author Randell Jesup <>
Tue, 27 Dec 2016 19:41:02 -0500
changeset 327415 126348e718d03dec640b30b5def70fce8aa71527
parent 123443 8f83edc05fa4d458d82668a95853c907544d0b8d
child 419598 fba6f974041a3d3c22ef95c44f6e4262e8e5c52f
permissions -rwxr-xr-x
Bug 1250356: rollup of changes for webrtc after applying v49 update r=pkerr,ng,pehrsons,etc See ssh:// for the patch development history.

#!/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

#   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
from __future__ import print_function
import sys
import os
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 | {0} | {1}"
          .format(search_string, expected_count))

    print("TEST-UNEXPECTED-FAIL | {0} | "
    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))
        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, 

    print("Expected: {0}; found: {1}".format(expected_count, count))
    for k in sorted(details):
        print("Found {0} occurences in {1}".format(details[k],k))