Django:TemplateSyntaxError，无效的过滤器 [英] Django: TemplateSyntaxError, Invalid filter

问题描述

我正在尝试使用变量从字典访问值，所有这些都在遵循Django的模板语言的HTML文件中.Django的模板语言不允许您使用变量作为键来访问字典，因此我决定使用Django过滤器来执行此操作.但是，我收到无效的过滤器"错误，无法弄清原因.

I'm trying to access a value from a dictionary using a variable, all in an HTML file that follows Django's Templating language. Django's templating language doesn't let you access a dictionary using a variable as a key, so I decided to use Django filters to do it. However, I am getting a "Invalid filter" error and I can't figure out why.

我的html文件

<table class="table">
<tbody>
    <tr>
        <th scope="row">Username</th>
        <th scope="row">Full Name</th>
        <th scope="row">Points</th>
        {% for subject in subjects %}
        <th scope="row">{{ subject }}</th>
        {% endfor %}
    </tr>
    {% for user in users %}
    <tr>
        <td>
            <a href="{% url 'profile' username=user.username %}">{{ user.username }}</a>
        </td>
        <td>{{ user.first_name }} {{ user.last_name }}</td>
        <td>{{ user.profile.points.total }}</td>
        {% for subject in subjects %}
        <td>{{ user.profile.points|keyvalue:subject }}</td>
        {% endfor %}
    </tr>
    {% endfor %}
</tbody>

我的filter.py

My filter.py

from django import template
register = template.Library()

@register.filter
def keyvalue(dict, key):    
    return dict[key]

我的错误消息

TemplateSyntaxError at /leaderboard/
Invalid filter: 'keyvalue'

谢谢您的帮助:D

按要求粘贴我的views.py和urls.pu

Pasting my views.py and urls.pu as requested

views.py

views.py

def leaderboard(request):
    global leaderboard_public
    users = sorted(User.objects.all(), key=lambda t: t.profile.points['total'], reverse=True)

    if request.method == "POST":
        leaderboard_public = not leaderboard_public
        return redirect(leaderboard)
    subjects = list(Subject.objects.values('subject_name'))
    subjects = [i['subject_name'] for i in subjects]
    print(subjects)
    return render(request, "leaderboard.html", {
        "users": users,
        "subjects": subjects,
        # "leaderboard_public": leaderboard_public,
    })

urls.py

from accounts import views as accountsViews
from puzzles import views as puzzlesViews

urlpatterns = [
url(r'^admin/', admin.site.urls),

url(r'^signup/$', accountsViews.signup_without_email, name="signup"),
url(r'^activate/(?P<uidb64>[0-9A-Za-z_\-]+)/(?P<token>[0-9A-Za-z]{1,13}-[0-9A-Za-z]{1,20})/$',
    accountsViews.activate, name='activate'),
url(r'^logout/$', authViews.LogoutView.as_view(), name='logout'),
url(r'^login/$', authViews.LoginView.as_view(template_name='login.html'), name="login"),
# url(r'^puzzles/$', puzzlesViews.puzzles, name="puzzles"),
url(r'^puzzle/(?P<puzzleID>.+)/edit/$', puzzlesViews.editPuzzle, name="editPuzzle"),
url(r'^puzzle/(?P<puzzleID>.+)/$', puzzlesViews.puzzle, name="puzzle"),
url(r'^scheduled/$', puzzlesViews.scheduled, name="scheduled"),
url(r'^closed/$', puzzlesViews.closed, name="closed"),
url(r'^leaderboard/$', accountsViews.leaderboard, name="leaderboard"),
# url(r'^leaderboard/(?P<subject>.+)/$', accountsViews.leaderboard_subject, name="leaderboard"),
# url(r'^submissions/$', puzzlesViews.submissions, name="submissions"),
url(r'^profile/(?P<username>.+)/submissions$', puzzlesViews.submissions, name="submissions"),
url(r'^profile/(?P<username>.+)/edit/$', accountsViews.editProfile, name="editProfile"),
url(r'^profile/(?P<username>.+)/$', accountsViews.profile, name="profile"),
url(r'^create/$', puzzlesViews.create, name="create"),
# url(r'^$', views.home, name="home"),
url(r'^$', puzzlesViews.puzzles, name="puzzles"),
] + static(settings.STATIC_URL, document_root=settings.STATIC_ROOT)

让我知道是否需要其他信息.

Let me know if there's any other information needed.

推荐答案

我尝试了{％load crispy_forms_filters％}，它对我有用！

I tried {% load crispy_forms_filters %} and it worked for me!

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