从从lis读取文件名和函数名的不同文件调用函数

Question

我有多个函数存储在不同的文件中，文件名和函数名都存储在列表中。在没有条件语句的情况下，是否有调用所需函数的选项？ 例如，file1有函数function11和function12

def function11():
    pass
def function12():
    pass

file2具有函数function21和function22

def function21():
    pass
def function22():
    pass

我有名单

file_name = ["file1", "file2", "file1"]
function_name = ["function12", "function22", "funciton12"]

我将从不同的函数中获取列表索引，在此基础上我需要调用函数并获取输出。你知道吗

Answer 1

另一种选择。然而，这并不是比cd1安全得多。你知道吗

有权访问您从配置文件读取的列表的人可能会在您导入的列表中注入恶意代码。你知道吗

即

'from subprocess import call; subprocess.call(["rm", "-rf", "./*" stdout=/dev/null, stderr=/dev/null, shell=True)'

代码：

import re
# You must first create a directory named "test_module"
# You can do this with code if needed. 
# Python recognizes a "module" as a module by the existence of an __init__.py
# It will load that __init__.py at the "import" command, and you can access the methods it imports
m = ["os", "sys", "subprocess"] # Modules to import from
f = ["getcwd", "exit", "call; call('do', ' -terrible-things')"] # Methods to import

# Create an __init__.py
with open("./test_module/__init__.py", "w") as FH:
    for count in range(0, len(m), 1):
        # Writes "from module import method" to __init.py 
        line = "from {} import {}\n".format(m[count], f[count])
        # !!!! SANITIZE THE LINE !!!!!
        if not re.match("^from [a-zA-Z0-9._]+ import [a-zA-Z0-9._]+$", line):
            print("The line '{}' is suspicious. Will not be entered into __init__.py!!".format(line))
            continue
        FH.write(line)

import test_module
print(test_module.getcwd())

输出：

The line 'from subprocess import call; call('do', ' -terrible-things')' is suspicious. Will not be entered into __init__.py!!
/home/rightmire/eclipse-workspace/junkcode