Python 用于循环的 Django 模板
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Django template for loop
提问by Nielsen Ramon
I have a template where I get certain variables back. One variable is instance.category which outputs: "words words words" which are values split by spaced.
我有一个模板，可以在其中获取某些变量。一个变量是 instance.category ，它输出：“words words words”，它们是按间距分割的值。
When I use the code below I get letter by letter back and not the words.
当我使用下面的代码时，我会逐个回复，而不是单词。
{% for icon in instance.category %}
<p>{{ icon }}</p>
{% endfor %}
Output
输出
<p>w</p>
<p>o</p>
<p>r</p>
<p>d</p>
<p>w</p>
....
I need:
我需要：
<p>word</p>
<p>word</p>
<p>word</p>
The Django plugin code
Django插件代码
from cmsplugin_filer_image.cms_plugins import FilerImagePlugin
from cms.plugin_pool import plugin_pool
from django.utils.translation import ugettext_lazy as _
from models import Item
class PortfolioItemPlugin(FilerImagePlugin):
model = Item
name = "Portfolio item"
render_template = "portfolio/item.html"
fieldsets = (
(None, {
'fields': ('title', 'category',)
}),
(None, {
'fields': (('image', 'image_url',), 'alt_text',)
}),
(_('Image resizing options'), {
'fields': (
'use_original_image',
('width', 'height', 'crop', 'upscale'),
'use_autoscale',
)
}),
(_('More'), {
'classes': ('collapse',),
'fields': (('free_link', 'page_link', 'file_link', 'original_link', 'target_blank'),)
}),
)
plugin_pool.register_plugin(PortfolioItemPlugin)
Any help is appreciated!
任何帮助表示赞赏！
采纳答案by augustomen
If your separator is always " "and categoryis a string, you don't actually need a custom template filter. You could simply call splitwith no parameters:
如果你的分隔符始终" "和category是一个字符串，你实际上并不需要自定义模板过滤器。您可以简单地split不带参数调用：
{% for icon in instance.category.split %}
<p>{{ icon }}</p>
{% endfor %}
回答by alecxe
You are passing a string instance.categoryinto the template and then iterating over its chars.
您将一个字符串传递instance.category到模板中，然后迭代它的字符。
Instead, pass a list to the template: instance.category.split()which will split your words words wordsstring into the list ['words', 'words', 'words']:
相反，将列表传递给模板：instance.category.split()这会将您的words words words字符串拆分为列表['words', 'words', 'words']：
>>> s = "words words words"
>>> s.split()
['words', 'words', 'words']
Or, you can define a custom filterthat will split a string into the list:
或者，您可以定义一个自定义过滤器，将字符串拆分为列表：
from django import template
register = template.Library()
@register.filter
def split(s, splitter=" "):
return s.split(splitter)
Then, use it in the template this way:
然后，以这种方式在模板中使用它：
{% for icon in instance.category|split %}
<p>{{ icon }}</p>
{% endfor %}
