This question already has answers here:
date format function to display date as "Jan 13 2014"
(3 answers)
Closed 7 years ago.
I writing a code in VBScript, but I cannot get the datetime part right.
I'm using FormatDateTime(now), but the gives not the best result like
FormatDateTime(now)
8-01-2016 9:05:12 becomes 01-08-2016 9:05:12.
28-01-2016 19:01:18 stays 28-01-2016 19:01:18.
Has to be:
8-01-2016 9:05:12
28-01-2016 19:01:18
Is there a way to get both the same?
Try using the format argument of FormatDateTime.
FormatDateTime(now, 2) 'This should return in mm/dd/yy format
More info here.
http://www.w3schools.com/asp/func_formatdatetime.asp
format (Optional) A value that specifies the date/time format to use
Can take the following values:
0 = vbGeneralDate - Default. Returns date: mm/dd/yy and time if
specified: hh:mm:ss PM/AM.
1 = vbLongDate - Returns date: weekday,
monthname, year
2 = vbShortDate - Returns date: mm/dd/yy
3 =
vbLongTime - Returns time: hh:mm:ss PM/AM
4 = vbShortTime - Return
time: hh:mm
This should do what you want.
od = "28-01-2016 19:01:18"
nd = FormatDateTime(od,0)
MsgBox(nd)
Output: 1/28/2016 7:01:18 PM
Related
This question already has answers here:
UTC time stamp format - Split in parts a string containing the UTC time date (classic asp/vbscript)
(2 answers)
Closed 1 year ago.
I've ready numerous posts here about the mismatch error but every format I try I still get the same error.
Dim calendarDate, sDate, sFinal
calendarDate = "Sun Apr 05 00:00:00 CDT 2020"
sYear = Right(calendarDate, 4)
sDay = Mid(calendarDate,9,2)
sMonth = Mid(calendarDate,5,3)
If sMonth = "Apr" Then sMonth = "04" Else sMonth = sMonth
End if
sDate = sYear & "-" & sMonth & "-" & sDay
sFinal = CDate(sDate)
I get the type mismatch CDate error but the format should work? I have also tried MM/DD/YYYY.
And I have tried sFinal = DateSerial(sYear,sMonth,SDay) also does not work. But if you don't use the variables...
sFinal = DateSerial("2020","04","05") this works. I don't understand why my SYear, SMonth, SDay would not work as they are the same numbers."
My end goal here is to subtract 14 days from the calendar date but i can't even get my variable into a date format to subtract it...so maybe there is something simpler I should be doing here?
Thank you all for any help you can give much appreciated! Long time reader, first time posting.
The simplest way to manipulate will be to use Split() as each value separated by a Space (Chr(32)).
What you end up with is an array containing each element that made up the string split by the Space character (Chr(32)), so by concatenating values back together we can construct our date-time how CDate() expects, excluding the first and sixth element which will be Sun and CDT respectively, as well as re-ordering the year in the constructed string that is passed to CDate().
Dim input: input = "Sun Apr 05 00:00:00 CDT 2020"
Dim data: data = Split(input, Chr(32))
'Ignore first and sixth element in the array and build our date value
Dim output: output = CDate(data(1) & " " & data(2) & " " & data(5)) & " " & TimeValue(data(3))
Call Wscript.Echo(output)
Output:
05/04/2020 00:00:00
Note: Result will be based on the user's regional settings as CDate() uses this while parsing a Date string.
Useful Linkss
UTC time stamp format - Split in parts a string containing the UTC time date (classic asp/vbscript) (Example of parsing date string).
I am trying to convert the string to the type of 'datetime' in python. My data match the format, but still get the
'ValueError: time data 11 11 doesn't match format specified'
I am not sure where does the "11 11" in the error come from.
My code is
train_df['date_captured1'] = pd.to_datetime(train_df['date_captured'], format="%Y-%m-%d %H:%M:%S")
Head of data is
print (train_df.date_captured.head())
0 2011-05-13 23:43:18
1 2012-03-17 03:48:44
2 2014-05-11 11:56:46
3 2013-10-06 02:00:00
4 2011-07-12 13:11:16
Name: date_captured, dtype: object
I tried the following by just selecting the first string and running the code with same datetime format. They all work without problem.
dt=train_df['date_captured']
dt1=dt[0]
date = datetime.datetime.strptime(dt1, "%Y-%m-%d %H:%M:%S")
print(date)
2011-05-13 23:43:18
and
dt1=pd.to_datetime(dt1, format='%Y-%m-%d %H:%M:%S')
print (dt1)
2011-05-13 23:43:18
But why wen I using the same format in pd.to_datetime to convert all the data in the column, it comes up with the error above?
Thank you.
I solved it.
train_df['date_time'] = pd.to_datetime(train_df['date_captured'], errors='coerce')
print (train_df[train_df.date_time.isnull()])
I found in line 100372, the date_captured value is '11 11'
category_id date_captured ... height date_time
100372 10 11 11 ... 747 NaT
So the code with errors='coerce' will replace the invalid parsing with NaN.
Thank you.
I need to format subtract 2 long date time result to "hh.mm.ss"
open date closed date subtracted
01/01/2012 16:04 04/01/2012 17:07 3.01:02:58
02/01/2012 08:52 02/01/2012 17:03 08:10:27
using
closeddate.subtract(opendate)
return result in format D.hh:mm:ss
and using
DateDiff("h", OpenDate, closeDate)
return hours only (no minutes or seconds)
Also tried
closeddate.subtract(opendate).tostring("hh:mm:ss")
not working
as many Microsoft forums answered there is no way to do that
so i used
DateDiff("s", OpenDate, closeDate)
and passed result to helper function return calculated time in format hh:mm:ss
I have a ruby script that checks a provided date, to make sure it is today's date. This is not working when the date provided doesn't have a 2 digit padding for the month. Is there anyway to get ruby to see that as equal? The example is that it says "Date Processed 3/13/2014 is not today's date 03/13/2014!" the difference is in the month - 3 vs 03. Below is the code. ev_val is provided from a csv and it is m/d/yyyy format. It is not provided with a 0 padding, though. Any thoughts?
Thanks!
tnow = Time.now
if ev_val != tnow.strftime("%m/%d/%Y")
log_linemsg = "Date Processed #{ev_val} is not today's date #{tnow.strftime("%m/%d/%Y")}! Processing date must be today's Date!!!\nSTOPPING SCRIPT!!!"
log_line = ["#{$cname}","#{log_linemsg}","","",]
puts log_linemsg
insert_logitems(connection, table_namelog, log_line)
exit
end
require "date"
date_val = Date.parse ev_val
today = Date.today
if today != date_val
log_linemsg = "Date Processed #{ev_val} is not today's date #{today}! Processing date must be today's Date!!!\nSTOPPING SCRIPT!!!"
end
Since you only care about the date portion, I would use Date instead of Time.
Take your input string and parse it into a Date object, then compare it to today's date.
?> date_val = Date.parse('3/13/2014')
=> Thu, 13 Mar 2014
>> date_val == Date.today
=> true
In your example Date.parse(ev_val) != Date.today should work for the comparison.
I have a text_field :birthday_line in my user form, that I need to parse into the user's birthday attribute.
So I'm doing something like this in my User class.
attr_accessor :birthday_line
before_save :set_birthday
def set_birthday
self.birthday = Date.strptime(birthday_line, I18n.translate("date.formats.default")
end
But the problem is that for some reason it gives me an error saying Invalid date when I try to pass in a string 27 января 1987 г. wich should be parsed to 1987-01-27.
The format and month names in my config/locales/ru.yml
ru:
date:
formats:
default: "%d %B %Y г."
month_names: [~, января, февраля, марта, апреля, мая, июня, июля, августа, сентября, октября, ноября, декабря]
seem to be correct.
Date.parse also doesn't help, it just parses the day number (27) and puts the month and year to todays date (so it'll be September 27 2013 instead of January 27 1987).
I had the same problem and what I can suggest:
string_with_cyrillic_date = '27 Января 1987'
1)create array of arrays like this
months = [["января", "Jan"], ["февраля", "Feb"], ["марта", "Mar"], ["апреля", "Apr"], ["мая", "May"], ["июня", "Jun"], ["июля", "Jul"], ["августа", "Aug"], ["сентября", "Sep"], ["октября", "Oct"], ["ноября", "Nov"], ["декабря", "Dec"]]
2) Now you can iterate this and find your cyrillic month:
months.each do |cyrillic_month, latin_month|
if string_with_cyrillic_date.match cyrillic_month
DateTime.parse string_with_cyrillic_date.gsub!(/#{cyrillic_month}/, latin_month)
end
end
And now you will receive the date that you expect
27 Jan 1987