I'm trying to parse out a date in google bigquery and running into this issue.
Column is SessionDateTime and is a string.
PARSE_DATE("%m%d%y", SessionDateTime) as parsed_column
Returns
Failed to parse input string "3/1/2022 03:06 PM"
Any suggestions? Is it the AM/PM causing the issue?
Able to find the answer.
PARSE_DATETIME( '%m/%d/%Y %I:%M %p', SessionDateTime )
Related
I have a table field in below format in snowflake. While trying to_date(), to_timestamp() function, its erring out with error message as - Timestamp '8/05/2018 9:03:53 PM' is not recognized.
Format - '8/05/2018 9:03:53 PM'
Specify the format in the TO_TIMESTAMP as follows:
to_timestamp('8/05/2018 9:03:53 PM','MM/DD/YYYY HH12:MI:SS AM') -- assumes MM/DD/YYYY
or
to_timestamp('8/05/2018 9:03:53 PM','DD/MM/YYYY HH12:MI:SS AM') -- assumes DD/MM/YYYY
The format you are using is ambiguous, is it 8th May or August 5th?
Due to the above we don't support that date format.
For more information have a look here
Can't format the below date using moment.js, the below statement returns Invalid Date
moment('20171206T062406927Z').format('D-MMM-YYYY');
Please help me on this.
You need to tell moment which format your date string is in:
moment('20171206T062406927Z', 'YYYYMMDD[T]HHmmssSSSZ', true).format('D-MMM-YYYY');
Edit: updated as per #VincenzoC comment to ensure the timestamp is parsed in UTC
Also fix: use HH for 24-hour format (not hh), and pass a third true parameter to ensure the timestamp is parsed in strict mode.
I am getting string result in my date and time. I want to know in which format it is coming.
String:
2015-08-14T22:26:41.9975398Z
Format tried:
yyyy-MM-dd'T'HH:mm:ss.sTZD
yyyy-MM-dd'T'HH:mm:ss.SSSZ
I think it was IsoString just read this docs : isostring . :)
I need to get current time in R in this particular format:
2014-01-07T14:57:55+05:30
Sys.time() seems to return in a different format than this. How do I exactly get this ?
Link to the format : https://en.wikipedia.org/wiki/ISO_8601
The function for converting/formatting a time string is as.POSIXct or as.POSIXlt. The documentation for these points to the docs for strptime for format symbols. This reference indicates %F is the correct symbol for ISO-8601 however, implementing that results in a format different from what you suggest.
> as.POSIXct(Sys.time(),format="%F")
[1] "2016-10-02 18:57:58 EDT"
I suspect looking at strptime you will find the combination necessary to output the exact format you need.
Is this what your looking for?
format(Sys.time(), format="%Y-%m-%dT%H:%M:%S+01:00")
format(Sys.time(), format="%Y-%m-%dT%H:%M:%S%z")
The meaning of the letters you find a the documentation of strptime() function
Given a DateTime of
10-OCT-2015 07:10 PM
How can one convert this? I've got the first part correct but it's the Time I'm having difficulty with.
ParseDateTime([column name],"dd-MMM-yyyy hh:mm")
How can AM/PM be represented?
Figured it out!
ParseDateTime([Diary Date],"dd-MMM-yyyy hh:mm tt","en-US")