如何在python中动态加载不同位置的类并获取类中的方法名称

Question

我有wriiten简单的class.py文件，其中有类 班级雇员:

empCount = 0

def __init__(self, name, salary):
    self.name = name
    self.salary = salary
    Employee.empCount += 1

def displayCount(self,salary):
    print "Total Employee %d" % Employee.empCount

def displayEmployee(self):
    print "Name : ", self.name,  ", Salary: ", self.salary

现在我已经编写了另一个脚本来导入文件并获取类中的方法，但我无法获取方法

import inspect
import sys
pat="E://pythonscripts"
sys.path.append(pat)

#pat="E:/pythonscripts/Simpleclass"
__import__('Simpleclass', globals={})

for name, method in inspect.getmembers('Simpleclass', inspect.ismethod):
    print name
    (args, varargs, varkw, defaults) = inspect.getargspec(method)
    for arg in args:
        print arg

运行inspect脚本时，我收到以下错误

Traceback (most recent call last):
  File "C:/Python27/stack.py", line 9, in <module>
    for name, method in inspect.getmembers(Employee, inspect.ismethod):
NameError: name 'Employee' is not defined

Answer 1

添加路径的 employee.py 模块在 sys.path

import sys
sys.path.append(<directory of employee.py module>)

对于导入动态模块使用 __import__

__import__('employee', globals={})

或

进口；进口

您可以使用 inspect 对于introspecting python类

import inspect

for name, method in inspect.getmembers(<your class>, inspect.ismethod):
    print name
    (args, varargs, varkw, defaults) = inspect.getargspec(method)
    for arg in args:
        print arg