Question

我在linux中使用python运行一个命令，我有几种不同的基于系统类型的验证方法。我的问题是我可以使验证功能获取功能列表而不是创建所有组合。

目前我有：

def verify_A(*args):
    checks the commands was successful using method A

def verify_B(*args):
    checks the commands was successful using method B

def run_command(*args)
    runs the commands on linux terminal, no checking

def run_and_verify_A(*args):
     run_command(*args)
     verify_A(*args)

def run_and_verify_B(*args):
    run_command(*args)
    verify_B(*args)

def run_and_verify_All(*args):
    run_command(*args)
    if not verify_A(*args):
        verify_B(*args)

我想要的是：

def run_command(*args)
    runs the commands on linux terminal, no checking


verify_list=['verify_A','verify_B']

def run_and_verify(verify_list):
     run_command(*args)
     for func in verify_list:
         if eval(func):
             print "verification passed"
             return True
        else:
             print "verification is failed... running next verify method"

但是我的run_and verify函数没有按预期工作..

Answer 1

您可以将功能放在列表中：

def foo():
    return something

def bar():
    return stuff

lst = [foo, bar]

for func in lst:
    if func()
        print('passed...')
        break  # or return, however you decide to set it up
    else:
        print('failed...')

Answer 2

考虑到所有函数具有相同的签名，请创建函数列表并使用any按指定的顺序进行评估。 any的短路属性基本上可以帮助您忽略过去成功的所有功能

示例

verify = [verify_A, verify_B, run_and_verify_A, run_and_verify_B, run_and_verify_Allrun_and_verify_All] def run_and_verify(verify_list, *args): run_command(*args) if any(fn(*args) for fn in verify_list): print "verification passed" else: print "verification is failed... running next verify method"

python运行函数列表，直到一个通过

2 个答案: