如何使用Django中的if语句编写更好的模板逻辑？ [英] How do I write better template logic with if statements in Django?

问题描述

如何在Django模板中写得更有效率，这样就不会是一个巨大的if语句？我想禁用一个元素到request.path中的所有页面和一些页面的一些例外。这是我到目前为止

  {％if/ create-account /in request.path或/ lists / request.path中的request.path或request.path中的/ contact-us /或request.path％中的/ news /
 {％else％} 
 {％include'includes / element。 html'％} 
 {％endif％} 
  

必须有更好的方法。

在X，Y或Z之外的所有页面上显示此元素。

解决方案

我可以想到的最短的时间是在需要显示的那些url的上下文数据中包含一个上下文变量它

 ＃views 
 {'ignore_element'：':)'} 
 
＃模板
 {％if not ignore_element％} 
 {％include'includes / element.html'％} 
 {％endif％} 
  pre> 
 
 
这将工作，因为它只为那些不包括这个上下文值的人显示。
 
 
 
任何更好的解决方案将取决于为什么这4个应该免除显示的逻辑背后
 
How can I write this more efficient in Django templates so it won't be a huge if statement? I want to disable an element to ALL pages in request.path and some exceptions for some pages. This is what I got so far
{% if "/create-account/" in request.path or "/lists/" in request.path or "/contact-us/" in request.path or "/news/" in request.path %}
{% else %}
    {% include 'includes/element.html' %}
{% endif %}
There must be a better way. To clarify my needs:

Display this element on all pages except X, Y or Z.
解决方案
The shortest way I can think of at the minute is to include a context variable in the context data for those urls that do need to show it
# views
{ 'ignore_element': ':)' }

# template
{% if not ignore_element %}
    {% include 'includes/element.html' %}
{% endif %}
This would work since its only shown for those that don't include this context value.

Any better solution options would depend on the logic behind why these 4 should be exempt from showing it

                        
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