我的脚本是python,cassandra是data stax社区版。
TypeError: A str or unicode value was expected, but int was received instead (3902503)这是我在尝试插入到cassandra柱族时遇到的错误。
代码如下:
for x in feed:
cf.insert(uuid.uuid4(), x)X是一个简单的数组,形式为"{key:value}“
错误日志显示:
Traceback (most recent call last):
File "C:\Users\me\Desktop\pro1\src\pro1.py", line 73, in <module>
str("swf"): str("aws")
File "c:\Python27\lib\site-packages\pycassa\columnfamily.py", line 969, in insert
mut_list = self._make_mutation_list(columns, timestamp, ttl)
File "c:\Python27\lib\site-packages\pycassa\columnfamily.py", line 504, in _make_mutation_list
columns.iteritems())
File "c:\Python27\lib\site-packages\pycassa\columnfamily.py", line 503, in <lambda>
return map(lambda (c, v): Mutation(self._make_cosc(_pack_name(c), _pack_value(v, c), timestamp, ttl)),
File "c:\Python27\lib\site-packages\pycassa\columnfamily.py", line 462, in _pack_value
return packer(value)
File "c:\Python27\lib\site-packages\pycassa\marshal.py", line 231, in pack_bytes
% (v.__class__.__name__, str(v)))
TypeError: A str or unicode value was expected, but int was received instead (3902503) 我似乎错过了一件很重要的事情...这就是我来这里请教专家的原因!
发布于 2012-09-16 18:32:38
确保您的值与列族类型相匹配。看起来您的列族要么是一个BytesType,要么没有与之关联的类型,因此pycassa将只接受字符串值。您可以将所有值映射到带有嵌套字典理解的列表理解的str中(后者需要Python2.7或更高版本):
cf.insert(uuid.uuid4(), [{k: str(v) for k, v in d.iteritems()} for d in x])https://stackoverflow.com/questions/12446033
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