如何在修饰器中使用命名参数？

问题描述

如果我有以下函数：

def intercept(func):
  # do something here

@intercept(arg1=20)
def whatever(arg1,arg2):
  # do something here

我希望仅当arg1为20时才触发截取。我希望能够将命名参数传递给函数。我如何才能做到这一点？

这里有一个小代码示例：

def intercept(func):
    def intercepting_func(*args,**kargs):
        print "whatever"
        return func(*args,**kargs)
    return intercepting_func

@intercept(a="g")
def test(a,b):
    print "test with %s %s" %(a,b)

test("g","d")

这引发以下异常类型错误：intercept()获取意外的关键字参数‘a’

---

解决方案

from functools import wraps

def intercept(target,**trigger):
    def decorator(func):
        names = getattr(func,'_names',None)
        if names is None:
            code = func.func_code
            names = code.co_varnames[:code.co_argcount]
        @wraps(func)
        def decorated(*args,**kwargs):
            all_args = kwargs.copy()
            for n,v in zip(names,args):
                all_args[n] = v
            for k,v in trigger.iteritems():
                if k in all_args and all_args[k] != v:
                    break
            else:
                return target(all_args)
            return func(*args,**kwargs)
        decorated._names = names
        return decorated
    return decorator

示例：

def interceptor1(kwargs):
    print 'Intercepted by #1!'

def interceptor2(kwargs):
    print 'Intercepted by #2!'

def interceptor3(kwargs):
    print 'Intercepted by #3!'

@intercept(interceptor1,arg1=20,arg2=5) # if arg1 == 20 and arg2 == 5
@intercept(interceptor2,arg1=20)        # elif arg1 == 20
@intercept(interceptor3,arg2=5)         # elif arg2 == 5
def foo(arg1,arg2):
    return arg1+arg2

>>> foo(3,4)
7
>>> foo(20,4)
Intercepted by #2!
>>> foo(3,5)
Intercepted by #3!
>>> foo(20,5)
Intercepted by #1!
>>>

functools.wraps做维基上的简单修饰符；更新修饰符的__doc__、__name__等属性。