django admin - how to get the current url parameters in python

Question

Here I made an example of what I have in my admin.py:

@admin.register(Car)
class CarAdmin(CustomAdmin):
    list_display = ('get_color',)
    def get_color(self, obj):
        return mark_safe('<a href="/admin/myapp/car/?color={}">{}</a>'.format(obj.color, obj.color))

I have produced a link that can be used to display cars with a specific color. Let's say the page is currently showing cars that cost less than $20k and I want this link to preserve the current filter. Any ideas how to do that? Maybe there is a way to get the current url from python.

Note that I know I can write a javascript to modify the links after the page is loaded, but that is a terrible solution.

ahmed · Accepted Answer · 2022-11-30 18:10:50Z

9

You can save the full path or request before the get_color is called, something like:

class CarAdmin(CustomAdmin):
    list_display = ('get_color',)

    def get_queryset(self, request):
        self.full_path = request.get_full_path()
        return super().get_queryset(request)

    def get_color(self, obj):
        # TODO: Handle empty query parameters 
        return mark_safe(f'<a href="{self.full_path}&color={obj.color}">{obj.color}</a>')

edited Nov 30, 2022 at 18:10

answered Dec 12, 2017 at 19:55

ahmed

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