在python脚本中启动Fab File [英] Launch Fab File inside a python script
问题描述
我有一个fab脚本，它工作正常.要启动它，我执行:
I have a fab script, it works fine. To start it, I execute:
fab -H 192.168.xx.xx deployFiles deployConfiguration:'master',7
deployFiles和deployConfiguration都是我的fabfile.py中的功能. 'master'和7是我的deployConfiguration
deployFiles and deployConfiguration are both functions in my fabfile.py. 'master' and 7 are my parameters for deployConfiguration
我还有另一个Python脚本，我想启动他内部的上一个fab命令.
I have another Python script and I want to launch, the previous fab command inside him.
如何使用Python脚本中的这些参数执行我的fabfile?
How can I execute, my fabfile with these parameters from a Python script?
推荐答案
您只需导入它们，然后对其进行调用.使用设置上下文管理器或在fabric.api.env
You just import them, and call them. Using either the settings context manager or setting the relevant settings on fabric.api.env
from fabric.context_managers import settings
from fabfile import deployFiles, deployConfiguration
with settings(host_string='user@192.168.xx.xx'):
deployFiles()
deployConfiguration('master', 7)
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