要获得 (0,9999) 之间的 range_key,我可以这样做吗?
conn = boto.connect_dynamodb()
table = conn.get_table("mytable")
...
result = table.query(
hash_key = "66",
range_key_condition = {"0":"GE", "9999":"LE"}
)
使用 boto v2.2.2-dev,我总是得到空结果
编辑:这是另一个错误示例:
In [218]: qa = taa.query(hash_key = "1")
In [219]: qa.next()
Out[219]: {u'attra': u'this is attra', u'key': u'1', u'range': 1.1}
上面没有"range_key_condition"也可以
In [220]: qa = taa.query(hash_key = "1", range_key_condition = {0.1: "GE"})
In [221]: qa.next()
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
/home/user/python/enva/<ipython-input-221-dba0a498b6e1> in <module>()
----> 1 qa.next()
/home/user/python/enva/local/lib/python2.7/site-packages/boto-2.2.2_dev-py2.7.egg/boto/dynamodb/layer2.pyc
in query(self, table, hash_key, range_key_condition,
attributes_to_get, request_limit, max_results, consistent_read,
scan_index_forward, exclusive_start_key, item_class)
559 """
560 if range_key_condition:
--> 561 rkc = self.dynamize_range_key_condition(range_key_condition)
562 else:
563 rkc = None
/home/user/python/enva/local/lib/python2.7/site-packages/boto-2.2.2_dev-py2.7.egg/boto/dynamodb/layer2.pyc
in dynamize_range_key_condition(self, range_key_condition)
83 structure required by Layer1.
84 """
---> 85 return range_key_condition.to_dict()
86
87 def dynamize_scan_filter(self, scan_filter):
AttributeError: 'dict' object has no attribute 'to_dict'
最佳答案
如果您使用的是最新版本的 boto(看起来您是),条件的方式已从以前的版本更改,以尝试使查询更具可读性。试试这个:
from boto.dynamodb.condition import *
conn = boto.connect_dynamodb()
table = conn.get_table("mytable")
...
result = table.query(
hash_key = "66",
range_key_condition = BETWEEN(0, 9999))
虽然您必须更新您的 boto 代码,但它应该可以工作,因为我在调查此问题时刚刚在 BETWEEN 中发现了一个错误(请参阅 https://github.com/boto/boto/issues/620)。
关于python - 如何使用 "range_key_condition"通过 boto 查询 DynamoDB 表?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9584730/