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获取函数的局部变量

locals decorator unix python

Martijn · 技术社区 · 14 年前

我正在尝试从一个修饰器中获取局部变量。一个例子:

def needs_privilege(privilege, project=None):
    """Check whether the logged-in user is authorised based on the
    given privilege

    @type privilege: Privilege object, id, or str
    @param privilege: The requested privilege"""

    def validate(func, self, *args, **kwargs):
        """Validator of needs_privillige"""
        try: check(self.user, privilege, project)
        except AccessDenied:
            return abort(status_code=401)
        else: 
            return func(self, *args, **kwargs)

    return decorator(validate)

在装饰函数之后,如下所示:

 @needs_privilege("some_privilege")
 def some_function():
     pass

我想从某个函数中检索“privilige”变量(validate()使用)。搜索了一个多小时后,我觉得很迷茫。这有可能吗?

编辑 : 让我更详细地描述一下我的问题:我能得到“一些”字符串吗? 没有执行一些函数?类似:

a = getattr(module, 'somefunction')
print a.decorator_arguments

?谢谢你到目前为止对我的帮助!

4 回复 | 直到 14 年前

mouad 14 年前

您的修饰器基本上检查用户是否有执行给定函数的权限,我不理解您为什么要检索(附加)要包装的函数的特权,但是您可以这样做,而不必为所有函数添加另一个参数。

def needs_privilege(privilege, project=None):
    """Check whether the logged-in user is authorised based on the
    given privilege

    @type privilege: Privilege object, id, or str
    @param privilege: The requested privilege"""

    def validate(func, self, *args, **kwargs):
        """Validator of needs_privillige"""
        try: check(self.user, privilege, project)
        except AccessDenied:
            return abort(status_code=401)
        else:
            return func(self, *args, **kwargs)
    validate.privelege = privelege
    return decorator(validate)

顺便说一下,你的装饰师应该是这样的:

def needs_privilege(privilege, project=None):
    def validate(func):
        def new_func(self, *args, **kwargs):
            try: 
                check(self.user, privilege, project)
            except AccessDenied:
                return abort(status_code=401)
            else:
                return func(self, *args, **kwargs)
        new_func.privilege = privilege
        return new_func
    return validate

Radomir Dopieralski 14 年前

可以将其作为参数传递:

def needs_privilege(privilege, project=None):
    """Check whether the logged-in user is authorised based on the
    given privilege

    @type privilege: Privilege object, id, or str
    @param privilege: The requested privilege"""

    def validate(func, self, *args, **kwargs):
        """Validator of needs_privillige"""
        try: check(self.user, privilege, project)
        except AccessDenied:
            return abort(status_code=401)
        else: 
            return func(self, privilege, *args, **kwargs)

    return decorator(validate)

@needs_privilege("some_privilege")
def some_function(privilege):
    pass

unutbu 14 年前

如果你不需要 decorator 模块。如果你不需要 装饰者 模块,您可以这样编写装饰器:

def needs_privilege(privilege, project=None):
    def validate(func):
        def _validate(self, *args, **kwargs):
            return func(self, *args, **kwargs)
        _validate.decorator_args=(privilege,project)
        return _validate
    return validate

@needs_privilege("some_privilege")
def some_function(self):
    pass

a = some_function
print(a.decorator_args)
# ('some_privilege', None)

Robie Basak 14 年前

函数也是可以有属性的对象。您可以在装饰器中设置属性。下面是一个例子:

class TestClass(object):
    def needs_privilege(privilege, project=None):
        def wrapper(func):
            def validate(self, *args, **kwargs):
                """Validator of needs_privillige"""
                print 'validator check for %s' % privilege
                return func(*args, **kwargs)
            validate.privilege = privilege
            return validate

        return wrapper

    @needs_privilege("foo")
    def bar():
        print "called"

>>> test.TestClass().bar()
validator check for foo
called
>>> test.TestClass.bar.privilege
'foo'
>>> test.TestClass().bar.privilege
'foo'