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authorMaarten <mzaanen@users.noreply.github.com>2015-09-05 12:35:58 +0200
committerTim Graham <timograham@gmail.com>2015-09-05 10:19:38 -0400
commitfe58d96e50e33b05f2a45f1493eca39ec9b3d030 (patch)
tree12e3fb6a6ff04ba8309b62b600ce8b2cd1af29e1
parente687794f6b04566ec1bf6d19a620c165d61f3c79 (diff)
Fixed #25355 -- Made two tweaks to docs/topics/db/aggregation.txt.
-rw-r--r--docs/topics/db/aggregation.txt4
1 files changed, 2 insertions, 2 deletions
diff --git a/docs/topics/db/aggregation.txt b/docs/topics/db/aggregation.txt
index 409f58aa4c..65a1eb7ee6 100644
--- a/docs/topics/db/aggregation.txt
+++ b/docs/topics/db/aggregation.txt
@@ -73,7 +73,7 @@ In a hurry? Here's how to do common aggregate queries, assuming the models above
{'price_per_page': 0.4470664529184653}
# All the following queries involve traversing the Book<->Publisher
- # many-to-many relationship backward
+ # foreign key relationship backwards.
# Each publisher, each with a count of books as a "num_books" attribute.
>>> from django.db.models import Count
@@ -242,7 +242,7 @@ price field of the book model to produce a minimum and maximum value.
The same rules apply to the ``aggregate()`` clause. If you wanted to
know the lowest and highest price of any book that is available for sale
-in a store, you could use the aggregate::
+in any of the stores, you could use the aggregate::
>>> Store.objects.aggregate(min_price=Min('books__price'), max_price=Max('books__price'))