How to compare 2 timestamps (date + time + timezone) [duplicate]
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This question already has an answer here:
Compare two timestamps in Python
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I got 2 dates from my database and I need to compare if one date is greater than another one.
I got the date in this format:
2018-11-07 18:00:40.679087+00:00
2018-11-14 00:00:17.908676+00:00
I tried a lot of solutions, but I didn't figured it out how it works fine.
ps.: I got a lot of questions that looks like this one, but in my case I have the timezone to consider that's the difference.
python
asked Nov 19 at 11:59
mr.abdo
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marked as duplicate by jpp python
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Nov 19 at 12:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
-1
down vote
favorite
This question already has an answer here:
Compare two timestamps in Python
2 answers
I got 2 dates from my database and I need to compare if one date is greater than another one.
I got the date in this format:
2018-11-07 18:00:40.679087+00:00
2018-11-14 00:00:17.908676+00:00
I tried a lot of solutions, but I didn't figured it out how it works fine.
ps.: I got a lot of questions that looks like this one, but in my case I have the timezone to consider that's the difference.
python
asked Nov 19 at 11:59
mr.abdo
74
marked as duplicate by jpp python
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Nov 19 at 12:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Take a look here.
– Vasilis G.
Nov 19 at 12:02
I tried this one, but in my case the format of date contains timezone
– mr.abdo
Nov 19 at 12:08
Read the docs!%zUTC offset in the form ±HHMM[SS[.ffffff]]. later on: Changed in version 3.7: UTC offsets can have a colon as a separator between hours, minutes and seconds
– Christian König
Nov 19 at 12:11
That's the point @ChristianKönig. Tku so much.
– mr.abdo
Nov 19 at 12:15
@Networker in my case I have the timezone to consider, that's the difference. I tried a lot of things that I found here, but none of them works
– mr.abdo
Nov 19 at 12:39
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
This question already has an answer here:
Compare two timestamps in Python
2 answers
I got 2 dates from my database and I need to compare if one date is greater than another one.
I got the date in this format:
2018-11-07 18:00:40.679087+00:00
2018-11-14 00:00:17.908676+00:00
I tried a lot of solutions, but I didn't figured it out how it works fine.
ps.: I got a lot of questions that looks like this one, but in my case I have the timezone to consider that's the difference.
python
asked Nov 19 at 11:59
mr.abdo
74
This question already has an answer here:
Compare two timestamps in Python
2 answers
I got 2 dates from my database and I need to compare if one date is greater than another one.
I got the date in this format:
2018-11-07 18:00:40.679087+00:00
2018-11-14 00:00:17.908676+00:00
I tried a lot of solutions, but I didn't figured it out how it works fine.
ps.: I got a lot of questions that looks like this one, but in my case I have the timezone to consider that's the difference.
This question already has an answer here:
Compare two timestamps in Python
2 answers
python
python
asked Nov 19 at 11:59
mr.abdo
74
asked Nov 19 at 11:59
mr.abdo
74
edited Nov 19 at 12:17
asked Nov 19 at 11:59
mr.abdo
74
asked Nov 19 at 11:59
mr.abdo
74
asked Nov 19 at 11:59
mr.abdo
74
74
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Nov 19 at 12:37
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Nov 19 at 12:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Take a look here.
– Vasilis G.
Nov 19 at 12:02
I tried this one, but in my case the format of date contains timezone
– mr.abdo
Nov 19 at 12:08
Read the docs!%zUTC offset in the form ±HHMM[SS[.ffffff]]. later on: Changed in version 3.7: UTC offsets can have a colon as a separator between hours, minutes and seconds
– Christian König
Nov 19 at 12:11
That's the point @ChristianKönig. Tku so much.
– mr.abdo
Nov 19 at 12:15
@Networker in my case I have the timezone to consider, that's the difference. I tried a lot of things that I found here, but none of them works
– mr.abdo
Nov 19 at 12:39
add a comment |
Take a look here.
– Vasilis G.
Nov 19 at 12:02
I tried this one, but in my case the format of date contains timezone
– mr.abdo
Nov 19 at 12:08
Read the docs!%zUTC offset in the form ±HHMM[SS[.ffffff]]. later on: Changed in version 3.7: UTC offsets can have a colon as a separator between hours, minutes and seconds
– Christian König
Nov 19 at 12:11
That's the point @ChristianKönig. Tku so much.
– mr.abdo
Nov 19 at 12:15
@Networker in my case I have the timezone to consider, that's the difference. I tried a lot of things that I found here, but none of them works
– mr.abdo
Nov 19 at 12:39
Take a look here.
– Vasilis G.
Nov 19 at 12:02
Take a look here.
– Vasilis G.
Nov 19 at 12:02
I tried this one, but in my case the format of date contains timezone
– mr.abdo
Nov 19 at 12:08
I tried this one, but in my case the format of date contains timezone
– mr.abdo
Nov 19 at 12:08
Read the docs!
%zUTC offset in the form ±HHMM[SS[.ffffff]]. later on: Changed in version 3.7: UTC offsets can have a colon as a separator between hours, minutes and seconds– Christian König
Nov 19 at 12:11
Read the docs!
%zUTC offset in the form ±HHMM[SS[.ffffff]]. later on: Changed in version 3.7: UTC offsets can have a colon as a separator between hours, minutes and seconds– Christian König
Nov 19 at 12:11
That's the point @ChristianKönig. Tku so much.
– mr.abdo
Nov 19 at 12:15
That's the point @ChristianKönig. Tku so much.
– mr.abdo
Nov 19 at 12:15
@Networker in my case I have the timezone to consider, that's the difference. I tried a lot of things that I found here, but none of them works
– mr.abdo
Nov 19 at 12:39
@Networker in my case I have the timezone to consider, that's the difference. I tried a lot of things that I found here, but none of them works
– mr.abdo
Nov 19 at 12:39
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Try this
from dateutil import parser
date1='2018-11-07 18:00:40.679087+00:00'
date2='2018-11-14 00:00:17.908676+00:00'
print parser.parse(date2) - parser.parse(date1)
Result:6 days, 5:59:37.229589
answered Nov 19 at 12:16
Narendra Lucky
568
That's an elegant way to solve it. Tku
– mr.abdo
Nov 19 at 12:27
Glad to hear this helps u ;)
– Narendra Lucky
Nov 19 at 12:30
add a comment |
up vote
0
down vote
Try this one (using built-in Python module):
from datetime import datetime
d1 = "2018-11-07 18:00:40.679087+00:00"
d2 = "2018-11-14 00:00:17.908676+00:00"
d1 = d1[:d1.rfind(":")] + d1[d1.rfind(":")+1:]
d2 = d2[:d2.rfind(":")] + d2[d2.rfind(":")+1:]
t1 = datetime.strptime(d1, "%Y-%m-%d %H:%M:%S.%f%z")
t2 = datetime.strptime(d2, "%Y-%m-%d %H:%M:%S.%f%z")
diff = t2 - t1
print(diff)
Output:
6 days, 5:59:37.229589
answered Nov 19 at 12:13
Vasilis G.
2,9402721
1
no need to insert/remove the:for Python 3.7
– Christian König
Nov 19 at 12:26
I am running Python 3.6 and I am gettingValueErrorif I don't remove it.
– Vasilis G.
Nov 19 at 12:28
add a comment |
up vote
0
down vote
The : is bugging you, you can parse it like:
d = '2018-11-07 18:00:40.679087+00:00'
d = d[:-3]+d[-2:]
datetime.strptime(d, '%Y-%m-%d %H:%M:%S.%f%z')
datetime.datetime(2018, 11, 7, 18, 0, 40, 679087, tzinfo=datetime.timezone.utc)
answered Nov 19 at 12:26
Franco Piccolo
1,325611
1
no need to insert/remove the:for Python 3.7
– Christian König
Nov 19 at 12:27
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Try this
from dateutil import parser
date1='2018-11-07 18:00:40.679087+00:00'
date2='2018-11-14 00:00:17.908676+00:00'
print parser.parse(date2) - parser.parse(date1)
Result:6 days, 5:59:37.229589
answered Nov 19 at 12:16
Narendra Lucky
568
That's an elegant way to solve it. Tku
– mr.abdo
Nov 19 at 12:27
Glad to hear this helps u ;)
– Narendra Lucky
Nov 19 at 12:30
add a comment |
up vote
2
down vote
accepted
Try this
from dateutil import parser
date1='2018-11-07 18:00:40.679087+00:00'
date2='2018-11-14 00:00:17.908676+00:00'
print parser.parse(date2) - parser.parse(date1)
Result:6 days, 5:59:37.229589
answered Nov 19 at 12:16
Narendra Lucky
568
That's an elegant way to solve it. Tku
– mr.abdo
Nov 19 at 12:27
Glad to hear this helps u ;)
– Narendra Lucky
Nov 19 at 12:30
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Try this
from dateutil import parser
date1='2018-11-07 18:00:40.679087+00:00'
date2='2018-11-14 00:00:17.908676+00:00'
print parser.parse(date2) - parser.parse(date1)
Result:6 days, 5:59:37.229589
answered Nov 19 at 12:16
Narendra Lucky
568
Try this
from dateutil import parser
date1='2018-11-07 18:00:40.679087+00:00'
date2='2018-11-14 00:00:17.908676+00:00'
print parser.parse(date2) - parser.parse(date1)
Result:6 days, 5:59:37.229589
answered Nov 19 at 12:16
Narendra Lucky
568
answered Nov 19 at 12:16
Narendra Lucky
568
answered Nov 19 at 12:16
Narendra Lucky
568
answered Nov 19 at 12:16
Narendra Lucky
568
568
That's an elegant way to solve it. Tku
– mr.abdo
Nov 19 at 12:27
Glad to hear this helps u ;)
– Narendra Lucky
Nov 19 at 12:30
add a comment |
That's an elegant way to solve it. Tku
– mr.abdo
Nov 19 at 12:27
Glad to hear this helps u ;)
– Narendra Lucky
Nov 19 at 12:30
That's an elegant way to solve it. Tku
– mr.abdo
Nov 19 at 12:27
That's an elegant way to solve it. Tku
– mr.abdo
Nov 19 at 12:27
Glad to hear this helps u ;)
– Narendra Lucky
Nov 19 at 12:30
Glad to hear this helps u ;)
– Narendra Lucky
Nov 19 at 12:30
add a comment |
up vote
0
down vote
Try this one (using built-in Python module):
from datetime import datetime
d1 = "2018-11-07 18:00:40.679087+00:00"
d2 = "2018-11-14 00:00:17.908676+00:00"
d1 = d1[:d1.rfind(":")] + d1[d1.rfind(":")+1:]
d2 = d2[:d2.rfind(":")] + d2[d2.rfind(":")+1:]
t1 = datetime.strptime(d1, "%Y-%m-%d %H:%M:%S.%f%z")
t2 = datetime.strptime(d2, "%Y-%m-%d %H:%M:%S.%f%z")
diff = t2 - t1
print(diff)
Output:
6 days, 5:59:37.229589
answered Nov 19 at 12:13
Vasilis G.
2,9402721
1
no need to insert/remove the:for Python 3.7
– Christian König
Nov 19 at 12:26
I am running Python 3.6 and I am gettingValueErrorif I don't remove it.
– Vasilis G.
Nov 19 at 12:28
add a comment |
up vote
0
down vote
Try this one (using built-in Python module):
from datetime import datetime
d1 = "2018-11-07 18:00:40.679087+00:00"
d2 = "2018-11-14 00:00:17.908676+00:00"
d1 = d1[:d1.rfind(":")] + d1[d1.rfind(":")+1:]
d2 = d2[:d2.rfind(":")] + d2[d2.rfind(":")+1:]
t1 = datetime.strptime(d1, "%Y-%m-%d %H:%M:%S.%f%z")
t2 = datetime.strptime(d2, "%Y-%m-%d %H:%M:%S.%f%z")
diff = t2 - t1
print(diff)
Output:
6 days, 5:59:37.229589
answered Nov 19 at 12:13
Vasilis G.
2,9402721
1
no need to insert/remove the:for Python 3.7
– Christian König
Nov 19 at 12:26
I am running Python 3.6 and I am gettingValueErrorif I don't remove it.
– Vasilis G.
Nov 19 at 12:28
add a comment |
up vote
0
down vote
up vote
0
down vote
Try this one (using built-in Python module):
from datetime import datetime
d1 = "2018-11-07 18:00:40.679087+00:00"
d2 = "2018-11-14 00:00:17.908676+00:00"
d1 = d1[:d1.rfind(":")] + d1[d1.rfind(":")+1:]
d2 = d2[:d2.rfind(":")] + d2[d2.rfind(":")+1:]
t1 = datetime.strptime(d1, "%Y-%m-%d %H:%M:%S.%f%z")
t2 = datetime.strptime(d2, "%Y-%m-%d %H:%M:%S.%f%z")
diff = t2 - t1
print(diff)
Output:
6 days, 5:59:37.229589
answered Nov 19 at 12:13
Vasilis G.
2,9402721
Try this one (using built-in Python module):
from datetime import datetime
d1 = "2018-11-07 18:00:40.679087+00:00"
d2 = "2018-11-14 00:00:17.908676+00:00"
d1 = d1[:d1.rfind(":")] + d1[d1.rfind(":")+1:]
d2 = d2[:d2.rfind(":")] + d2[d2.rfind(":")+1:]
t1 = datetime.strptime(d1, "%Y-%m-%d %H:%M:%S.%f%z")
t2 = datetime.strptime(d2, "%Y-%m-%d %H:%M:%S.%f%z")
diff = t2 - t1
print(diff)
Output:
6 days, 5:59:37.229589
answered Nov 19 at 12:13
Vasilis G.
2,9402721
edited Nov 19 at 12:20
answered Nov 19 at 12:13
Vasilis G.
2,9402721
answered Nov 19 at 12:13
Vasilis G.
2,9402721
answered Nov 19 at 12:13
Vasilis G.
2,9402721
2,9402721
1
no need to insert/remove the:for Python 3.7
– Christian König
Nov 19 at 12:26
I am running Python 3.6 and I am gettingValueErrorif I don't remove it.
– Vasilis G.
Nov 19 at 12:28
add a comment |
1
no need to insert/remove the:for Python 3.7
– Christian König
Nov 19 at 12:26
I am running Python 3.6 and I am gettingValueErrorif I don't remove it.
– Vasilis G.
Nov 19 at 12:28
1
1
no need to insert/remove the
: for Python 3.7– Christian König
Nov 19 at 12:26
no need to insert/remove the
: for Python 3.7– Christian König
Nov 19 at 12:26
I am running Python 3.6 and I am getting
ValueError if I don't remove it.– Vasilis G.
Nov 19 at 12:28
I am running Python 3.6 and I am getting
ValueError if I don't remove it.– Vasilis G.
Nov 19 at 12:28
add a comment |
up vote
0
down vote
The : is bugging you, you can parse it like:
d = '2018-11-07 18:00:40.679087+00:00'
d = d[:-3]+d[-2:]
datetime.strptime(d, '%Y-%m-%d %H:%M:%S.%f%z')
datetime.datetime(2018, 11, 7, 18, 0, 40, 679087, tzinfo=datetime.timezone.utc)
answered Nov 19 at 12:26
Franco Piccolo
1,325611
1
no need to insert/remove the:for Python 3.7
– Christian König
Nov 19 at 12:27
add a comment |
up vote
0
down vote
The : is bugging you, you can parse it like:
d = '2018-11-07 18:00:40.679087+00:00'
d = d[:-3]+d[-2:]
datetime.strptime(d, '%Y-%m-%d %H:%M:%S.%f%z')
datetime.datetime(2018, 11, 7, 18, 0, 40, 679087, tzinfo=datetime.timezone.utc)
answered Nov 19 at 12:26
Franco Piccolo
1,325611
1
no need to insert/remove the:for Python 3.7
– Christian König
Nov 19 at 12:27
add a comment |
up vote
0
down vote
up vote
0
down vote
The : is bugging you, you can parse it like:
d = '2018-11-07 18:00:40.679087+00:00'
d = d[:-3]+d[-2:]
datetime.strptime(d, '%Y-%m-%d %H:%M:%S.%f%z')
datetime.datetime(2018, 11, 7, 18, 0, 40, 679087, tzinfo=datetime.timezone.utc)
answered Nov 19 at 12:26
Franco Piccolo
1,325611
The : is bugging you, you can parse it like:
d = '2018-11-07 18:00:40.679087+00:00'
d = d[:-3]+d[-2:]
datetime.strptime(d, '%Y-%m-%d %H:%M:%S.%f%z')
datetime.datetime(2018, 11, 7, 18, 0, 40, 679087, tzinfo=datetime.timezone.utc)
answered Nov 19 at 12:26
Franco Piccolo
1,325611
answered Nov 19 at 12:26
Franco Piccolo
1,325611
answered Nov 19 at 12:26
Franco Piccolo
1,325611
answered Nov 19 at 12:26
Franco Piccolo
1,325611
1,325611
1
no need to insert/remove the:for Python 3.7
– Christian König
Nov 19 at 12:27
add a comment |
1
no need to insert/remove the:for Python 3.7
– Christian König
Nov 19 at 12:27
1
1
no need to insert/remove the
: for Python 3.7– Christian König
Nov 19 at 12:27
no need to insert/remove the
: for Python 3.7– Christian König
Nov 19 at 12:27
add a comment |
Take a look here.
– Vasilis G.
Nov 19 at 12:02
I tried this one, but in my case the format of date contains timezone
– mr.abdo
Nov 19 at 12:08
Read the docs!
%zUTC offset in the form ±HHMM[SS[.ffffff]]. later on: Changed in version 3.7: UTC offsets can have a colon as a separator between hours, minutes and seconds– Christian König
Nov 19 at 12:11
That's the point @ChristianKönig. Tku so much.
– mr.abdo
Nov 19 at 12:15
@Networker in my case I have the timezone to consider, that's the difference. I tried a lot of things that I found here, but none of them works
– mr.abdo
Nov 19 at 12:39