获取Python JSON序列化日期时间 [英] Get Python JSON to serialize datetime
问题描述
datetime
s。当我尝试序列化这些对象时,Python抛出一个异常:
TypeError:datetime.datetime(2012,6,5,17,49 ,35,672115)不是JSON可序列化
使用Python 2.7,有没有办法告诉json serializer:当你看到datetime,不要烦人,抛出异常,只是序列化使用: property.strftime('%Y-%m-%d%I:%M%p')
谢谢!
一个帮助函数,将序列化 datetime
对象,并使用默认
kwarg json.dump 或json.dumps。
此外,您还需要考虑是否支持时区感知 datetime
对象。以及是否要在序列化期间保留时区,或者仅在序列化之前转换为UTC。
以下是一个示例,假设您要在序列化之前转换为UTC。它依赖于 python-dateutil 库:
from dateutil.tz import tzutc
UTC = tzutc()
def serialize_date(dt):
将日期/时间值序列化为ISO8601文本表示
调整(如果需要)到UTC时区
例如:
>>> serialize_date( datetime(2012,4,10,22,38,20,604391))
'2012-04-10T22:38:20.604391Z'
如果dt.tzinfo:
dt = dt.astimezone(UTC).replace(tzinfo = None)
返回dt.isoformat()+'Z'
Have some nested objects that I'd like serialize using JSON. The problem is that some of the properties contain datetime
s. When I try to serialize these pbjects, Python throws an exception:
TypeError: datetime.datetime(2012, 6, 5, 17, 49, 35, 672115) is not JSON serializable
Using Python 2.7, is there a way to tell the json serializer: "When you see a datetime, don't be annoying and throw an exception, just serialize using: property.strftime('%Y-%m-%d %I:%M%p')
"
Thanks!
You'll want to define a helper function that will serialize datetime
objects, and use default
kwarg of json.dump or json.dumps. See the comments with links to the duplicate answers.
Also, you will want to consider whether to support or not to support timezone-aware datetime
objects. And whether you want to preserve the timezone during the serialization or just convert to UTC prior to serialization.
Here's an example that assumes you want to convert to UTC before serialization. It relies upon python-dateutil library:
from dateutil.tz import tzutc
UTC = tzutc()
def serialize_date(dt):
"""
Serialize a date/time value into an ISO8601 text representation
adjusted (if needed) to UTC timezone.
For instance:
>>> serialize_date(datetime(2012, 4, 10, 22, 38, 20, 604391))
'2012-04-10T22:38:20.604391Z'
"""
if dt.tzinfo:
dt = dt.astimezone(UTC).replace(tzinfo=None)
return dt.isoformat() + 'Z'
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