场景: 我有个档案项目.py其任务是导航到另一个文件夹以获取存储在.yml文件中的环境变量。这里有两种情况,一种工作是导航到一个目录,另一种工作是导航到一个egg文件中的目录,我使用了zipfile模块。如何将这两个案例都包含在项目.py并编写一个if-else条件,根据目录类型在这两个函数之间进行选择。下面是代码。我对编程相当陌生,非常感谢您的帮助。你知道吗
类别项目(dict):
def __init__(self, envmt, app_path):
self.config_dict = {}
self.envmt = envmt
with zipfile.Zipfile("C:/Users/project/poc/dist/project-1.0.0-py3.6", 'r') as myzip:
config = load(myzip.open("base.yml_path"))
for key, value in config.items():
configvals = value
for key in configvals:
self.config_dict[key]= configvals[key]
self.base = config
with zipfile.Zipfile("C:/Users/project/poc/dist/project-1.0.0-py3.6", 'r') as myzip:
config = load(myzip.open("env.yml_path"))
for key, value in config.items():
configvals = value
for key in configvals:
self.config_dict[key]= configvals[key]
self.env = config
def __init__(self, envmt, app_path):
self.config_dict = {}
self.envmt = envmt
with open(os.path.join(app_path, 'config', 'base.yml'), 'r') as cfile:
config = load(cfile)
for key, value in config.items():
configvals = value
for key in configvals:
self.config_dict[key]= configvals[key]
self.base = config
with open(os.path.join(app_path, 'config', envmt+'.yml'), 'r') as cfile:
config = load(cfile)
for key, value in config.items():
configvals = value
for key in configvals:
self.config_dict[key]= configvals[key]
self.env = config
如何编写一个条件来在上述两个函数之间进行选择。你知道吗
如果两个
__init__
属于同一个class
,则只使用最后一个,请考虑编写如下类:相关问题 更多 >
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