gecko-dev/config/check_source_count.py
Justin Wood baa089903e Bug 1559975 - Make config/ lint with 'py2' and 'py3' r=catlee
Depends on D36091

Differential Revision: https://phabricator.services.mozilla.com/D36092

--HG--
extra : moz-landing-system : lando
2019-07-08 17:30:34 +00:00

57 lines
2.0 KiB
Python
Executable File

#!/usr/bin/env python
# This Source Code Form is subject to the terms of the Mozilla Public
# License, v. 2.0. If a copy of the MPL was not distributed with this
# file, You can obtain one at http://mozilla.org/MPL/2.0/.
# Usage: check_source_count.py SEARCH_TERM COUNT ERROR_LOCATION REPLACEMENT [FILES...]
# Checks that FILES contains exactly COUNT matches of SEARCH_TERM. If it does
# not, an error message is printed, quoting ERROR_LOCATION, which should
# probably be the filename and line number of the erroneous call to
# check_source_count.py.
from __future__ import absolute_import
from __future__ import print_function
import sys
import re
search_string = sys.argv[1]
expected_count = int(sys.argv[2])
error_location = sys.argv[3]
replacement = sys.argv[4]
files = sys.argv[5:]
details = {}
count = 0
for f in files:
text = file(f).read()
match = re.findall(search_string, text)
if match:
num = len(match)
count += num
details[f] = num
if count == expected_count:
print("TEST-PASS | check_source_count.py {0} | {1}"
.format(search_string, expected_count))
else:
print("TEST-UNEXPECTED-FAIL | check_source_count.py {0} | "
.format(search_string),
end='')
if count < expected_count:
print("There are fewer occurrences of /{0}/ than expected. "
"This may mean that you have removed some, but forgotten to "
"account for it {1}.".format(search_string, error_location))
else:
print("There are more occurrences of /{0}/ than expected. We're trying "
"to prevent an increase in the number of {1}'s, using {2} if "
"possible. If it is unavoidable, you should update the expected "
"count {3}.".format(search_string, search_string, replacement,
error_location))
print("Expected: {0}; found: {1}".format(expected_count, count))
for k in sorted(details):
print("Found {0} occurences in {1}".format(details[k], k))
sys.exit(-1)