read in string as datetime object with 3 digits for millisecond

2

I am reading in a string that is in a datetime format. However, instead of having 6 digits for the millisecond, it has only 3 with a letter Z at the end. How do I read in this string and make it a datetime object, and add 1 day, and write out this as a string in the above format i.e. 3 digits for the millisecond and a letter Z at the end. I tried the following code but not successful:

old_date= "2018-06-06T23:59:59.999Z"

new_date = datetime.datetime.strptime(old_date, '%Y-%m-%d %H:%M:%S.%f%Z') + datetime.timedelta(days=1)

print(new_date)

python python-3.x datetime

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asked Nov 21 '18 at 3:02

duckmanduckman

15713

add a comment | 

2

I am reading in a string that is in a datetime format. However, instead of having 6 digits for the millisecond, it has only 3 with a letter Z at the end. How do I read in this string and make it a datetime object, and add 1 day, and write out this as a string in the above format i.e. 3 digits for the millisecond and a letter Z at the end. I tried the following code but not successful:

old_date= "2018-06-06T23:59:59.999Z"

new_date = datetime.datetime.strptime(old_date, '%Y-%m-%d %H:%M:%S.%f%Z') + datetime.timedelta(days=1)

print(new_date)

python python-3.x datetime

share|improve this question

asked Nov 21 '18 at 3:02

duckmanduckman

15713

add a comment | 

2

2

2

I am reading in a string that is in a datetime format. However, instead of having 6 digits for the millisecond, it has only 3 with a letter Z at the end. How do I read in this string and make it a datetime object, and add 1 day, and write out this as a string in the above format i.e. 3 digits for the millisecond and a letter Z at the end. I tried the following code but not successful:

old_date= "2018-06-06T23:59:59.999Z"

new_date = datetime.datetime.strptime(old_date, '%Y-%m-%d %H:%M:%S.%f%Z') + datetime.timedelta(days=1)

print(new_date)

python python-3.x datetime

share|improve this question

asked Nov 21 '18 at 3:02

duckmanduckman

15713

I am reading in a string that is in a datetime format. However, instead of having 6 digits for the millisecond, it has only 3 with a letter Z at the end. How do I read in this string and make it a datetime object, and add 1 day, and write out this as a string in the above format i.e. 3 digits for the millisecond and a letter Z at the end. I tried the following code but not successful:

old_date= "2018-06-06T23:59:59.999Z"

new_date = datetime.datetime.strptime(old_date, '%Y-%m-%d %H:%M:%S.%f%Z') + datetime.timedelta(days=1)

print(new_date)

python python-3.x datetime

python python-3.x datetime

share|improve this question

asked Nov 21 '18 at 3:02

duckmanduckman

15713

share|improve this question

asked Nov 21 '18 at 3:02

duckmanduckman

15713

share|improve this question

share|improve this question

asked Nov 21 '18 at 3:02

duckmanduckman

15713

asked Nov 21 '18 at 3:02

duckmanduckman

15713

asked Nov 21 '18 at 3:02

duckmanduckman

15713

15713

add a comment | 

add a comment | 

1 Answer
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You should using this format

datetime.datetime.strptime(old_date, '%Y-%m-%dT%H:%M:%S.%fZ')

Out[180]: datetime.datetime(2018, 6, 6, 23, 59, 59, 999000)

Update

dt1=datetime.datetime.strptime(old_date, '%Y-%m-%dT%H:%M:%S.%fZ')+datetime.timedelta(days=1)

dt1.strftime('%Y-%m-%d %H:%M:%S.%f')[:-3]+'Z'

Out[196]: '2018-06-07 23:59:59.999Z'

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edited Nov 21 '18 at 3:22

answered Nov 21 '18 at 3:05

Wen-BenWen-Ben

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks. how about writing this as a string with 3 digits for the milisecondpart and Z at the end?
  
  – duckman
  Nov 21 '18 at 3:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @duckman check the update
  
  – Wen-Ben
  Nov 21 '18 at 3:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  you are the champ!
  
  – duckman
  Nov 21 '18 at 3:23
  

  
  
  
  

add a comment | 

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1 Answer
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1 Answer
1

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oldest

votes

1

You should using this format

datetime.datetime.strptime(old_date, '%Y-%m-%dT%H:%M:%S.%fZ')

Out[180]: datetime.datetime(2018, 6, 6, 23, 59, 59, 999000)

Update

dt1=datetime.datetime.strptime(old_date, '%Y-%m-%dT%H:%M:%S.%fZ')+datetime.timedelta(days=1)

dt1.strftime('%Y-%m-%d %H:%M:%S.%f')[:-3]+'Z'

Out[196]: '2018-06-07 23:59:59.999Z'

share|improve this answer

edited Nov 21 '18 at 3:22

answered Nov 21 '18 at 3:05

Wen-BenWen-Ben

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks. how about writing this as a string with 3 digits for the milisecondpart and Z at the end?
  
  – duckman
  Nov 21 '18 at 3:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @duckman check the update
  
  – Wen-Ben
  Nov 21 '18 at 3:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  you are the champ!
  
  – duckman
  Nov 21 '18 at 3:23
  

  
  
  
  

add a comment | 

1

You should using this format

datetime.datetime.strptime(old_date, '%Y-%m-%dT%H:%M:%S.%fZ')

Out[180]: datetime.datetime(2018, 6, 6, 23, 59, 59, 999000)

Update

dt1=datetime.datetime.strptime(old_date, '%Y-%m-%dT%H:%M:%S.%fZ')+datetime.timedelta(days=1)

dt1.strftime('%Y-%m-%d %H:%M:%S.%f')[:-3]+'Z'

Out[196]: '2018-06-07 23:59:59.999Z'

share|improve this answer

edited Nov 21 '18 at 3:22

answered Nov 21 '18 at 3:05

Wen-BenWen-Ben

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks. how about writing this as a string with 3 digits for the milisecondpart and Z at the end?
  
  – duckman
  Nov 21 '18 at 3:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @duckman check the update
  
  – Wen-Ben
  Nov 21 '18 at 3:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  you are the champ!
  
  – duckman
  Nov 21 '18 at 3:23
  

  
  
  
  

add a comment | 

1

1

1

You should using this format

datetime.datetime.strptime(old_date, '%Y-%m-%dT%H:%M:%S.%fZ')

Out[180]: datetime.datetime(2018, 6, 6, 23, 59, 59, 999000)

Update

dt1=datetime.datetime.strptime(old_date, '%Y-%m-%dT%H:%M:%S.%fZ')+datetime.timedelta(days=1)

dt1.strftime('%Y-%m-%d %H:%M:%S.%f')[:-3]+'Z'

Out[196]: '2018-06-07 23:59:59.999Z'

share|improve this answer

edited Nov 21 '18 at 3:22

answered Nov 21 '18 at 3:05

Wen-BenWen-Ben

1

You should using this format

datetime.datetime.strptime(old_date, '%Y-%m-%dT%H:%M:%S.%fZ')

Out[180]: datetime.datetime(2018, 6, 6, 23, 59, 59, 999000)

Update

dt1=datetime.datetime.strptime(old_date, '%Y-%m-%dT%H:%M:%S.%fZ')+datetime.timedelta(days=1)

dt1.strftime('%Y-%m-%d %H:%M:%S.%f')[:-3]+'Z'

Out[196]: '2018-06-07 23:59:59.999Z'

share|improve this answer

edited Nov 21 '18 at 3:22

answered Nov 21 '18 at 3:05

Wen-BenWen-Ben

1

share|improve this answer

share|improve this answer

edited Nov 21 '18 at 3:22

edited Nov 21 '18 at 3:22

edited Nov 21 '18 at 3:22

answered Nov 21 '18 at 3:05

Wen-BenWen-Ben

1

answered Nov 21 '18 at 3:05

Wen-BenWen-Ben

1

answered Nov 21 '18 at 3:05

Wen-BenWen-Ben

1

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks. how about writing this as a string with 3 digits for the milisecondpart and Z at the end?
  
  – duckman
  Nov 21 '18 at 3:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @duckman check the update
  
  – Wen-Ben
  Nov 21 '18 at 3:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  you are the champ!
  
  – duckman
  Nov 21 '18 at 3:23
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks. how about writing this as a string with 3 digits for the milisecondpart and Z at the end?
  
  – duckman
  Nov 21 '18 at 3:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @duckman check the update
  
  – Wen-Ben
  Nov 21 '18 at 3:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  you are the champ!
  
  – duckman
  Nov 21 '18 at 3:23
  

  
  
  
  

Thanks. how about writing this as a string with 3 digits for the milisecondpart and Z at the end?

– duckman
Nov 21 '18 at 3:15

Thanks. how about writing this as a string with 3 digits for the milisecondpart and Z at the end?

– duckman
Nov 21 '18 at 3:15

@duckman check the update

– Wen-Ben
Nov 21 '18 at 3:22

@duckman check the update

– Wen-Ben
Nov 21 '18 at 3:22

you are the champ!

– duckman
Nov 21 '18 at 3:23

you are the champ!

– duckman
Nov 21 '18 at 3:23

add a comment | 

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