Django - 使用模板标签和'带'？

Question

我有一个自定义模板标签:

def uploads_for_user(user):
    uploads = Uploads.objects.filter(uploaded_by=user, problem_upload=False)
    num_uploads = uploads.count()
    return num_uploads

我想做这样的事情,所以我可以正确地复数:

{% with uploads_for_user leader as upload_count %}
    {{ upload_count }} upload{{ upload_count|pluralize }}
{% endwith %}

但是,uploads_for_user leader在此上下文中不起作用,因为'with'标记需要单个值 - Django返回:

TemplateSyntaxError at /upload/
u'with' expected format is 'value as name'

知道如何绕过这个吗？

Answer 1

你可以把它变成一个过滤器:

{% with user|uploads_for as upload_count %}

Answer 2

虽然过滤器仍然有效,但这个问题的当前答案是使用Django 1.4中引入的赋值标记.

所以解决方案与您原来的尝试非常相似:

{% uploads_for_user leader as upload_count %}
{{ upload_count }} upload{{ upload_count|pluralize }}

更新:根据文档分配标签自Django 1.9以来已被弃用(simple_tag现在可以将结果存储在模板变量中,应该使用它)

Answer 3

在Django 1.9中,django.template.Library.assignment_tag()被删除:simple_tag现在可以将结果存储在模板变量中,应该使用它.

所以,现在简单的标签,我们可以使用如下:

可以将标记结果存储在模板变量中,而不是直接输出它.这是通过使用as参数后跟变量名来完成的.这样做可以让您在自己认为合适的地方输出内容:

{% get_current_time "%Y-%m-%d %I:%M %p" as the_time %}
<p>The time is {{ the_time }}.</p>