管理不同的python脚本并调用它们
为了便于维护,我将python程序分为不同的文件 以下是python脚本文件名:管理不同的python脚本并调用它们,python,python-2.7,Python,Python 2.7,为了便于维护,我将python程序分为不同的文件 以下是python脚本文件名: 1.master.py 2.setup.py 3.core1.py 4.core2.py 5.standard.py master.py: import setup import standard do something import setup import standard if __name__ == '__main__': do something s
1.master.py
2.setup.py
3.core1.py
4.core2.py
5.standard.py
master.py:
import setup
import standard
do something
import setup
import standard
if __name__ == '__main__':
do something
setup.py:
import standard
a = 5
b = 5
if standard.d == 10:
do something
import standard
def a():
return 5
def b():
return 5
if __name__ == '__main__':
if standard.d == 10:
do something
standard.py:
import setup
if setup.a == 5:
d = 10
do something
import setup
def d():
if setup.a == 5:
return 10
else:
print 'setup.a != 5'
if __name__ == '__main__':
if setup.a == 5:
do something
当我们运行主脚本时,它将导入setup.py,setup.py将导入standard.py。我们将获得错误消息设置。未定义
我有两种选择来解决这个问题
a = 5
b = 5
import standard
if standard.d == 10:
do something
任何建议都会有很大帮助。我是编程界的新手。因此我需要专家的建议,他们将如何处理这种情况 我建议您将其他脚本应该使用的对象分离为函数(甚至类),然后编写脚本的私有主体 代码如下所示: master.py:
import setup
import standard
do something
import setup
import standard
if __name__ == '__main__':
do something
setup.py:
import standard
a = 5
b = 5
if standard.d == 10:
do something
import standard
def a():
return 5
def b():
return 5
if __name__ == '__main__':
if standard.d == 10:
do something
standard.py:
import setup
if setup.a == 5:
d = 10
do something
import setup
def d():
if setup.a == 5:
return 10
else:
print 'setup.a != 5'
if __name__ == '__main__':
if setup.a == 5:
do something
不要使用循环导入。这两个文件正在无休止地相互导入