我有一个名为test1.py的脚本,它不在模块中。它只有在脚本本身运行时应该执行的代码。没有函数、类、方法等。我有另一个脚本作为服务运行。我想从作为服务运行的脚本调用test1.py。

例如:

文件test1.py:

print "I am a test"
print "see! I do nothing productive."

文件service.py:

# Lots of stuff here
test1.py # do whatever is in test1.py

我知道有一种方法是打开文件,读取内容,然后对它求值。我想应该有更好的方法。至少我希望如此。


当前回答

通常的做法是这样的。

test1.py

def some_func():
    print 'in test 1, unproductive'

if __name__ == '__main__':
    # test1.py executed as script
    # do something
    some_func()

service.py

import test1

def service_func():
    print 'service func'

if __name__ == '__main__':
    # service.py executed as script
    # do something
    service_func()
    test1.some_func()

其他回答

这个过程有点非传统,但适用于所有的python版本,

假设你想在一个'if'条件中执行一个名为' recommendation .py'的脚本,然后使用,

if condition:
       import recommend

技术是不同的,但有效!

根据上面的例子,这是最好的方法:

# test1.py

def foo():
    print("hellow")
# test2.py
from test1 import foo # might be different if in different folder.
foo()

但根据标题,使用os.startfile(“path”)是最好的方法,因为它很小,而且它可以工作。这将执行指定的文件。我的python版本是3。x +。

这可以在Python 2中使用

execfile("test2.py")

如果对您的情况很重要,请参阅文档了解名称空间的处理。

在Python 3中,可以使用(感谢@fantastory)

exec(open("test2.py").read())

然而,你应该考虑使用不同的方法;(在我看来)你的想法不太清楚。

I found runpy standard library most convenient. Why? You have to consider case when error raised in test1.py script, and with runpy you are able to handle this in service.py code. Both traceback text (to write error in log file for future investigation) and error object (to handle error depends on its type): when with subprocess library I wasn't able to promote error object from test1.py to service.py, only traceback output. Also, comparing to "import test1.py as a module" solution, runpy is better cause you have no need to wrap code of test1.py into def main(): function.

代码片段为例,使用traceback模块捕获最后一个错误文本:

import traceback
import runpy #https://www.tutorialspoint.com/locating-and-executing-python-modules-runpy

from datetime import datetime


try:
    runpy.run_path("./E4P_PPP_2.py")
except Exception as e:
    print("Error occurred during execution at " + str(datetime.now().date()) + " {}".format(datetime.now().time()))
    print(traceback.format_exc())
    print(e)

如果你想让test1.py保持可执行的功能,就像在service.py内部调用它一样,然后做如下的事情:

test1.py

def main():
    print "I am a test"
    print "see! I do nothing productive."

if __name__ == "__main__":
    main()

service.py

import test1
# lots of stuff here
test1.main() # do whatever is in test1.py