def test_positive_list_filters_with_pagination(self):
"""Make sure filters list can be displayed with different items per
page value
:id: b9c7c6c1-70c2-4d7f-8d36-fa8613acc865
:BZ: 1428516
:expectedresults: `per-page` correctly sets amount of items displayed
per page, different `per-page` values divide a list into correct
number of pages
:CaseImportance: Critical
"""
role = make_role()
res_types = iter(PERMISSIONS.keys())
permissions = []
# Collect more than 20 different permissions
while len(permissions) <= 20:
permissions += [
permission['name']
for permission in Filter.available_permissions(
{'resource-type': next(res_types)})
]
# Create a filter for each permission
for perm in permissions:
make_filter({
'role': role['name'],
'permissions': perm,
})
# Test different `per-page` values
for per_page in (1, 5, 20):
with self.subTest(per_page):
# Verify the first page contains exactly the same items count
# as `per-page` value
filters = Role.filters({
'name': role['name'],
'per-page': per_page,
})
self.assertEqual(len(filters), per_page)
# Verify pagination and total amount of pages by checking the
# items count on the last page
last_page = (len(permissions) / per_page
+ int(len(permissions) % per_page != 0))
filters = Role.filters({
'name': role['name'],
'page': last_page,
'per-page': per_page,
})
self.assertEqual(
len(filters), len(permissions) % per_page or per_page)
def test_positive_list_filters_with_pagination(self):
"""Make sure filters list can be displayed with different items per
page value
:id: b9c7c6c1-70c2-4d7f-8d36-fa8613acc865
:BZ: 1428516
:expectedresults: `per-page` correctly sets amount of items displayed
per page, different `per-page` values divide a list into correct
number of pages
:CaseImportance: Critical
"""
role = make_role()
res_types = iter(PERMISSIONS.keys())
permissions = []
# Collect more than 20 different permissions
while len(permissions) <= 20:
permissions += [
permission['name']
for permission in Filter.available_permissions(
{'resource-type': next(res_types)})
]
# Create a filter for each permission
for perm in permissions:
make_filter({
'role': role['name'],
'permissions': perm,
})
# Test different `per-page` values
for per_page in (1, 5, 20):
with self.subTest(per_page):
# Verify the first page contains exactly the same items count
# as `per-page` value
filters = Role.filters({
'name': role['name'],
'per-page': per_page,
})
self.assertEqual(len(filters), per_page)
# Verify pagination and total amount of pages by checking the
# items count on the last page
last_page = (len(permissions) / per_page +
int(len(permissions) % per_page != 0))
filters = Role.filters({
'name': role['name'],
'page': last_page,
'per-page': per_page,
})
self.assertEqual(len(filters),
len(permissions) % per_page or per_page)
def make_role_with_permissions(self):
"""Create new role with a filter"""
role = make_role()
res_types = iter(PERMISSIONS.keys())
permissions = []
# Collect more than 20 different permissions
while len(permissions) <= 20:
permissions += [
permission['name']
for permission in Filter.available_permissions(
{"search": f"resource_type={next(res_types)}"})
]
# Create a filter for each permission
for perm in permissions:
make_filter({'role': role['name'], 'permissions': perm})
return {
'role': role,
'permissions': permissions,
}
