How to create a Django reverse url with a parameter

Question

I am having problems constructing a reverse url with a parameter

I have tried the answers to this question but with no luck

urls.py

urlpatterns = [
    path('publication-item/<str:publication_title>', PublicationItemView.as_view(), name='publication-item'),
]

xxx.html

<a href="{% url 'publication-item {{ publication.publication_title | urlencode }}' %}">

gives the error

Reverse for 'publication-item {{ publication.publication_title | urlencode }}' not found. 'publication-item {{ publication.publication_title | urlencode }}' is not a valid view function or pattern name.

Whereas

xxx.html

<a href="{% url 'publication-item' %}?publication_title={{ publication.publication_title | urlencode }}">

produces

Reverse for 'publication-item' with no arguments not found. 1 pattern(s) tried: ['publicationspublication\-item/(?P<publication_title>[^/]+)$']

The following works, but of course it doesn't pass the parameter

urls.py

urlpatterns = [
    path('publication-item', PublicationItemView.as_view(), name='publication-item'),
]

xxx.html

<a href="{% url 'publication-item' %}" class="btn btn-primary">

Where am I going wrong?

willeM_ Van Onsem · Accepted Answer · 2021-02-26 17:17:34Z

3

You should not use double curly brackets, you are already in a template tag, hence Django will interpret the variable:

<a href="{% url 'publication-item' publication.publication_title %}">

You do not need to URL encode yourself, Django will automatically URL encode the parameters.

answered Feb 26, 2021 at 17:17

willeM_ Van Onsem

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