在Python中,当两个模块试图相互导入时会发生什么?更一般地说,如果多个模块试图在一个循环中导入会发生什么?
另见我能做什么关于“ImportError:不能导入名称X”或“AttributeError:…”(很可能是由于循环导入)”?关于可能导致的常见问题,以及如何重写代码以避免此类导入的建议。参见为什么循环导入看起来在调用堆栈中更上一层,但随后在更下一层引发ImportError ?有关问题发生的原因和方式的技术细节。
在Python中,当两个模块试图相互导入时会发生什么?更一般地说,如果多个模块试图在一个循环中导入会发生什么?
另见我能做什么关于“ImportError:不能导入名称X”或“AttributeError:…”(很可能是由于循环导入)”?关于可能导致的常见问题,以及如何重写代码以避免此类导入的建议。参见为什么循环导入看起来在调用堆栈中更上一层,但随后在更下一层引发ImportError ?有关问题发生的原因和方式的技术细节。
当前回答
模块a.py:
import b
print("This is from module a")
模块b.py
import a
print("This is from module b")
运行“Module a”将输出:
>>>
'This is from module a'
'This is from module b'
'This is from module a'
>>>
它输出了这3行,而由于循环导入,它应该输出不定式。 这里列出了运行“模块a”时逐行发生的事情:
The first line is import b. so it will visit module b The first line at module b is import a. so it will visit module a The first line at module a is import b but note that this line won't be executed again anymore, because every file in python execute an import line just for once, it does not matter where or when it is executed. so it will pass to the next line and print "This is from module a". After finish visiting whole module a from module b, we are still at module b. so the next line will print "This is from module b" Module b lines are executed completely. so we will go back to module a where we started module b. import b line have been executed already and won't be executed again. the next line will print "This is from module a" and program will be finished.
其他回答
去年在comp.lang.python上对此进行了很好的讨论。它很彻底地回答了你的问题。
Imports are pretty straightforward really. Just remember the following: 'import' and 'from xxx import yyy' are executable statements. They execute when the running program reaches that line. If a module is not in sys.modules, then an import creates the new module entry in sys.modules and then executes the code in the module. It does not return control to the calling module until the execution has completed. If a module does exist in sys.modules then an import simply returns that module whether or not it has completed executing. That is the reason why cyclic imports may return modules which appear to be partly empty. Finally, the executing script runs in a module named __main__, importing the script under its own name will create a new module unrelated to __main__. Take that lot together and you shouldn't get any surprises when importing modules.
模块a.py:
import b
print("This is from module a")
模块b.py
import a
print("This is from module b")
运行“Module a”将输出:
>>>
'This is from module a'
'This is from module b'
'This is from module a'
>>>
它输出了这3行,而由于循环导入,它应该输出不定式。 这里列出了运行“模块a”时逐行发生的事情:
The first line is import b. so it will visit module b The first line at module b is import a. so it will visit module a The first line at module a is import b but note that this line won't be executed again anymore, because every file in python execute an import line just for once, it does not matter where or when it is executed. so it will pass to the next line and print "This is from module a". After finish visiting whole module a from module b, we are still at module b. so the next line will print "This is from module b" Module b lines are executed completely. so we will go back to module a where we started module b. import b line have been executed already and won't be executed again. the next line will print "This is from module a" and program will be finished.
我完全同意pythoneer的回答。但是我偶然发现了一些代码,它们在循环导入时存在缺陷,并在尝试添加单元测试时引起了问题。因此,为了快速修补它而不改变一切,你可以通过动态导入来解决这个问题。
# Hack to import something without circular import issue
def load_module(name):
"""Load module using imp.find_module"""
names = name.split(".")
path = None
for name in names:
f, path, info = imp.find_module(name, path)
path = [path]
return imp.load_module(name, f, path[0], info)
constants = load_module("app.constants")
同样,这不是一个永久性的修复,但可以帮助那些希望在不修改太多代码的情况下修复导入错误的人。
干杯!
这里有个例子让我震惊!
foo.py
import bar
class gX(object):
g = 10
bar.py
from foo import gX
o = gX()
main.py
import foo
import bar
print "all done"
在命令行:$ python main.py
Traceback (most recent call last):
File "m.py", line 1, in <module>
import foo
File "/home/xolve/foo.py", line 1, in <module>
import bar
File "/home/xolve/bar.py", line 1, in <module>
from foo import gX
ImportError: cannot import name gX
我用下面的方法解决了这个问题,没有任何错误。 考虑两个文件a.py和b.py。
我把这个添加到a.py,它工作了。
if __name__ == "__main__":
main ()
a.py:
import b
y = 2
def main():
print ("a out")
print (b.x)
if __name__ == "__main__":
main ()
b.py:
import a
print ("b out")
x = 3 + a.y
得到的输出是
>>> b out
>>> a out
>>> 5