Python pickle 似乎在类内部中断，但在命令行脚本中却没有

Question

我一直在尝试从数据库中提取一些词典。我已经恢复使用 marshal 模块，但仍然想知道为什么 pickle 在反序列化某些数据时如此困难。这是一个命令行 python session ，基本上显示了我正在尝试做的事情:

>>> a = {'service': 'amazon', 'protocol': 'stream', 'key': 'lajdfoau09424jojf.flv'}
>>> import pickle; import base64
>>> pickled = base64.b64encode(pickle.dumps(a))
>>> pickled
'KGRwMApTJ3Byb3RvY29sJwpwMQpTJ3N0cmVhbScKcDIKc1Mna2V5JwpwMwpTJ2xhamRmb2F1MDk0MjRqb2pmLmZsdicKcDQKc1Mnc2VydmljZScKcDUKUydhbWF6b24nCnA2CnMu'
>>> unpickled = pickle.loads(base64.b64decode(pickled))
>>> unpickled
{'protocol': 'stream', 'service': 'amazon', 'key': 'lajdfoau09424jojf.flv'}
>>> unpickled['service']
'amazon'

一切正常，但是当我在一个类的工厂方法中尝试这个时，似乎 pickle.loads 部分出错了。我尝试加载的字符串的 pickle 方式与上述相同。我什至尝试复制在上面的命令行 session 中 pickle 的确切字符串，然后尝试取消 pickle 它，但没有成功。这是后一种尝试的代码:

class Resource:

    _service = 'unknown'
    _protocol = 'unknown'
    _key = 'unknown'

    '''
    Factory method that creates an appropriate instance of one of Resource’s subclasses based on 
    the type of data provided (the data being a serialized dictionary with at least the keys 'service', 
    'protocol', and 'key'). 
    @param resource_data (string) -- the data used to create the new Resource instance. 
    '''
    @staticmethod
    def resource_factory(resource_data):
        # Unpack the raw resource data and then create the appropriate Resource instance and return. 
        resource_data = "KGRwMApTJ3Byb3RvY29sJwpwMQpTJ3N0cmVhbScKcDIKc1Mna2V5JwpwMwpTJ2xhamRmb2F1MDk0MjRqb2pmLmZsdicKcDQKc1Mnc2VydmljZScKcDUKUydhbWF6b24nCnA2CnMu" #hack to just see if we can unpickle this string
        logging.debug("Creating resource: " + resource_data)
        unencoded = base64.b64decode(resource_data)
        logging.debug("Unencoded is: " + unencoded)
        unpacked = pickle.loads(unencoded)
        logging.debug("Unpacked: " + unpacked)
        service = unpacked['service']
        protocol = unpacked['protocol']
        key = unpacked['key']

        if (service == 'amazon'):
            return AmazonResource(service=service, protocol=protocol, key=key)
        elif (service == 'fs'):
            return FSResource(service=service, protocol=protocol, key=key)

Answer 1

您的代码有效。你如何测试它？

import logging
import base64
import pickle
class Resource:
    @staticmethod
    def resource_factory(resource_data):
        resource_data = "KGRwMApTJ3Byb3RvY29sJwpwMQpTJ3N0cmVhbScKcDIKc1Mna2V5JwpwMwpTJ2xhamRmb2F1MDk0MjRqb2pmLmZsdicKcDQKc1Mnc2VydmljZScKcDUKUydhbWF6b24nCnA2CnMu" #hack to just see if we can unpickle this string
        # logging.debug("Creating resource: " + resource_data)
        unencoded = base64.b64decode(resource_data)
        # logging.debug("Unencoded is: " + unencoded)
        unpacked = pickle.loads(unencoded)
        logging.debug("Unpacked: " + repr(unpacked))
        service = unpacked['service']
        protocol = unpacked['protocol']
        key = unpacked['key']

logging.basicConfig(level=logging.DEBUG)
Resource.resource_factory('')

产量

# DEBUG:root:Unpacked: {'protocol': 'stream', 'service': 'amazon', 'key': 'lajdfoau09424jojf.flv'}