How can I convert epoch datetime in python?

Question

I'm calling an api and getting the datetime field in this format /Date(1500462505221)/ . My motive is to convert it to this format "2018-05-11 23:25:47" . How do I accomplish this ?

score 6 · Answer 1 · edited Jan 13 '19 at 04:12

6

You can use datetime library to make life simple:

from datetime import datetime

datetime.fromtimestamp(1485714600).strftime("%Y-%m-%d %I:%M:%S")

edited Jan 13 '19 at 04:12

nick

1,090
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answered Jul 03 '18 at 13:22

kanishk

91
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I forgot to mention. Is it possible without doing any string operations ? Any library that takes the string as is and converts to datetime ? – Nilesh Kumar Guria Jul 03 '18 at 13:52
can you give me one example of string which you want to convert in datetime? – kanishk Jul 03 '18 at 13:59
'/Date(1506342842815)/' – Nilesh Kumar Guria Jul 03 '18 at 14:02
import time time.strftime('%Y-%m-%d %H:%M:%S', time.gmtime(1506342842815/1000.)) – kanishk Jul 03 '18 at 18:17
So basically I would need to trim off "/Date(" and ")/" ? – Nilesh Kumar Guria Jul 04 '18 at 10:42
Why you need to trim, you directly give number rather than '/Date(1506342842815)/'. – kanishk Jul 04 '18 at 13:18

Deepak Verma · Answer 2 · 2018-07-03T13:24:58.447

1

So here is what i got to convert time in human readable format, maybe this can help:

import time
time.strftime("%a, %d %b %Y %H:%M:%S %Z", time.localtime(1500462505221))

or

time.strftime('%Y-%m-%d %H:%M:%S', time.localtime(1500462505221))

and one more thing i tried this example by removing last three digit of as follows and got this as output:

>>time.strftime('%Y-%m-%d %H:%M:%S', time.localtime(1500462505))
>>'2017-07-19 11:08:25'

You can also refer this link: [https://www.systutorials.com/241698/how-to-convert-epoch-timestamp-to-human-readable-date-format-in-python/

edited Jul 03 '18 at 13:24

answered Jul 03 '18 at 13:19

Deepak Verma

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Is there any solution without doing any string manipulation/modifications ? – Nilesh Kumar Guria Jul 03 '18 at 13:51
i tried it without any string modification but that's not working fine. In fact trimming the string would be a good option. – Deepak Verma Jul 04 '18 at 11:38

Chiheb Nexus · Answer 3 · 2018-07-03T13:42:34.050

You can do something like this using time and re modules:

import re, time

epoch_str = "/Date(1500462505221)/"
date = re.match(r'/Date\((\d+)\)/', epoch_str).groups()[0]
epoch_fmt = time.strftime('%Y-%M-%d %H:%M:%S', time.gmtime(float(date)))
# Or you can do
# date_ = time.gmtime(float(date))
# epoch_fmt = '{}-{}-{} {}:{}:{}'.format(date_.tm_year, date_.tm_mon, date_.tm_mday, date_.tm_hour, date_.tm_min, date_.tm_sec) 
print(epoch_fmt)
# 49517-20-07 04:20:21

How can I convert epoch datetime in python?

3 Answers3