How do I access an object's properties by name?

Question

class a(type):
    def __str__(self):
        return 'aaa'
    def __new__(cls, name, bases, attrs):
        attrs['cool']='cool!!!!'
        new_class = super(a,cls).__new__(cls, name, bases, attrs)
                #if 'media' not in attrs:
                    #new_class.media ='media'
        return new_class

class b(object):
    __metaclass__=a
    def __str__(self):
        return 'bbb'

print b
print b()['cool']#how can i print 'cool!!!!'

Just in case people don't read past the title, you should try to make it reflect the rest of your question. — Josh Lee
– Josh Lee, Commented Jan 19, 2010 at 8:15
Printing Foo or Bar(en.wikipedia.org/wiki/Foobar) is cooler! — Nikita Fedyashev
– Nikita Fedyashev, Commented Jan 19, 2010 at 9:18

Josh Lee · Accepted Answer · 2010-01-19 08:15:25Z

5

print b().cool

attrs in your __new__ method becomes the object's dictionary. Properties of Python objects are referenced with the . syntax.

answered Jan 19, 2010 at 8:15

Josh Lee

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Comments

lutz · Accepted Answer · 2010-01-19 08:13:24Z

1

print "cool!!!"

Or did I miss something?

answered Jan 19, 2010 at 8:13

lutz

1 Comment

Erik Over a year ago

I got a laugh outta your answer

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