嵌套循环中的Django模板计数器 [英] Django template counter in nested loops
问题描述
base_parts = [
{'important item':43},
{'less item':22,'less item':3,'less item':45}
]
在我的模板我可以这样做:
{%for base_part in base_parts%}
{%for k,v in base_part.items%}
{#... do stuff#}
{#我尝试获得运行总计的项目作为ID#}
内部ID:{%forloop.counter0%}< BR />
外部ID:{%forloop.parentloop.counter0%}< BR />
{%endfor%}
{%endfor%}
你可以看到,我想要的是我已经迭代的项目总数的总计,但是我包括的两种方法都返回重复。我知道我可以连接这些循环,但是我使用的是一个formset,并且真的希望id被索引0,1,2 ...等。
是否有一种在模板中实现这种类型计数的方法?
任何帮助非常感谢。
编辑
此时的输出如下所示:
outerID:0< br />
innerID:0< br />
outerID:0< br />
innerID:1< br />
outerID:1< br />
innerID:0< br />
outerID:1< br />
innerID:1< br />
outerID:1< br />
innerID:2< br />
我想要:
totalID:0< br />
totalID:1< br />
totalID:2< br />
totalID:3< br />
totalID:4< br />
totalID:5< br />
totalID:6< br />
totalID:7< br />
totalID:8< br />
totalID:9< br />
我找到了一个更好的解决方案, itertools
。 (比我以前的答案更好)
您可以将循环的当前状态设置为发送到视图上下文的itertools变量。
这次我尝试一个虚拟的Django项目,它的作用就像一个魅力。
views.py:
from django.shortcuts import render_to_response
import itertools
def home(request):
iterator = itertools.count()
base_parts = [
{'important item':43},
{'less item1':22,'less item2':3,'less item3':45},
{'最重要的项目':55}
]
返回render_to_response('index.html',
{'base_parts':base_parts,'iterator':iterator})
index.html:
{base for base_parts%中的%%
{%for k,v in base_part.items%}
{{iterator.next}} - {{v}}< br />
{%endfor%}
{%endfor%}
HTML输出:
0 - 43
1 - 22
2 - 45
3 - 3
4 - 55
排序值:
(这部分不是实际问题的答案,更像是在玩)
您可以使用Django的 SortedDict ,而不是Python的内置字典来保留项目订单。
views.py
from django.shortcuts import render_to_response
导入itertools
从django.utils.datastructures import SortedDict
def home(request):
iterator = itertools.count()
base_parts = [
SortedDict([('important item',43)]),
SortedDict([('less item1',22),
('less item2',3),
('less item3',45)]),
SortedDict([('最重要的项目',55)])
]
print base_parts [1]
return render_to_response 'index.html',
{'base_parts':base_parts,'iterator':iterator})
HTML输出:
0 - 43
1 - 22
2 - 3
3 - 45
4 - 55
编辑2014-May-25
您还可以使用 collections.OrderedDict ,而不是Django的SortedDict。
编辑2016- 6月28日
调用 iterator.next
在Python 3中不起作用。您可以创建你自己的迭代器类,继承自itertools.count:
import itertools
class TemplateIterator(itertools.count )
def next(self):
return next(self)
Hi I have a list of two dictionaries I am passing to a Django template:
base_parts = [
{'important item': 43},
{'lesser item': 22, 'lesser item': 3, 'lesser item': 45}
]
in my template I can do this:
{% for base_part in base_parts %}
{% for k, v in base_part.items %}
{# ...do stuff #}
{# I try to get a running total of items to use as an ID #}
inner ID: {% forloop.counter0 %}< br/>
outer ID: {% forloop.parentloop.counter0 %}< br/>
{% endfor %}
{% endfor %}
As you can see, what I want is a running total of the total number of items I have iterated through, but both methods I have included return duplicates. I know I could concatenate the loops, but I am using a formset and really would like the ids to be indexed 0,1,2...etc.
Is there a way to achieve this type of count in the template?
Any help much appreciated.
EDIT
output at the moment looks like:
outerID: 0<br />
innerID: 0<br />
outerID: 0<br />
innerID: 1<br />
outerID: 1<br />
innerID: 0<br />
outerID: 1<br />
innerID: 1<br />
outerID: 1<br />
innerID: 2<br />
I want:
totalID: 0<br />
totalID: 1<br />
totalID: 2<br />
totalID: 3<br />
totalID: 4<br />
totalID: 5<br />
totalID: 6<br />
totalID: 7<br />
totalID: 8<br />
totalID: 9<br />
I found a better solution with itertools
. (Better than my previous answer)
You can set current state of the loop to the itertools variable sent to the view context.
This time i tried on a dummy Django project and it works like a charm.
views.py:
from django.shortcuts import render_to_response
import itertools
def home(request):
iterator=itertools.count()
base_parts = [
{'important item': 43},
{'lesser item1': 22, 'lesser item2': 3, 'lesser item3': 45},
{'most important item': 55}
]
return render_to_response('index.html',
{'base_parts': base_parts, 'iterator':iterator})
index.html:
{% for base_part in base_parts %}
{% for k, v in base_part.items %}
{{ iterator.next }} - {{ v }}<br/>
{% endfor %}
{% endfor %}
HTML Output:
0 - 43
1 - 22
2 - 45
3 - 3
4 - 55
Sorted values:
(This part is not an answer to the actual question. It's more like I'm playing around)
You can use Django's SortedDict instead of Python's built-in dictionary to keep items order.
views.py
from django.shortcuts import render_to_response
import itertools
from django.utils.datastructures import SortedDict
def home(request):
iterator=itertools.count()
base_parts = [
SortedDict([('important item', 43)]),
SortedDict([('lesser item1', 22),
('lesser item2', 3),
('lesser item3', 45)]),
SortedDict([('most important item', 55)])
]
print base_parts[1]
return render_to_response('index.html',
{'base_parts': base_parts, 'iterator':iterator})
HTML Output:
0 - 43
1 - 22
2 - 3
3 - 45
4 - 55
Edit 2014-May-25
You can also use collections.OrderedDict instead of Django's SortedDict.
Edit 2016-June-28
Calling iterator.next
doesn't work in Python 3. You can create your own iterator class, inheriting from itertools.count:
import itertools
class TemplateIterator(itertools.count):
def next(self):
return next(self)
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