.models import Post
def listposts(request):
a = Post.objects.all()
a[0].title = 'hihi'
a[0].content = 'haha'
print a[0].title
print a[0].content
return RenderToResponse('blog/showlist.html',{'postlist': a});
blog/showlist.html
---
Hello
{% for post in postlist.all %}
{{ post.title }}
{{ post.content|safe }}
{% endfor %}
As you may have guest, I expected the first post title to be 'hihi'
and it's content to be 'haha'. Well... it's not. I still get the same
post I have in my database. Yet the "print" lines give me the right
post content in the server debug messages.
Any ideas?
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