you code is good.
i find another one but not work for me(forget author name, sorry):
"""
Template tags for working with lists.
You'll use these in templates thusly::
{% load listutil %}
{% for sublist in mylist|parition:"3" %}
{% for item in mylist %}
do something with {{ item }}
{% endfor %}
{% endfor %}
"""
from django import template
register = template.Library()
@register.filter
def partition(thelist, n):
"""
Break a list into ``n`` pieces. The last list may be larger than
the rest if
the list doesn't break cleanly. That is::
>>> l = range(10)
>>> partition(l, 2)
[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9]]
>>> partition(l, 3)
[[0, 1, 2], [3, 4, 5], [6, 7, 8, 9]]
>>> partition(l, 4)
[[0, 1], [2, 3], [4, 5], [6, 7, 8, 9]]
>>> partition(l, 5)
[[0, 1], [2, 3], [4, 5], [6, 7], [8, 9]]
"""
try:
n = int(n)
thelist = list(thelist)
except (ValueError, TypeError):
return [thelist]
p = len(thelist) / n
return [thelist[p*i:p*(i+1)] for i in range(n - 1)] + [thelist[p*(i
+1):]]
@register.filter
def partition_horizontal(thelist, n):
"""
Break a list into ``n`` peices, but "horizontally." That is,
``partition_horizontal(range(10), 3)`` gives::
[[1, 2, 3],
[4, 5, 6],
[7, 8, 9],
[10]]
Clear as mud?
"""
try:
n = int(n)
thelist = list(thelist)
except (ValueError, TypeError):
return [thelist]
newlists = [list() for i in range(n)]
for i, val in enumerate(thelist):
newlists[i%n].append(val)
return newlists
On 9月6日, 下午11时01分, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]>
wrote:
> I wrote this tag a couple months ago that should be of some
> help.http://www.djangosnippets.org/snippets/296/
> You can at least use it as a starting point for your own tag.
>
> On Sep 6, 8:23 am, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote:
>
> > need your help, thx
>
> > On 9月6日, 下午3时21分, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote:> sorry, i
> > am a newbie to python.
> > > i have a queryset with 6 record, i want to show in 3 col and 2 row,
> > > how to do this, thx!!!
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