I'm having a problem with Coldfusion's DateDiff(). I'm trying to get the difference between two dates with times, like the following examples:
fromdate=06/11/2017 22:10
todate =16/11/2017 23:20
should return:
10 days, 1 hour and 10 minutes
fromdate=06/11/2017 22:10
todate =16/11/2017 20:20
should return:
9 days, 22 hours, 10 minutes
Any help?
Code:
<cfset dtFrom = "11/06/2017 22:10" />
<cfset dtTo = "11/16/2017 23:20" />
<cfoutput>
#DateDiff( "d", dtFrom, dtTo)# Days,
#DateDiff( "h", dtFrom, dtTo) % 24# Hours
#DateDiff( "n", dtFrom, dtTo) % 24 % 60# Minutes
</cfoutput>
In addition to the previous suggestion, DateDiff() isn't going to understand those specific strings or that "06/11/2017" should mean November 6. The result will be:
158 days 1 hours 10 minutes
For it to work as expected, you must convert the strings into date objects first. For example use LSParseDateTime with the right Locale.
fromDate = lsParseDateTime("06/11/2017 22:10", "English (UK)", "dd/MM/yyyy hh:mm");
toDate = lsParseDateTime("16/11/2017 23:20", "English (UK)", "dd/MM/yyyy hh:mm");
or possibly:
fromDate = lsParseDateTime("06/11/2017 22:10", "English (UK)");
toDate = lsParseDateTime("16/11/2017 23:20", "English (UK)");
Here is one way.
totalMinutes = datediff("n", fromDate, toDate);
days = int(totalMinutes /(24 * 60)) ;
minutesRemaining = totalMinutes - (days * 24 * 60);
hours = int(minutesRemaining / 60);
minutes = minutesRemaining mod 60;
writeoutput(days & ' days ' & hours & ' hours ' & minutes & ' minutes');
Related
I want to convert seconds to days, hours and minutes
Currently, it works just for hours and minutes but not for days. Can you please support me tell me what I did wrong:
<cfscript>
seconds = '87400';
midnight = CreateTime(0,0,0);
time = DateAdd("s", seconds, variables.midnight);
date= xxxxxxxxxxxxxxxxxxxxx???
</cfscript>
<cfoutput>
#DateFormat(variables.date, 'd')# not working
#TimeFormat(variables.time, 'HH:mm')#
</cfoutput>
For the value 87400 the expected result is
1 Days, 0 hours, 16 minutes
If I take 94152 seconds it will be:
1 days, 3 hours, 22 minutes
The only issue i have is to get the correct days ... hours and minutes are diplayed but not the correct days
thank you for all the support
A simple way to calculate the intervals is by taking advantage of the modulus operator:
totalSeconds = 94152;
days = int(totalSeconds / 86400);
hours = totalSeconds / 3600 % 24;
minutes = totalSeconds / 60 % 60;
seconds = totalSeconds % 60;
For 94152 seconds, the results would be:
Interval
Value
DAYS
1
HOURS
2
MINUTES
9
SECONDS
12
TOTALSECONDS
94152
demo trycf.com
I understand from your question that you don't need to get a certain date and time along a timeline, but convert a total amount of seconds in days, hours and minutes. To do that you don't necessary need to use cfml time and date functions like CreateTime() or DateAdd(). You just may need these in order to get a reference point of time or date along a timeline, which doesn't seem to be the case, otherwise you would know the value of your starting date variable. Thus, you can solve this with plain rule of three. There may be simpler methods, so I'm posting an alternative only.
We know that:
60 seconds is equivalent to 1 minute
60 minutes is equivalent to 1 hour
24 hours is equivalent to 1 day
Thus, your calcualtion within cfml could be like so:
<cfscript>
//Constants for calculation
secondsPerDay= 60*60*24;
secondsPerHour= 60*60;
secondsPerMinute= 60;
//Seconds to convert
secondsTotal=87400;
// temp variable
secondsRemain= secondsTotal;
days= int( secondsRemain / secondsPerDay);
secondsRemain= secondsRemain - days * secondsPerDay;
hours= int( secondsRemain / secondsPerHour);
secondsRemain= secondsRemain - hours * secondsPerHour;
minutes= int( secondsRemain / secondsPerMinute);
secondsRemain= secondsRemain - minutes * secondsPerMinute;
writeoutput( "#secondsTotal# seconds are: #days# days, #hours# hours, #minutes# minutes and #secondsRemain# seconds." );
</cfscript>
That outputs:
87400 seconds are: 1 days, 0 hours, 16 minutes and 40 seconds.
How to calculate Duration of Time between two date time through VBscript
Date1 = 2021-01-22 11:43:38.000
Date2 = 2021-01-22 14:32:38.000
result should be HH:MM:SS
TimeSerial and FormatDateTime return a date or time that will take the regional settings of the computer into account. On my European computer no AM extension is shown because we use the 24h time format.
An additional problem with TimeSerial is that it will overflow once there are more than 32767 seconds.
A different approach could be to calculate the values for hours, minutes and seconds separately. A possible solution could be:
secValue = DateDiff("s",Date1,Date2)
hours = secValue \ 3600
hh = hours
if hours < 10 then
hh = Right("0" & hours, 2)
end if
mm = Right("0" & (secValue - hours * 3600) \ 60, 2)
ss = Right("0" & secValue mod 60, 2)
diff = hh & ":" & mm & ":" & ss
wscript.echo diff
Finally i answered to my Question Feeling great
Date1 = alA_Filling(0)
Date2 = alA_Filling(1)
secValue = DateDiff("s",Date1,Date2)
ts = TimeSerial(0, 0, secValue)
Duration = FormatDateTime(ts, vbLongTime)
But i got output like 2:49:00 AM why added AM to this may be this vbLongTime
how can i remove that AM
Sorry if similar questions have been asked too many times, but it seems that there's one or more issues with every answer I find.
I have a date in the form of a String: Ex.: "04112005"
This is a date. 4th of November, 2005.
I want to get the difference, in years and days, between the current date and this date.
The code I have so far gets the year and just substracts them:
fun getAlderFraFodselsdato(bDate: String): String {
val bYr: Int = getBirthYearFromBirthDate(bDate)
var cYr: Int = Integer.parseInt(SimpleDateFormat("yyyy").format(Date()))
return (cYr-bYr).toString()
}
However, naturally, this is quite innacurate, since the month and days aren't included.
I've tried several approaches to create Date, LocalDate, SimpleDate etc. objects and using these to calcualate the difference. But for some reason I haven't gotten any of them to work.
I need to create a Date (or similar) object of the current year, month and day. Then I need to create the same object from a string containing say, month and year (""04112005""). Then I need to get the difference between these, in years, months and days.
All hints are appreciated.
I would use java.time.LocalDate for parsing and today along with a java.time.Period that calculates the period between two LocalDates for you.
See this example:
fun main(args: Array<String>) {
// parse the date with a suitable formatter
val from = LocalDate.parse("04112005", DateTimeFormatter.ofPattern("ddMMyyyy"))
// get today's date
val today = LocalDate.now()
// calculate the period between those two
var period = Period.between(from, today)
// and print it in a human-readable way
println("The difference between " + from.format(DateTimeFormatter.ISO_LOCAL_DATE)
+ " and " + today.format(DateTimeFormatter.ISO_LOCAL_DATE) + " is "
+ period.getYears() + " years, " + period.getMonths() + " months and "
+ period.getDays() + " days")
}
The output for a today of 2020-02-21 is
The difference between 2005-11-04 and 2020-02-21 is 14 years, 3 months and 17 days
It Works Below 26 API level
There are too many formates of dates you just enter the format of date and required start date and end date. It will show you result. You just see different date formate hare and here if you need.
tvDifferenceDateResult.text = getDateDifference(
"12 November, 2008",
"31 August, 2021",
"dd MMMM, yyyy")
General method to calculate date difference
fun getDateDifference(fromDate: String, toDate: String, formater: String):String{
val fmt: DateTimeFormatter = DateTimeFormat.forPattern(formater)
val mDate1: DateTime = fmt.parseDateTime(fromDate)
val mDate2: DateTime = fmt.parseDateTime(toDate)
val period = Period(mDate1, mDate2)
// period give us Year, Month, Week and Days
// days are between 0 to 6
// if you want to calculate days not weeks
//you just add 1 and multiply weeks by 7
val mDays:Int = period.days + (period.weeks*7) + 1
return "Year: ${period.years}\nMonth: ${period.months}\nDay: $mDays"
}
For legacy Date functions below api 26 without running desugaring with Gradle plugin 4.0, java.time.* use:
fun getLegacyDateDifference(fromDate: String, toDate: String, formatter: String= "yyyy-MM-dd HH:mm:ss" , locale: Locale = Locale.getDefault()): Map<String, Long> {
val fmt = SimpleDateFormat(formatter, locale)
val bgn = fmt.parse(fromDate)
val end = fmt.parse(toDate)
val milliseconds = end.time - bgn.time
val days = milliseconds / 1000 / 3600 / 24
val hours = milliseconds / 1000 / 3600
val minutes = milliseconds / 1000 / 3600
val seconds = milliseconds / 1000
val weeks = days.div(7)
return mapOf("days" to days, "hours" to hours, "minutes" to minutes, "seconds" to seconds, "weeks" to weeks)
}
The above answers using java.time.* api is much cleaner and accurate though.
I want to convert the given year, month and min information to day of year info.
For eg lets say
year 2004, month 2, day 2 = 33rd day of year
how can I do it in matlab?
Get the datenum for Jan 1 of that year, and subtract it from the given yy/mm/dd. For example, today's day of the year:
jan1 = datenum(datestr(now,'yy'),'yy')
now - jan1 + 1
Check the above against here.
For a specific date,
>> yy = 2004; mm = 2; dd = 2;
>> doty = datenum(yy,mm,dd) - datenum(yy,1,0)
doty =
33
How can I get the Today -2 days (the last 2 working days from now)? but skipping the weekend?
Example #1: Today is February 25, I want February 21
Example #2: Today is February 26, I want February 24
PS: Date format is DD/MM/YYYY
I have this, but the result is going forward, should I use datediff or what?:
<%
Dim d
d = DateAdd("m", 1, Now)
d = "01/" & Month(d) & "/" & Year(d)
d = DateAdd("d", -1, d)
If Weekday(d) = 7 Then
d = DateAdd("d", -1, d)
ElseIf Weekday(d) = 1 Then
d = DateAdd("d", -2, d)
End If
Response.Write "Day: " & d
%>
To get your desired result you need to subtract 3 days on Saturdays, 4 days on Sundays and Mondays, and 2 days on all other days. This can be achieved with something like this:
today = Now
num = Weekday(today, vbWednesday)
d = today - (2 + num\5 + num\6)
response.write "Two working days back: " & d
The Weekday function returns a numeric value for each weekday. By basing the week on Wednesday you can calculate the additional number of days you need to subtract from the current date with integer divisions:
num\5 returns 1 for Saturday, Sunday and Monday, and 0 otherwise.
num\6 returns 1 for Sunday and Monday, and 0 otherwise.
Thus the term 2 + num\5 + num\6 becomes 3 for Saturdays, 4 for Sundays and Mondays, and 2 for all other days.
This might be overkill for what you need but here are two routines I use in my scripts to add or subtract workdays while considering weekends and holidays.
Function AddWorkingDays(dtStart, intDays)
' Start/Default case...
AddWorkingDays = CDate(dtStart)
' If positive days, step forward, otherwise step backward...
Dim intStep, intCount
If intDays > 0 Then intStep = 1 Else intStep = -1
Do While intCount <> intDays
AddWorkingDays = AddWorkingDays + intStep
If IsValidDate(AddWorkingDays) Then intCount = intCount + intStep
Loop
End Function
Function IsValidDate(d)
Dim intWeekday, intMonth, intDay
intWeekday = Weekday(d)
intMonth = Month(d)
intDay = Day(d)
' Weekend dates are not acceptable...
If intWeekday = vbSaturday Or intWeekday = vbSunday Then Exit Function
' Holidays are also not acceptable...
If intMonth = 01 Then If intDay = 01 Then Exit Function ' New Year's Day
If intMonth = 07 Then If intDay = 04 Then Exit Function ' Independence Day
If intMonth = 12 Then If intDay = 25 Then Exit Function ' Christmas Day
' Memorial Day is the last Monday in May...
If intWeekday = vbMonday Then If intMonth = 05 Then If intDay >= 25 Then Exit Function
' ... (Thanksgiving, others) ...
' All tests passed. Date is a valid workday...
IsValidDate = True
End Function