I have the below data which is an object type variable named $timestamps
Sat Jan 15 16:21:24
Sat Jan 15 01:31:22
Fri Jan 14 20:58:09
Fri Jan 14 20:51:02
I'm having trouble converting it to Datetime object because of the weird date format. How would you handle this?
I would like it as a datetime object because I plan to convert from current (UTC) to EST.
TIA
You can use the the ParseExact() method provided by the [datetime] class for this:
[datetime]::ParseExact('Fri Jan 14 20:58:09','ddd MMM dd HH:mm:ss',$null)
# returns a - datetime - object of:
# Friday, January 14, 2022 8:58:09 PM
dd - for the day.
MM - for the month.
HH - for the hour - Capitalized for the 24 hour time format.
mm - for the minutes.
ss - for the seconds.
Edit: as suggested by mklement0, we can use [cultureinfo]::InvariantCulture to make the parsing specific to an English date time format. Also, changing dd to d as a more robust solution for days without 2 digits; which should cover both singular, and double digit days.
Seeing $timestamps is an array of strings, you can use a loop (of your choice - in this case the Foreach-Object cmdlet) to iterate through each string parsing the text to return a datetime object:
$timestamps | ForEach-Object {
$culture = [cultureinfo]::InvariantCulture
$format = 'ddd MMM d HH:mm:ss'
$date = [datetime]::ParseExact($_,$format,$culture,'AssumeUniversal, AdjustToUniversal')
[System.TimeZoneInfo]::ConvertTimeBySystemTimeZoneId($date, 'Eastern Standard Time')
}
Using 'AssumeUniversal, AdjustToUniversal' ensures a UTC output.
Assuming from your comment that you'd like to do a conversion to Eastern Time, passing the newly created datetime object to [System.TimeZoneInfo]::ConvertTimeBySystemTimeZoneId() with an argument of the desired time zone, you can get your result in the new time zone.
When using $null, the CultureInfo object that corresponds to the current culture is used.
The DateTime.ParseExact() method is probably what you're looking for.
PS C:\TEMP>$timestamp = 'Sat Jan 15 16:21:24'
PS C:\TEMP>$format = 'ddd MMM dd HH:mm:ss'
PS C:\TEMP>[datetime]::ParseExact($timestamp, $format, $null)
Saturday, January 15, 2022 04:21:24 PM
PS C:\TEMP>
Related
I am having a string which contains 24 hr format.
I am trying to convert it to 24hrs format but it is not changing. below is my code.
$fromtime = '2018-03-28,23:37:50'
[datetime]$fromtime24hrFormat = ([datetime]$fromtime).ToString("yyyy-MM-dd,HH:mm:ss")
$fromtime24hrFormat
Wednesday, March 28, 2018 11:37:50 PM
It shows in PM which is correct. But is it not possible to show it in 24 hr format?
You're so close!
$fromtime = '2018-03-28,23:37:50'
[datetime]$fromtime24hrFormat = ([datetime]$fromtime)
$fromtime24hrFormat.ToString("yyyy-MM-dd,HH:mm:ss")
The problem is that your last line was effectively just dumping the datetime object to the output; which will use a default formatting.
Did you want to add AM/PM to the original input? Try:
$fromtime = '2018-03-28,23:37:50'
$fromtime24hrFormat = [datetime]$fromtime
$fromtime24hrFormat.ToString("yyyy-MM-dd,HH:mm:ss tt")
2018-03-28,23:37:50 p.m.
"p.m." vs PM is caused by my Norwegian regional settings
As today is Wednesday with date June 8, 2016. how can i write a code to get the day of given dates:
like what day is Nov 29
I'm trying to create a struct with
date
day
month
with month and date as input
Use the builtin weekday() function:
>> [num, name] = weekday('08-Jun-2016')
num =
4
name =
Wed
>> [num, name] = weekday('29-Nov-2016')
num =
3
name =
Tue
In addition to the weekday function, you can use the DDD or DDDD formats in the datestr function, like this:
datestr('08-Jun-2016','DDD') %Returns the string 'Wed'
datestr('08-Jun-2016','DDDD') %Returns the string 'Wednesday'
Or, to use a more practical format
datestr('08-Jun-2016','DDDD, mmmm DD, yyyy')
% Returns the string: 'Wednesday, June 08, 2016'
I want to convert a date to this format "dd Mon yyyy".
I have this code which works:
$date = [DateTime]::Parse("21/11/2014")
$dateFormatted = $date.GetDateTimeFormats()[12]
#$dateFormatted displays 21 November 2014
Is there a way to convert it using something like this?:
$dateFormatted = $date.ToString("dd Mon yyyy")
At the moment this returns "21 11on 2014"
I worked it out:
$dateFormatted = $date.ToString("dd MMMM yyyy")
I can't get my head around how formatting a datetime variable inside a string works in PowerShell.
$startTime = Get-Date
Write-Host "The script was started $startTime"
# ...Do stuff...
$endTime = Get-Date
Write-Host "Done at $endTime. Time for the full run was: $( New-TimeSpan $startTime $endTime)."
gives me the US date format while I want ISO 8601.
I could use
$(Get-Date -Format u)
but I want to use $endTime to make the calculation of the timespan correct.
I have tried all permutations of $, (, ), endTime, -format, u, .ToString(...) and .ToShortDate(), but the one that works.
"This is my string with date in specified format $($theDate.ToString('u'))"
or
"This is my string with date in specified format $(Get-Date -format 'u')"
The sub-expression ($(...)) can include arbitrary expressions calls.
Microsoft Documents both standard and custom DateTime format strings.
You can use the -f operator
$a = "{0:D}" -f (get-date)
$a = "{0:dddd}" -f (get-date)
Spécificator Type Example (with [datetime]::now)
d Short date 26/09/2002
D Long date jeudi 26 septembre 2002
t Short Hour 16:49
T Long Hour 16:49:31
f Date and hour jeudi 26 septembre 2002 16:50
F Long Date and hour jeudi 26 septembre 2002 16:50:51
g Default Date 26/09/2002 16:52
G Long default Date and hour 26/09/2009 16:52:12
M Month Symbol 26 septembre
r Date string RFC1123 Sat, 26 Sep 2009 16:54:50 GMT
s Sortable string date 2009-09-26T16:55:58
u Sortable string date universal local hour 2009-09-26 16:56:49Z
U Sortable string date universal GMT hour samedi 26 septembre 2009 14:57:22 (oups)
Y Year symbol septembre 2002
Spécificator Type Example Output Example
dd Jour {0:dd} 10
ddd Name of the day {0:ddd} Jeu.
dddd Complet name of the day {0:dddd} Jeudi
f, ff, … Fractions of seconds {0:fff} 932
gg, … position {0:gg} ap. J.-C.
hh Hour two digits {0:hh} 10
HH Hour two digits (24 hours) {0:HH} 22
mm Minuts 00-59 {0:mm} 38
MM Month 01-12 {0:MM} 12
MMM Month shortcut {0:MMM} Sep.
MMMM complet name of the month {0:MMMM} Septembre
ss Seconds 00-59 {0:ss} 46
tt AM or PM {0:tt} ““
yy Years, 2 digits {0:yy} 02
yyyy Years {0:yyyy} 2002
zz Time zone, 2 digits {0:zz} +02
zzz Complete Time zone {0:zzz} +02:00
: Separator {0:hh:mm:ss} 10:43:20
/ Separator {0:dd/MM/yyyy} 10/12/2002
Instead of using string interpolation you could simply format the DateTime using the ToString("u") method and concatenate that with the rest of the string:
$startTime = Get-Date
Write-Host "The script was started " + $startTime.ToString("u")
im trying to formatting the date field 'created_at' from Twitter API response with Zend_Date. I want output the date like this:
21 of July of 2009, 12:30:00 (for example)
What format is this?:
Fri Oct 23 15:47:42 +0000 2009
thanks a lot
I've had the best luck just doing
$d = new Zend_Date(strtotime($input));
$twitter_format_out = $d->toString('EEE MMM dd HH:mm:ss Z YYY');
These date are not looking a standard format. Therefore, you have to create a format with the right constants (see them here).
Your first example (21 of July of 2009, 12:30:00):
$format = "d ' of ' MMMM ' of ' YYYY, h:mm:ss";
Your second example (Fri Oct 23 15:47:42 +0000 2009):
$format = "EEE MMM d h:mm:ss Z YYYY";
This formats you can use both for importing a date
$date = new Zend_Date($string, $format);
Or for outputting
$date->toString($format);
Look in the manual for locale support etc.