嵌套循环中的 Django 模板计数器 [英] Django template counter in nested loops

问题描述

我有两个字典的列表，我要传递给 Django 模板:

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}
]

在我的模板中，我可以这样做:

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 %}

如您所见，我想要的是我迭代过的项目总数的运行总和，但是我包含的两种方法都返回重复项.我知道我可以连接循环，但我使用的是表单集并且真的希望将 id 索引为 0、1、2...等.

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.

有没有办法在模板中实现这种类型的计数?

Is there a way to achieve this type of count in the template?

非常感谢任何帮助.

编辑

此刻的输出看起来像:

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 />

推荐答案

我用 itertools 找到了更好的解决方案.(比我之前的答案更好)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.这次我尝试了一个虚拟的 Django 项目，它的效果非常好.

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:

index.html:

{% for base_part in base_parts %}
    {% 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

<小时>

排序值:

(这部分不是对实际问题的回答.更像是我在玩)

(This part is not an answer to the actual question. It's more like I'm playing around)

您可以使用 Django 的 SortedDict 而不是 Python 的内置字典来保持项目顺序.

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 输出:

0 - 43
1 - 22
2 - 3
3 - 45
4 - 55

<小时>

编辑 2014 年 5 月 25 日

您也可以使用 collections.OrderedDict 而不是Django 的 SortedDict.

You can also use collections.OrderedDict instead of Django's SortedDict.

编辑 2016 年 6 月 28 日

调用 iterator.next 在 Python 3 中不起作用.您可以创建自己的迭代器类，继承自 itertools.count:

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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