Django URL Reversal Issue with Literal Strings

The issue you're encountering is due to a misunderstanding of how Django's URL patterns and the url template tag work together.

The Problem

In your URL pattern:

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path("<str:space_name>/exhibitions/", views.exhibition_list, name="exhibition_list")

You've defined only one parameter: space_name. The /exhibitions/ part is a literal string in the URL pattern, not a parameter.

However, in your template, you're trying to pass two arguments to the url tag:

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{% url 'exhibition_list' space.name 'exhibitions' %}

This is incorrect because:

1. The URL pattern only expects one parameter (space_name)
2. You're trying to pass 'exhibitions' as a second parameter, but it's not a parameter in your URL pattern - it's a fixed part of the URL

The Solution

You should only pass the parameters that are defined in your URL pattern. In this case, just space_name:

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{% url 'exhibition_list' space.name %}

This will correctly generate a URL like /artwork/public/exhibitions/ where "public" is the value of space.name.

Why Your Error Occurs

The error message shows:

Reverse for 'exhibition_list' with arguments '('', 'exhibitions')' not found

This indicates two issues:

1. You're passing two arguments when the URL pattern only expects one
2. The first argument (space.name) appears to be empty (''), which might be a separate issue to investigate

Additional Note

If space.name is empty, that would also cause issues. Make sure that space.name contains a valid value before using it in the URL tag.