加入 Gitee
与超过 1200万 开发者一起发现、参与优秀开源项目,私有仓库也完全免费 :)
免费加入
文件
克隆/下载
check_sub_module.py 987 Bytes
一键复制 编辑 原始数据 按行查看 历史
zourongchun 提交于 2022-03-30 15:50 . chmod check_sub_module.py
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
#
# Copyright (c) 2022 Huawei Device Co., Ltd.
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
#
# http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
#
import os
import sys
def main():
if sys.argv[1] == "" or sys.argv[2] == "" :
print("false")
return 1
sub_module = os.path.join(sys.argv[1], sys.argv[2])
if os.path.exists(sub_module) == True :
print("true")
return 0
print("false")
return 0
if __name__ == '__main__':
sys.exit(main())
马建仓 AI 助手
尝试更多
代码解读
代码找茬
代码优化