wscall: don't notify by default (#19662)

This commit is contained in:
Lauréline Guérin 2020-09-18 10:49:20 +02:00
parent a4c013548e
commit 508dc53c19
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2 changed files with 2 additions and 2 deletions

View File

@ -72,7 +72,7 @@ def test_wscalls_new(pub, value):
# go to the page and add a webservice call
resp = app.get('/backoffice/settings/wscalls/')
resp = resp.click('New webservice call')
assert resp.form['notify_on_errors'].value == 'yes'
assert resp.form['notify_on_errors'].value is None
assert resp.form['record_on_errors'].value == 'yes'
resp.form['name'] = 'a new webservice call'
resp.form['description'] = 'description'

View File

@ -57,7 +57,7 @@ class NamedWsCallUI(object):
CheckboxWidget,
'notify_on_errors',
title=_('Notify on errors'),
value=self.wscall.notify_on_errors if self.wscall.slug else True)
value=self.wscall.notify_on_errors)
form.add(
CheckboxWidget,
'record_on_errors',