Am 05.02.2013 um 22:16 schrieb Conor Pollock:
> I'm currently trying to figure this out as well... I can pass a page query
> in the url and everything works as expected, but I can't figure out how to
> render the pagination controls in my template file.
>
> Did you
Hi,
If you have the request context processor installed, you can do this:
{% bootstrap_paginate object_list range=request.GET.range %}
Collin
On Thursday, March 5, 2015 at 4:35:33 AM UTC-5, SHINTO PETER wrote:
>
> {% load bootstrap_pagination %}
> {% bootstrap_paginate object_list range=10 %
here is my views/
from django.contrib.auth.models import User
from django.core.paginator import Paginator,EmptyPage, PageNotAnInteger
from friends.models import Friendship,UserProfile
from django.http import HttpResponse, HttpResponseRedirect, Http404
from django.shortcuts import render
def das
their practice account.
To maintain user inputted checkboxes on multiple pages, I must use POST. But,
pagination only uses GET. So, a hack on pagination is to wrap the Next/Previous
buttons in a sub-form and POST using the buttons. This works for pagination,
but then my normal “I’m done, submit
Hi!
I need to implement list (grid) with pagination, sorting and simple
search interface. I've looked at djangosnippets, generic list etc, but
so far I didn't find something that has all of these things. So my
question is if there are any ready components that allows to simply
create
Greetings,
Is it not possible to use Newforms with GET requests? I'm trying to
figure out how to paginate results from a search form, and I'm passing
GET values to the pagination view, instantiating an instance of the
Form object with request.GET as an argument (instead of t
on the platform
> with hundreds of sellers.
>
> I was randomizing the product category page when loaded so that items from
> all sellers could appear on the page.
>
> I was not using paginations and I have thousands of products so the page
> was obviously loading very slow.
>
on the page.
>>
>> I was not using paginations and I have thousands of products so the page
>> was obviously loading very slow.
>>
>> I added pagination and let python manage the query instead of it
>> happening on the server-side. This increased the page load but
Hi folks,
i want to build a searchform (with GET parameters) and also want to
use pagination. But i can't find an easy solution for this common
case.
My solution has a searchform, a view and a template. Can anybody tell
me if this is a good solution? I'm not sure because i fi
> I have many pages to be exported to the excel sheet. I use django and
> javascript.
> i use pagination. Can any one help me out ?
Just create a custom view that returns apropos content/headers.
There are several ways of exporting files that Excel can read.
CSV files:
==
+eas
hi,
I'm using Django pagination with jQuery. I can serialize the objects
list of the pagination object, but I'd like to serialize the whole
object to get more data (page number, total number of pages...). How
can I serialize the whole pagination object?
Thanks
***javascript***
https://docs.djangoproject.com/en/dev/topics/pagination/?from=olddocs#using-paginator-in-a-view
This
is a perfect example. You have to replace de model by the model of your
comments.
On Sunday, September 23, 2012 11:12:10 AM UTC-3, Scarl wrote:
>
> I wonder is it possible to disp
Should I rewrite django.contrib.comments.models? Or I just need to rewrite
my_django_project.myapp.models?
在 2012年9月23日星期日UTC+8下午10时38分41秒，Pablo Sanfilippo写道：
>
>
> https://docs.djangoproject.com/en/dev/topics/pagination/?from=olddocs#using-paginator-in-a-view
> This
> is a
Sorry, i thought that views handle comments. I see now that comments are
handled by the framework itself.
El domingo, 23 de septiembre de 2012 11:12:10 UTC-3, Scarl escribió:
>
> I wonder is it possible to display 10 comments on one page? If it can, how
> to do it? Using the django.core.paginat
Thanks for the answer!
I have found something on django's official website.
https://code.djangoproject.com/ticket/18143
It seems that the comments component are not paginateable with django's
own paginator...
在 2012年9月24日星期一UTC+8上午1时28分45秒，Pablo Sanfilippo写道：
>
> Sorry, i thought that views ha
On Monday, February 13, 2017 at 1:03:23 AM UTC+5:30, Kazi Atik wrote:
>
> here is my views/
>
> from django.contrib.auth.models import User
>
> from django.core.paginator import Paginator,EmptyPage, PageNotAnInteger
> from friends.models import Friendship,UserProfile
> from django.http import Htt
r the
full category or individual flashcards to be added to their practice
account.
To maintain user inputted checkboxes on multiple pages, I must use POST.
But, pagination only uses GET. So, a hack on pagination is to wrap the
Next/Previous buttons in a sub-form and POST using the buttons. This
semi-static forms flow path (form1, form2, …, formlast). Generally, when you do
pagination, there is no preset number of pages - it depends upon the data being
requested or available. So, I think that option would not work. But, if I’m
wrong, I’d love to know. Thanks.
And, as for using Java just so
27;s how you're supposed to program in JS? Been doing it that
way for years. ;-)
I may be mistaken, but I think the wizard link below requires a “static” or
semi-static forms flow path (form1, form2, …, formlast). Generally, when
you do pagination, there is no preset number of pages - it depends
ken, but I think the wizard link below requires a “static” or
semi-static forms flow path (form1, form2, …, formlast). Generally, when you do
pagination, there is no preset number of pages - it depends upon the data being
requested or available. So, I think that option would not work. But, if I’m
Hello,
I use a view in an application with which I can present the 20 pages
of a table at once.
from queues.models import Queue
def list(request, page=0, message=" "):
page = int(page)
q_list = ObjectPaginator(Queue.objects.all(), 20)
has_previous = q_list.has_previous_page(page)
I don't think there are pre-made components that will provide you with
all of those things, and being somewhat new to django I can only point
you to one bit of documentation and code that I found helpful for
pagination:
http://code.djangoproject.com/wiki/PaginatorTag
Regards,
Syd
On Jul
> I don't think there are pre-made components that will provide you with
> all of those things, and being somewhat new to django I can only point
> you to one bit of documentation and code that I found helpful for
> pagination:
>
> http://code.djangoproject.com/wiki/Pagin
-
> sort, filter and pagination should
> know about others parameters, eg. while sorting you shouldn't lost
> filter parameters etc.
Maybe you could put it at djangosnippets for others? I would be more
then happy to take a look at it :)
Przemek
--
AIKIDO TANREN DOJO - Poland - Warsaw
ch form, and I'm passing
> GET values to the pagination view, instantiating an instance of the
> Form object with request.GET as an argument (instead of the typical
> request.POST). I can't find any mention of GET support in the
> documentation, is this something that will eventu
it any "dictionary-like" object instance as its
> submitted data (request.GET and request.POST are both dictionary-like
> objects.)
>
> > I'm trying to
> > figure out how to paginate results from a search form, and I'm passing
> > GET values to the
anges) and
> > Sortable Headers (from djangosnippets) with some changes too, newforms
> > to createsearchform and generic list.
> > Changes I had to do were necessary because each of those components -
> > sort, filter and pagination should
> > know about others parame
I have a problem with generic.list_detail.object_list, a limiting
queryset, and pagination variables.
For some reason when I use the code below, results are restricted to
the sliced 100, but all the pagination variables show results as if the
slice wasn't there.
Example:
{{ pages }} = 12
As part of the official Django doc on Pagination
<https://docs.djangoproject.com/en/2.2/topics/pagination/#using-paginator-in-a-view>,
take a look at the listing function:
def listing(request):
contact_list = Contacts.objects.all()
paginator = Paginator(contact_list, 25) # S
Dear Team,
I have 4 cards on a single html page. I want to display data on each one
but since the data has many rows I want to apply pagination on each card.
How can I achieve this? I was thinking of using ListView but this displays
pagination on the whole page. I did not try any other codes
On Wed, Aug 5, 2009 at 10:03 AM, tom wrote:
>
> Hi folks,
>
> i want to build a searchform (with GET parameters) and also want to
> use pagination. But i can't find an easy solution for this common
> case.
> My solution has a searchform, a view and a template. Can anyb
The following code is for a search on q and attaches tuples from the
search on another model. The search results are appended together to
create the final data results "list".
def Bypub(request):
query = request.POST['q']
if query:
qset = (
Q(pubtitlestrip__icontains=q
Hello,
I have following view in views.py:
.
class ArticleArchiveIndexView(ArticleViewAbstractClass,
ArchiveIndexView):
queryset = Article.live.all()
date_field = 'pub_date'
context_object_name = 'articles_list'
paginate_by = ARTICLES_PER_PAGE
..
How do I use 'paginate_by'
I'm trying to implement Django's built in pagination feature with a raw
query set. I've researched the issue and the answer is I need to cast my
set as a list. Something like this:
paginator = Paginator(refg, 100) # Show 100 contacts per page
paginator._count = len(list(r
The code for rendering pagination controls in my app is the same everywhere:
{% if is_paginated %}
{% if page_obj.has_previous %}
Previous
{% endif %}
{% if page_obj.has_next %}
Previous
{% endif
Hello,
On nearly every page of the website I'm creating, I have a list of recent
articles in the sidebar that works as a complete archive. You can sort them
by newest/oldest, and page through the entire archive. The template code for
the pagination looks like this
{% for page_number in page
instead of
q_list = ObjectPaginator(Queue.objects.all(), 20)
Provide a order_by, if you would like them to be in descending order,
in order when they were created have,
q_list = ObjectPaginator(Queue.objects.all().order_by('-id'),
20)
Or better yet have a created_on field in the model, w
On Dec 3, 8:36 pm, "Silas" <[EMAIL PROTECTED]> wrote:
> I have a problem with generic.list_detail.object_list, a limiting
> queryset, and pagination variables.
>
> For some reason when I use the code below, results are restricted to
> the sliced 100, but all the pa
Thanks for the explanation Rajesh.
On Dec 4, 2:31 pm, "RajeshD" <[EMAIL PROTECTED]> wrote:
> On Dec 3, 8:36 pm, "Silas" <[EMAIL PROTECTED]> wrote:
>
> > I have a problem with generic.list_detail.object_list, a limiting
> > queryset, and paginati
8 AM drone4four wrote:
> As part of the official Django doc on Pagination
> <https://docs.djangoproject.com/en/2.2/topics/pagination/#using-paginator-in-a-view>,
> take a look at the listing function:
>
> def listing(request):
> contact_list = Contacts.objects.all()
>
nce the data has many rows I want to apply pagination on each card.
> How can I achieve this? I was thinking of using ListView but this displays
> pagination on the whole page. I did not try any other codes because I have
> no idea
>
> Kindly assist.
> --
> *Eugene*
>
>
>
ards, David
>
>> On Fri, Nov 26, 2021 at 6:41 AM Eugene TUYIZERE
>> wrote:
>> Dear Team,
>>
>> I have 4 cards on a single html page. I want to display data on each one but
>> since the data has many rows I want to apply pagination on each card. How
&
Jesse,
The error you are getting is from "q" not being in your post
QueryDictionary. This happens because when you click next you are not
rePOSTing the q variable. Either you can put the q variable into an input
box and POST it again by making your next link a submit button, or you could
change q
rieve with a
new view and use that view for the pagination. I can see that the
pagination works better with a simple retrieval. Can you tell me how
to create a new object from "list" that can be retrieved in a new
view?
Thanks.
On Mar 11, 10:20 pm, Micah Ransdell wrote:
> Jesse,
&g
Hello Micah,
I tried this in the template:
next
The browser URL is:
http://127.0.0.1:8000/Search/?q=?&page=2
and this in the template:
next
The browser URL is:
http://127.0.0.1:8000/Search/?q=harris&page=2,
which is correct for q, but the error for both is still:
Request Method: GE
Jesse,
You need to change your code in the view to look at the GET parameter rather
than POST. Change this line:
query = request.POST['q']
to:
query = request.GET['q']
Also, make sure when you are submitting to your search page that you are
submitting the page with GET in the method of your fo
Hello Micah,
I can get the q with GET, but I have too complicated of a search and I
need to use POST. I'm having much difficulty with my template code
with POST to work with paginator. I'll keep trying. Thanks for your
patience and help.
--~--~-~--~~~---~--~~
Uhm maybe this post will help its a tag that handles the
pagination on query objects. I've used it in a few of my projects and
its quite handy.
http://blog.awarelabs.com/?p=29
-Paul
On Mar 13, 12:13 pm, Jesse wrote:
> Hello Micah,
>
> I can get the q with GET, but I have t
Hello Paul,
Thanks!! I'll try it.
On Mar 13, 2:21 pm, pkenjora wrote:
> Uhm maybe this post will help its a tag that handles thepaginationon
> query objects. I've used it in a few of my projects and
> its quite handy.
>
> http://blog.awarelabs.com/?p=29
>
> -Paul
>
> On Mar 13, 12:13 pm,
Ok. page_obj is what I need.
On Oct 17, 10:34 pm, Andriyko wrote:
> Hello,
>
> I have following view in views.py:
> .
> class ArticleArchiveIndexView(ArticleViewAbstractClass,
> ArchiveIndexView):
> queryset = Article.live.all()
> date_field = 'pub_date'
> context_object_name = '
I need to paginate a list of images that is getting bigger and bigger and
is loading slowly. I tried the simple django-pagination module but even the
first page takes as long as the whole list to load. I was wondering where
to look but then thought maybe this pagination app only takes the whole
7;m trying to implement Django's built in pagination feature with a raw
> query set. I've researched the issue and the answer is I need to cast my
> set as a list. Something like this:
>
> paginator = Paginator(refg, 100) # Show 100 contacts per page
> paginator._count = len
Yep. My solution is a custom template tag
from django.template.defaulttags import register
from urllib.parse import urlencode
@register.simple_tag(takes_context=True)
def querystringmod(context, *args):
"""Modify current querystring:
{% querystringmod name value [name value [...]] %}
I've created a paginator for RawQuerySets.
https://github.com/seamusmb/django-paginator-rawqueryset
On Tuesday, March 26, 2013 5:51:41 AM UTC-4, chambe...@gmail.com wrote:
>
> I'm trying to implement Django's built in pagination feature with a raw
> query set. I've
ntire archive.
> The template code for the pagination looks like this
>
> {% for page_number in page_list %}
>
> {% ifnotequal page_number current_page %}
>
> href="?p={{ page_number }}&sort={{sort_order}}">{{page
lliam Battersea"
<[EMAIL PROTECTED]> wrote:
> Hello,
>
> On nearly every page of the website I'm creating, I have a list of recent
> articles in the sidebar that works as a complete archive. You can sort them
> by newest/oldest, and page through the entire archive. The temp
ote:
> > Hello,
> >
> > On nearly every page of the website I'm creating, I have a list of
> recent
> > articles in the sidebar that works as a complete archive. You can sort
> them
> > by newest/oldest, and page through the entire archive. The te
Take a look here too...
http://code.djangoproject.com/wiki/PaginatorTag
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You received this message because you are subscribed to the Google Groups
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To unsu
Ahh. I eventually did write an inclusion tag. I see I could have saved
myself some time, but it was nice to learn how to do it.
Thanks!
On Jan 10, 2008 9:53 AM, Paul Childs <[EMAIL PROTECTED]> wrote:
>
> Take a look here too...
>
> http://code.djangoproject.com/wiki/PaginatorTag
> >
>
--~--~---
I now how to use django pagination with ORM django after filtering the
required data and put in into *pagination div*
*queryset_list = employee.objects.all()*
*query=request.GET.get("q")*
*if query:*
*queryset_list=queryset_list.filter(*
* Q(name__icont
block content %}
Hello
{% end block %}
All I see when I render the page (go to the URL based on the slug I
entered in django-cms) is
FOO
I think it has to do with the pagination template tags it's wrapping
around my content before handing it off to the templat
Hi everybody:
I have a big problem with pagination using apache solr and haystack, this is
what it is happening: I have a site where the news are being indexing with
haystack and solr, but the problem is that when a I make the search the
pagination is showing me more match results that what really
Look for somewhere that you're evaluating the queryset. Pagination doesn't
need to evaluate it, since it can slice it, which turns into a start
and a limit on
the database side. Beware of using len() on a queryset (evaluates, IIRC), use
qs.count() instead (done on the DB).
On Wed, Oc
hi there,
I do want to use pagination, but I don't know how to solve a problem.
template-code:
{% for p in paginator.page_range %}
{{p}}
{% endfor %}
but obviously it should be:
{% for p in paginator.page_range %}
{{p}}
{% endfor %}
because in my-view I use the objectpaginator object
_staff`, `auth_user`.`is_active`, `auth_user`.`date_joined`
FROM `auth_user` WHERE `auth_user`.`username` LIKE BINARY %alpha%
1:
https://docs.djangoproject.com/en/2.2/topics/db/sql/#performing-raw-queries
--
Regards
Deep L Sukhwani
On Wed, 25 Sep 2019 at 00:47, leb dev wrote:
> I now how
On Sunday, 22 February 2009 21:00:24 Theme Park Photo, LLC wrote:
> {% block content %}
> Hello
> {% end block %}
It's not this extra space between end and block is it?
\d
--
Where I web: http://otherwise.relics.co.za/
Comics, tutorials, software and sundry
--~--~-~--~~
No! It was the fact that you really need all the stuff that's on the
sample template
On Feb 22, 2:29 pm, Donn wrote:
> On Sunday, 22 February 2009 21:00:24 Theme Park Photo, LLC wrote:
> > {% block content %}
> > Hello
> > {% end block %}
>
> It's not this extra space between end and block i
Hi All,
I have the following in a view:
objects = model.objects.all()
paginator = Paginator(objects,10)
return render_to_response(
'index.html',dict(
objects = paginator.page(page),
total_objects = len(objects),
)
)
Is that
What happens if you switch pagination off?
On Mon, Apr 19, 2010 at 19:29, Ariel wrote:
> Hi everybody:
> I have a big problem with pagination using apache solr and haystack, this is
> what it is happening: I have a site where the news are being indexing with
> haystack and s
How many results are returned when you search using the solr admin
interface?
Are you using the code that I posted for you previously?
Ray
On Apr 19, 6:29 pm, Ariel wrote:
> Hi everybody:
> I have a big problem with pagination using apache solr and haystack, this is
> what it is hap
using the code that I posted for you previously?
>
> Ray
>
> On Apr 19, 6:29 pm, Ariel wrote:
> > Hi everybody:
> > I have a big problem with pagination using apache solr and haystack, this
> is
> > what it is happening: I have a site where the news are being indexin
solr admin
> > interface?
>
> > Are you using the code that I posted for you previously?
>
> > Ray
>
> > On Apr 19, 6:29 pm, Ariel wrote:
> > > Hi everybody:
> > > I have a big problem with pagination using apache solr and haystack, this
> >
How many results are returned when you search using the solr admin
> > > interface?
> >
> > > Are you using the code that I posted for you previously?
> >
> > > Ray
> >
> > > On Apr 19, 6:29 pm, Ariel wrote:
> > > > Hi everybody:
> >
gt; the index.
>> > How could I know it ???
>> > Regards
>> > Ariel
>> >
>> >
>> >
>> >
>> >
>> > On Wed, Apr 21, 2010 at 4:40 AM, Ray McBride
>> wrote:
>> > > How many results are returned when you sear
t; > the index.
> > > How could I know it ???
> > > Regards
> > > Ariel
>
> > > On Wed, Apr 21, 2010 at 4:40 AM, Ray McBride
> > wrote:
> > > > How many results are returned when you search using the solr admin
> > > > interface?
gt; > Regards
> >> > Ariel
>
> >> > On Wed, Apr 21, 2010 at 4:40 AM, Ray McBride
> >> wrote:
> >> > > How many results are returned when you search using the solr admin
> >> > > interface?
>
> >> > > Are you using
On 2007-12-20 14:33:47 -0700, Julian <[EMAIL PROTECTED]> said:
>
> hi there,
>
> I do want to use pagination, but I don't know how to solve a problem.
>
> template-code:
>
> {% for p in paginator.page_range %}
> {{p}}
> {% endfor %}
>
&g
okay, thanks. i think this will bring me to a comfortable solution.
--~--~-~--~~~---~--~~
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
To unsubscri
I'm building a page with pagination and a filter form (2 GET requests). If
the URL includes both pagination and filter results, something like
`/questions/?page=2&all_questions=on`, it works fine. It also works if it
just has filter results, something like `/questions/?all_que
ecute(sql, params)
File "(...)\.env\lib\site-packages\django\db\utils.py", line 89, in
__exit__
raise dj_exc_value.with_traceback(traceback) from exc_value
File "(...)\.env\lib\site-packages\django\db\backends\utils.py", line 84,
in _execute
return self.cursor.
Chris Withers wrote:
> objects = model.objects.all()
> paginator = Paginator(objects,10)
> return render_to_response(
> 'index.html',dict(
> objects = paginator.page(page),
> total_objects = len(objects),
> )
> )
I'm guessing
.count is definitely the way to go. Although, I would probably pass it
to your template instead of determining it there.
On Sep 24, 10:32 am, Chris Withers wrote:
> Chris Withers wrote:
> > objects = model.objects.all()
> > paginator = Paginator(objects,10)
> > return render_to_re
Brian McKeever wrote:
> .count is definitely the way to go. Although, I would probably pass it
> to your template instead of determining it there.
What difference does it make?
Chris
--
Simplistix - Content Management, Batch Processing & Python Consulting
- http://www.simplistix.co
Chris Withers kirjoitti:
> Brian McKeever wrote:
>> .count is definitely the way to go. Although, I would probably pass it
>> to your template instead of determining it there.
>
> What difference does it make?
len(qs) evaluates queryset - thus pulling in _every_ object from DB to
Python - which
Jani Tiainen wrote:
> Chris Withers kirjoitti:
>> Brian McKeever wrote:
>>> .count is definitely the way to go. Although, I would probably pass it
>>> to your template instead of determining it there.
>> What difference does it make?
>
> len(qs) evaluates queryset - thus pulling in _every_ object
On Sep 25, 9:08 am, Jani Tiainen wrote:
> Chris Withers kirjoitti:
>
> > Brian McKeever wrote:
> >> .count is definitely the way to go. Although, I would probably pass it
> >> to your template instead of determining it there.
>
> > What difference does it make?
>
> len(qs) evaluates queryset - th
I can't think of a reason, if nothing else its a matter or taste/
preference. He said "I would probably...", so he may have been
implying that there was a technical reason but most likely he was just
stating his preference.
On Sep 25, 5:03 am, Chris Withers wrote:
> Jani Tiainen wrote:
> > Chri
.count uses a sql function that just counts the rows.
Doing len(model.objects.all()) pulls ALL of the objects from the
database including all of the associated data and then counts them.
It should be a significant performance difference for any large data
set.
On Sep 24, 11:12 am, Chris Withers
Oh, it's just a preference. I don't like calculating stuff in the
template. It violates the MVT pattern to some minor degree.
On Sep 25, 6:41 am, Bryan wrote:
> I can't think of a reason, if nothing else its a matter or taste/
> preference. He said "I would probably...", so he may have been
> i
6. When a user jumps from
page X to Y I would like to keep the pagination structure intact such that
the user may paginate back and forth between page 1 and 6 to see what is in
between. Basically this boils down to the following question:
how to find out what the page number would be of specific obje
in django, i am trying to list some queries of several objects like user lists,
categoies and Post list. the homepage will be contained couple of blocks or
boxes. each box will have different query list like Post list, User List,
category list. But only one context will have pagination and
ends\utils.py", line
> 76, in _execute_with_wrappers
> return executor(sql, params, many, context)
> File "(...)\.env\lib\site-packages\django\db\backends\utils.py", line
> 84, in _execute
> return self.cursor.execute(sql, params)
> File "(...)\.
ute
>> return self._execute_with_wrappers(sql, params, many=False,
>> executor=self._execute)
>> File "(...)\.env\lib\site-packages\django\db\backends\utils.py", line
>> 76, in _execute_with_wrappers
>> return executor(sql, params, many, context)
>
VidJa Hunter escribió:
>
> paginator.is_on_page(object) would return page 6 if object is element 31 of
> the given queryset.
>
I am leaving right now, but, what about something like this?
for page in paginator.page_range:
if object in paginator.page(page).object_list:
r
list like Post list,
> User List, category list. But only one context will have pagination and
> other won't and ofcourse the pagination will be working on Post list.
>
> here is the views.py:
>
> class BlogappListView(ListView):
> model = Category,CustomUs
lists, categoies and Post list. the homepage will be contained couple of
> blocks or boxes. each box will have different query list like Post list,
> User List, category list. But only one context will have pagination and
> other won't and ofcourse the pagination will be working on Post l
I'm using django and I have a search page that uses pagination. All
works well, but if I change the search term the first page is fine,
but if I go to the 2nd page I see results from the previous search. I
must manually reload the page to see the 2nd page for the new search
term. Is
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