Python解析日期与strptime

问题描述

我有url以这种格式返回日期

I have url that returns date in this format

url_date = "2015-01-12T08:43:02Z"

我不知道为什么有字符串，将它变得更简单为2015-01-1208：43：02使用

I don't know why there are strings, it would have been simpler to get it as "2015-01-1208:43:02" which would have been simpler to parse using

datetime.datetime.strptime(url_date , '%Y-%m-%d')

但它不工作。我尝试过

%Y-%m-%d
%Y-%m-%d-%H-%M-%S
%Y-%m-%d-%H-%M-%S-%Z

但是我不断收到错误，如时间数据2015-01-12T08：43：02Z不匹配...

But I keep getting errors like "time data 2015-01-12T08:43:02Z does not match ..."

推荐答案

您正在寻找的格式是 - '％Y-％m-％dT％H：％M：％SZ'。

The format you are looking for is - '%Y-%m-%dT%H:%M:%SZ' .

示例 -

>>> url_date = "2015-01-12T08:43:02Z"
>>> import datetime
>>> datetime.datetime.strptime(url_date , '%Y-%m-%dT%H:%M:%SZ')
datetime.datetime(2015, 1, 12, 8, 43, 2)

---

对于评论中的新要求 -

---

For the new requirement in comments -

您需要使用 .strftime（） c $ c> datetime.datetime 对象，格式为 - '％Y-％m-％d：％H：％M：％S'。示例 -

You would need to use .strftime() on the datetime.datetime object with the format - '%Y-%m-%d:%H:%M:%S'. Example -

>>> url_date = "2015-01-12T08:43:02Z"
>>> dt = datetime.datetime.strptime(url_date , '%Y-%m-%dT%H:%M:%SZ')
>>> dt.strftime('%Y-%m-%d:%H:%M:%S')
'2015-01-12:08:43:02'

如果您想要时间组件，可以使用 .time（）。示例 -

If you wanted the time component , you can use .time() for that. Example -

>>> dt = datetime.datetime.strptime(url_date , '%Y-%m-%dT%H:%M:%SZ')
>>> dt.time()
datetime.time(8, 43, 2)

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