From 566437adc9dd15dc18388bc256835a9448de0f6e Mon Sep 17 00:00:00 2001
From: Tim Graham <timograham@gmail.com>
Date: Thu, 2 Jun 2022 13:03:34 -0400
Subject: [PATCH] [4.1.x] Fixed #33757 -- Clarified Client.post() file upload
example.
Backport of 61badf1d58e79b84874afa6a1e00b79f20e786d1 from main
---
docs/topics/testing/tools.txt | 7 +++----
1 file changed, 3 insertions(+), 4 deletions(-)
diff --git a/docs/topics/testing/tools.txt b/docs/topics/testing/tools.txt
index 836dab54e4..dcbfefd295 100644
--- a/docs/topics/testing/tools.txt
+++ b/docs/topics/testing/tools.txt
@@ -241,15 +241,14 @@ Use the ``django.test.Client`` class to make requests.
Submitting files is a special case. To POST a file, you need only
provide the file field name as a key, and a file handle to the file you
- wish to upload as a value. For example::
+ wish to upload as a value. For example, if your form has fields
+ ``name`` and ``attachment``, the latter a
+ :class:`~django.forms.FileField`::
>>> c = Client()
>>> with open('wishlist.doc', 'rb') as fp:
... c.post('/customers/wishes/', {'name': 'fred', 'attachment': fp})
- (The name ``attachment`` here is not relevant; use whatever name your
- file-processing code expects.)
-
You may also provide any file-like object (e.g., :class:`~io.StringIO` or
:class:`~io.BytesIO`) as a file handle. If you're uploading to an
:class:`~django.db.models.ImageField`, the object needs a ``name``