I was wondering if someone could help me.
I'm very new at ASP I want to format the current date and time as follows:
yyyy-mm-dd hh:mm:ss
But all i can do is the following
Response.Write Date
Can someone help me out please.
Date formatting options are limited in Classic ASP by default, there is a function FormatDateTime() which can format your date is various ways based on the servers regional settings.
For more control over date formatting though there are built in date time functions
Year(date) - Returns a whole number representing the year. Passing Date() will give back the current year.
Month(date) - Returns a whole number between 1 and 12, inclusive, representing the month of the year. Passing Date() will return the current month of the year.
MonthName(month[, abbv]) - Returns a string indicating the specified month. Passing in Month(Date()) as the month will give back the current Month string. As suggested by #Martha
Day(date) - Returns a whole number between 1 and 31, inclusive, representing the day of the month. Passing Date() will return the current day of the month.
Hour(time) - Returns a whole number between 0 and 23, inclusive, representing the hour of the day. Passing Time() will return the current hour.
Minute(time) - Returns a whole number between 0 and 59, inclusive, representing the minute of the hour. Passing Time() will return the current minute.
Second(time) - Returns a whole number between 0 and 59, inclusive, representing the second of the minute. Passing Time() will return the current second.
IMPORTANT:
When formatting date / time values, always store the date / time value first. Also, any needed calculations (DateAdd() etc.) should be applied before attempting to format or you will get unexpected results.
The functions Month(), Day(), Hour(), Minute() and Second() all return whole numbers. Luckily there is an easy workaround that lets you pad these values quickly Right("00" & value, 2) what it does is append 00 to the front of the value then from the right take the first two characters. This ensures that all single digit values return prefixed with a 0.
Dim dd, mm, yy, hh, nn, ss
Dim datevalue, timevalue, dtsnow, dtsvalue
'Store DateTimeStamp once.
dtsnow = Now()
'Individual date components
dd = Right("00" & Day(dtsnow), 2)
mm = Right("00" & Month(dtsnow), 2)
yy = Year(dtsnow)
hh = Right("00" & Hour(dtsnow), 2)
nn = Right("00" & Minute(dtsnow), 2)
ss = Right("00" & Second(dtsnow), 2)
'Build the date string in the format yyyy-mm-dd
datevalue = yy & "-" & mm & "-" & dd
'Build the time string in the format hh:mm:ss
timevalue = hh & ":" & nn & ":" & ss
'Concatenate both together to build the timestamp yyyy-mm-dd hh:mm:ss
dtsvalue = datevalue & " " & timevalue
Call Response.Write(dtsvalue)
Note: You can build the date string in one call but decided to break it down into the three variables to make it easier to read.
How Can I Format Date
Example of Parsing a Date String (Answers provide approaches to taking a date string format and parsing it to a valid Date variable).
Format the date of the previous day format yyyymmdd with VBScript (Example of why storing date / time before performing formatting is important)
VBScript ISO8601 (Example of functions to construct an ISO 8601 compliant date string)
Related
I have column with cell format date or time (DD.MM.YYYY HH:MM:SS) and values like
03.12.2013 14:01:49
04.12.2013 10:19:27
04.12.2013 12:44:56
04.12.2013 14:20:12
04.12.2013 18:30:21
I need those values converted to unix epoch (seconds since 1970). Somehow it feels like the values are not recognized as dates, but rather as strings. I tried different formats, had little luck with dates without time.
Operations performed on date data should be automatic provided that the cells are formatted as as a user defined DD.MM.YYYY HH:MM:SS in the 'Format' > 'Cells' > 'Numbers' tab.
If you're using the standard settings, LibreOffice Calc uses 12/30/1899 as it's default date. So the first step is getting the number of days between 12/30/1899 and 1/1/1970:
=(DATE(1970,1,1) - DATE(1899,12,30)) = 25569
Number of seconds in a day:
=(60 * 60 * 24) = 86400
If, for example, in cell A2 you have the date 03.12.2013 14:01:49. I subtract the difference between Calc's default date and the Unix Epoch we just calculated, and multiply it by the number of seconds in a day:
=(A2 - 25569) * 86400
The result is a value of 1363096909 which is the Epoch time in seconds. If you need it in milliseconds, multiply the equation by 1000.
If it's something you use a lot, you can create a custom function that does this. Go to Tools > Macros > Edit Macros, and type out the following into whichever module comes up:
REM ***** BASIC *****
Function EPOCH(date_cell)
EPOCH = (date_cell - 25569)*86400
End Function
Close the macro IDE, and now you can use your EPOCH() like any other function!
This formula worked for me, where the others above did not:
= DATE( 1970, 1, 1 ) + ( A1 / 86400 )
I'm trying to compare a file's modified date with a specific date.
What I have is this:
If FormatDateTime(objFile.DateLastModified,vbShortDate) = specificDate Then
'Do something
End if
I've tried using IsDate and a variable with a value of #11/9/2015# but always returns false. I can't figure out how to set the variable "specificDate" to 11/9/2015.
If you are comparing the date only (but not the time), then you have to cut off the fractional part of the last modified value, since integer part represents days, and fractional part - hours, minutes and seconds. After any changes convert the values back to Date type with CDate(), as well as date containing string before the comparison.
Sub Test()
dtSpecificDate = CDate("11/9/2015")
With CreateObject("Scripting.FileSystemObject").GetFile("C:\Test\tmp.txt")
dtLastModified = CDate(Int(.DateLastModified))
End With
If dtLastModified = dtSpecificDate Then
' Do something
End If
End Sub
I'm attempting to pull some information out of text file that is updated after I query a piece of equipment. The text file contains lines such as shown here (abbreviated):
05-Nov-13 11:11:54.3496 ( -1 7020 10244) scpeng.exe:Automation Server...
05-Nov-13 14:10:54.3496 ( -1 7020 10244) scpeng.exe:Automation Server...
05-Nov-13 14:10:54.3496 ( -1 7020 10244) scpeng.exe:Automation Server...
05-Nov-13 14:10:56.3496 ( -1 7020 10244) scpeng.exe:CServer.cpp,....
The text file can contain up to several weeks of information. I have a subroutine that will run a few seconds after I query the equipment which should allow for the reply and the applicable line to be present in the text file. In the routine, I am trying to scroll through the lines examining the date to arrive at the date of the subroutine call followed by the time (or a time ~10 seconds prior the the current time) to arrive at the lines pertinent to where the information could be found.
do
msg = msgstream.ReadLine
logdate = mide(msg,1,9)
logday = Cdate(logdate)
loop while logday < date
do
msg = msgstream.Readline
logtime = mid(msg,12,8)
'logtime = CDate(logtime) This mod is not working
loop while logtime < time
The date loop appears to work however the time is giving me problems. It does not error out but I can't get it to run beyond one line of text. Can anyone suggest a fix or better option? I have read that the built-in Date function can include the time but I do not believe this version I'm using does. Also, the text file contains times in a 24 hour format where I believe the time function returns values in a 12 hr format ie "12:43:27 PM ST".
You're making this way too complicated. Simply parse the whole date string into a datetime value:
refdate = Now
Do
msg = msgstream.ReadLine
logdate = CDate(Mid(msg, 1, 19))
Loop While logdate < refdate
You can extract date and time portions from the value later, e.g. like this:
WScript.Echo DateValue(logdate)
WScript.Echo TimeValue(logdate)
Also, Time returns the current (unformatted) system time. Whether it's displayed in 12 hour or 24 hour format depends on your system's region settings. However, you can always get the hour (0-23) by using the Hour function.
Parse each line with a regex to get the correct date and time part. I prefer a regexp above string manipulation functions because you can separate format and code.
Reassemble the date from the two parts and see if the date is smaller than yesterday at this time.
Option Explicit
dim strTest, re, matches, myDatePart, myTimePart, logDate
' teststring
strTest = "08-Nov-13 14:10:56.3496 ( -1 7020 10244) scpeng.exe:CServer.cpp,...."
Set re = new regexp
' This pattern extracts two part, the date as (dd-www-dd) and the time as (hh:mm:ss)
re.pattern = "(\d{2}-\w{3}-\d{2}) +(\d{2}:\d{2}:\d{2})"
Set matches = re.Execute(strTest)
' Get the first and second submatch to define the date and time
myDatePart = matches(0).submatches(0)
myTimePart = matches(0).submatches(1)
' datevalue and timevalue automatically tranforms to Date type
logDate = datevalue(myDatePart) + timevalue(myTimePart)
' See if the date is smaller than yesterday exactly this time
msgbox (logDate < (DateAdd("d", -1, now))) ' Returns True, because 08 Nov is earlier than yesterday.
I have the following code
String test = "21/04/2013";
fmt = DateTimeFormat.getFormat("MM/dd/yyyy");
Date dateTest = fmt.parse(test);
Window.alert(fmt.format(dateTest));
And the alert box shows the date
09/04/2014
instead of
21/04/2013
Why?
As others already say, it's because of your pattern. What they don't say is why it behaves that way.
When parsing 21/04/2013 as MM/dd/yyyy, DateTimeFormat will decompose the date as:
Month Day of month Year
21 4 2013
and it'll then adjust things to make a valid date. To do that, the Month part is truncated at 12 (so that temporary date is Dec 4th, 2013) and the remainder (21 - 12 = 9) is then added, leading to Sept. 4th 2014, which according to your format displays as 09/04/2014.
You wanted to show 21/04/2013 but the format was MM/dd/yyyy.
It should be dd/MM/yyyy
So change it like this:
String test = "21/04/2013";
fmt = DateTimeFormat.getFormat("dd/MM/yyyy");
Date dateTest = fmt.parse(test);
Window.alert(fmt.format(dateTest));
You're reversing day and month.
String test = "21/04/2013";
fmt = DateTimeFormat.getFormat("dd/MM/yyyy");
Date dateTest = fmt.parse(test);
Window.alert(fmt.format(dateTest));
I have some data files with Unix timestamps (in this case, number of milliseconds since Jan 1, 1970 00:00 UTC). I would like to convert these to human-friendly date/time strings (e.g. 31-Aug-2012 11:36:24) in Matlab. Is there an easy way to do this in Matlab, or am I better off using an external library (e.g. java.text.SimpleDateFormat)?
How about
date = datestr(unix_time/86400 + datenum(1970,1,1))
if unix_time is given in seconds, unix_time/86400 will give the number of days since Jan. 1st 1970. Add to that the offset used by Matlab's datenum (datenum(0000,1,1) == 1), and you have the amount of days since Jan. 1st, 0000. This can be easily converted to human-readable form by Matlab's datestr.
If you have milliseconds, just use
date = datestr(unix_time/86400/1000 + datenum(1970,1,1))
Wrapped in functions, these would be
function dn = unixtime_to_datenum( unix_time )
dn = unix_time/86400 + 719529; %# == datenum(1970,1,1)
end
function dn = unixtime_in_ms_to_datenum( unix_time_ms )
dn = unix_time_ms/86400000 + 719529; %# == datenum(1970,1,1)
end
datestr( unixtime_to_datenum( unix_time ) )
Newer versions of MATLAB (verified in R2015a) have a datetime type that is useful for working with and formatting dates and times. You can convert UNIX timestamps into a MATLAB datetime with
dt = datetime( unix_time, 'ConvertFrom', 'posixtime' );
and then use datestr as before for formatting as a string
datestr( dt )