I get a Date in the following format from an api:
Mon Apr 29 14:40:17 2019
I try to parse it to a valid powershell Date with the following command:
$test = [DateTime]::ParseExact("Mon Apr 29 14:40:03 2019", "ddd MMM dd HH:mm:ss yyyy",$null)
Powershell returns "Exception calling "ParseExact" with "3" argument(s): "String was not recognized as a valid DateTime."
It seems that the problem occurs because of the abbreviated weekday format. If I remove "Mon" and "ddd" the parse works.
The information about the Format specifier is from: https://learn.microsoft.com/en-us/dotnet/standard/base-types/custom-date-and-time-format-strings#dddSpecifier
Anyone knows what causese the error?
Replacing $null with [System.Globalization.CultureInfo]::InvariantCulture solved the problem.
The working code is:
$test = [DateTime]::ParseExact("Mon Apr 29 14:40:03 2019", "ddd MMM dd HH:mm:ss yyyy",[System.Globalization.CultureInfo]::InvariantCulture)
Related
In my YAML Header I try this for date:
date: "`!expr format(Sys.time(), '%B %d, %Y')`"
date-format: dddd, D MMM YYYY
lang: de
But if I render I receive an error message:
Invalid Date
Did I use !expr in the correct way?
I have some saved dates in JavaScript using new Date() that looks like:
"Sun Feb 24 2019 14:44:20 GMT+0200 (Eastern European Standard Time)"
I'm trying to parse these to Elixir DateTime; I didn't find anything in "timex" that can help and I already know that I can use DateTime.from_iso8601 but for dates saved using new Date().toISOString() but what i need is to parse the above string.
Thanks in advance
You can use elixir binary pattern matching to extract the date parts and parse using Timex's RFC1123 format. The RFC1123 is the format e.g Tue, 05 Mar 2013 23:25:19 +0200. Run h Timex.Format.DateTime.Formatters.Default in iex to see other formats.
iex> date_string = "Sun Feb 24 2019 14:44:20 GMT+0200 (Eastern European Standard Time)"
iex> <<day_name::binary-3,_,month_name::binary-3,_,day::binary-2,_,year::binary-4,_,time::binary-8,_::binary-4,offset::binary-5,_,rest::binary>> = date_string
iex> Timex.parse("#{day_name}, #{day} #{month_name} #{year} #{time} #{offset}", "{RFC1123}")
iex> {:ok, #DateTime<2019-02-24 14:44:20+02:00 +02 Etc/GMT-2>}
Pattern matching:
The binary-size are in byte sizes. 1 byte == 1 character. For instance to get
3-character day_name the size is 3. Underscores (_) is used to pattern match the spaces in the date format
Updated answer to use binary-size rather than bitstring-size for simplicity
I didn't find anything in "timex" that can help
The Timex Parsing docs say that you can use strftime sequences, e.g %H:%M:%S, for parsing. Here's a list of strftime characters and what they match.
Here's a format string that I think should work on javascript Dates:
def parse_js_date() do
Timex.parse!("Sun Feb 24 2019 14:44:20 GMT+0200 (Eastern European Standard Time)",
"%a %b %d %Y %H:%M:%S GMT%z (%Z)",
:strftime)
end
Unfortunately, %Z doesn't want to match the time zone name, which causes Timex.parse!() to spit out an error. It looks like %Z in Elixir only matches one word, e.g. a timezone abbreviation EET. Therefore, my simple, clean solution is spoiled.
What you can do is chop off the time zone name before parsing the date string:
def parse_js_date_string() do
[date_str|_tz_name] = String.split(
"Sun Feb 24 2019 14:44:20 GMT+0200 (Eastern European Standard Time)",
" (",
parts: 2
)
Timex.parse!(date_str,
"%a %b %d %Y %H:%M:%S GMT%z",
:strftime)
end
In iex:
~/elixir_programs/my$ iex -S mix
Erlang/OTP 20 [erts-9.3] [source] [64-bit] [smp:4:4] [ds:4:4:10] [async-threads:10] [hipe] [kernel-poll:false]
Compiling 1 file (.ex)
Interactive Elixir (1.6.6) - press Ctrl+C to exit (type h() ENTER for help)
iex(1)> My.parse_js_date_string()
#DateTime<2019-02-24 14:44:20+02:00 +02 Etc/GMT-2>
iex(2)>
Im trying to use [datetime]::ParseExact
The string i need to convert to datetime is Wed Jun 27 08:50:00 2018 -0500
I couldnt figure out the correct format to convert it correctly.
Please help me out. Thanks
Try this as well:
$Date = 'Wed Jun 27 08:50:00 2018 -0500'
[datetime]::ParseExact($Date,"ddd MMM dd HH:mm:ss yyyy zzz",[CultureInfo]::InvariantCulture)
There are several options. To some extent it will depend on what you want to do with that time offset. Here is one way that ignores the offset:
$test = 'Wed Jun 27 08:50:00 2018 -0500'
$parts = $test.split(' ')
$date = get-date ('{0}{1}{2} {3}' -f $parts[2], $parts[1], $parts[4], $parts[3])
I'm trying to change date format from 08/03/2017
08 day
03 Month
2017 year
I I'm using
date("d F Y", strtotime($date));
the problem is that I Get 03 August 2017 instead of 08 March 2017
PS : I can't use any other then
dd/mm/yyyy
Try this
$date = DateTime::createFromFormat('d/m/Y', "08/03/2017");
echo $date->format('d F Y');
You can check it to http://php.net/manual/en/datetime.createfromformat.php
After Some researchs I found a solution
str_replace(' / ', '-',$date)
/*
You got August because you provide wrong format to this function. Function accept m/d/y and you provide d/m/y. you can try with this format m/d/y
or
you can try this with DateTime object.
*/
$date = "08/03/2017";
$dateObject = DateTime::createFromFormat('d/m/Y', $date);
echo $dateObject->format('d M Y');
I have some date other than current date in unix and I want to convert into a specific format
Original Format
D="Mon Dec 30 06:35:02 EST 2013"
New Format
E=20131230063502
E=`date +%Y%m%d%H%M%S`
this is the way to format the output of the date command and save it in the variable E
Using python:
def data(dstr):
m = {'Jan': '01', 'Feb':'02', 'Mar':'03', 'Apr':'04', 'May':'05', 'Jun':'06', 'Jul':'07', 'Aug':'08', 'Sep':'09', 'Oct':'10', 'Nov':'11', 'Dec':'12'}
val = dstr.split(' ')
month = m[val[1]]
time = val[3].split(':')
return '{}{}{}{}{}{}'.format(val[-1],month,val[2],time[0],time[1],time[2])
if __name__ == '__main__':
print data("Mon Dec 30 06:35:02 EST 2013")
In: Mon Dec 30 06:35:02 EST 2013
Out: 20131230063502