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authorDominik <ueqnn@student.kit.edu>2022-04-19 11:34:14 +0200
committerMariusz Felisiak <felisiak.mariusz@gmail.com>2022-04-19 20:22:09 +0200
commit2fc7cb9d395a5e90e25c41482c77b255a9f9d4dc (patch)
treede8fad7f81c187b72c16e3015c7280af4523395a
parent38f12b2a41e33e51d5636f162a578673e4e5a73b (diff)
downloaddjango-2fc7cb9d395a5e90e25c41482c77b255a9f9d4dc.tar.gz
[4.0.x] Fixed #33644 -- Corrected FAQ about displaying ManyToManyField in list_filter.
Backport of 7d26d5f8f17637a768f9d46e96547ae12e2418ae from main
-rw-r--r--docs/faq/admin.txt4
1 files changed, 2 insertions, 2 deletions
diff --git a/docs/faq/admin.txt b/docs/faq/admin.txt
index ea9690bbe6..7c6f7a12c7 100644
--- a/docs/faq/admin.txt
+++ b/docs/faq/admin.txt
@@ -49,10 +49,10 @@ My "list_filter" contains a ManyToManyField, but the filter doesn't display.
============================================================================
Django won't bother displaying the filter for a ``ManyToManyField`` if there
-are fewer than two related objects.
+are no related objects.
For example, if your :attr:`~django.contrib.admin.ModelAdmin.list_filter`
-includes :doc:`sites </ref/contrib/sites>`, and there's only one site in your
+includes :doc:`sites </ref/contrib/sites>`, and there are no sites in your
database, it won't display a "Site" filter. In that case, filtering by site
would be meaningless.