I am using a mailing list program which inserts a date into a web link that is "encoded" so it can't be changed or edited by users.
The format is described as follows:
An eight character string, AABBCCDD,where:
Year = 1980 + HexToInt(BB) / 3
Month = HexToInt(CC) / 7 - 21
Day = HexToInt(DD) / 7 - 5
There is also a checksum included to avoid casual modification:
AA = IntToHex(Year + Month + Day mod 200)
For example 2660BDAF would refer to 20 June, 2012.
Can you help me convert the following to Classic ASP:
CodedDateStr = Request.querystring("Exp")
AYear = 1980 + HexToInt(CodedDateStr[3] + CodedDateStr[4]) / 3
AMonth = HexToInt(CodedDateStr[5] + CodedDateStr[6]) / 7 - 21
ADay = HexToInt(CodedDateStr[7] + CodedDateStr[8]) / 7 - 5
ACheckSum = AYear + AMonth + ADay mod 200
if ACheckSum <> HexToInt(CodedDateStr[1] + CodedDateStr[2]) then
ValidDate = 0
else
ValidDate = 1
end if
AExpiryDate = EncodeDate(ADay, AMonth, AYear)
if Date() > AExpiryDate then
ExpiredOffer = 1
else
ExpiredOffer = 0
end if
....
It looks like the HexToInt equivalent is clng("&h" & hexnumber)
I'm not sure about EncodeDate, i hope it is not something cludgy like CDate(AMonth + "/" + ADay + "/" + AYear)
CLng("&h" & hexnumber) looks like a good method for HexToInt.
For EncodeDate, look at the DateSerial function, which takes a year, month, and day, and returns a Date value.
Related
I have a variable with a date table that looks like this
* table:
[day]
* number: 15
[year]
* number: 2015
[month]
* number: 2
How do I get the days between the current date and the date above? Many thanks!
You can use os.time() to convert your table to seconds and get the current time and then use os.difftime() to compute the difference. see Lua Wiki for more details.
reference = os.time{day=15, year=2015, month=2}
daysfrom = os.difftime(os.time(), reference) / (24 * 60 * 60) -- seconds in a day
wholedays = math.floor(daysfrom)
print(wholedays) -- today it prints "1"
as #barnes53 pointed out could be off by one day for a few seconds so it's not ideal, but it may be good enough for your needs.
You can use the algorithms gathered here:
chrono-Compatible Low-Level Date Algorithms
The algorithms are shown using C++, but they can be easily implemented in Lua if you like, or you can implement them in C or C++ and then just provide Lua bindings.
The basic idea using these algorithms is to compute a day number for the two dates and then just subtract them to give you the number of days.
--[[
http://howardhinnant.github.io/date_algorithms.html
Returns number of days since civil 1970-01-01. Negative values indicate
days prior to 1970-01-01.
Preconditions: y-m-d represents a date in the civil (Gregorian) calendar
m is in [1, 12]
d is in [1, last_day_of_month(y, m)]
y is "approximately" in
[numeric_limits<Int>::min()/366, numeric_limits<Int>::max()/366]
Exact range of validity is:
[civil_from_days(numeric_limits<Int>::min()),
civil_from_days(numeric_limits<Int>::max()-719468)]
]]
function days_from_civil(y, m, d)
if m <= 2 then
y = y - 1
m = m + 9
else
m = m - 3
end
local era = math.floor(y/400)
local yoe = y - era * 400 -- [0, 399]
local doy = math.modf((153*m + 2)/5) + d-1 -- [0, 365]
local doe = yoe * 365 + math.modf(yoe/4) - math.modf(yoe/100) + doy -- [0, 146096]
return era * 146097 + doe - 719468
end
local reference_date = {year=2001, month = 1, day = 1}
local date = os.date("*t")
local reference_days = days_from_civil(reference_date.year, reference_date.month, reference_date.day)
local days = days_from_civil(date.year, date.month, date.day)
print(string.format("Today is %d days into the 21st century.",days-reference_days))
os.time (under Windows, at least) is limited to years from 1970 and up. If, for example, you need a general solution to also find ages in days for people born before 1970, this won't work. You can use a julian date conversion and subtract between the two numbers (today and your target date).
A sample julian date function that will work for practically any date AD is given below (Lua v5.3 because of // but you could adapt to earlier versions):
local
function div(n,d)
local a, b = 1, 1
if n < 0 then a = -1 end
if d < 0 then b = -1 end
return a * b * (math.abs(n) // math.abs(d))
end
--------------------------------------------------------------------------------
-- Convert a YYMMDD date to Julian since 1/1/1900 (negative answer possible)
--------------------------------------------------------------------------------
function julian(year, month, day)
local temp
if (year < 0) or (month < 1) or (month > 12)
or (day < 1) or (day > 31) then
return
end
temp = div(month - 14, 12)
return (
day - 32075 +
div(1461 * (year + 4800 + temp), 4) +
div(367 * (month - 2 - temp * 12), 12) -
div(3 * div(year + 4900 + temp, 100), 4)
) - 2415021
end
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
Currently working on the Google Calendar API and basically my end will provide Start and End date for user to pick the date and time. But would like anyone to advise how to convert them from (DatePicker/TimePicker/input field) to RFC 3339 format (e.g. 2013-07-24T10:00:00.000-07:00).
var date = new Date();
var timeZone = date.getTimezoneOffset();
alert(date.getFullYear() + "-" + (date.getMonth() + 1) + "-" + date.getDate() + "T" + date.getHours() + ":" + date.getMinutes() + ":" + date.getSeconds() + "." + date.getMilliseconds() + (timeZone > 0 ? "-" : "+") + Math.floor(Math.abs(timeZone) / 60) + ":" + Math.abs(timeZone) % 60);
i have 2 date picker
Dim pday, eitday, otherday, tpenalty, difday, subpenalty As Integer
difday = Val(L1.Caption) - Val(L2.Caption)
pday = 7
eitday = 8
otherday = difday - eitday
tpenalty = 25
If difday <= pday Then
PENALTY.Caption = 0
ElseIf difday = eitday Then
PENALTY.Caption = tpenalty
ElseIf difday > eitday Then
For i = 0 To otherday - 1
subpenalty = subpenalty + 5
Next i
PENALTY.Caption = tpenalty + subpenalty
End If
the problem is when the month is change the calculation is invalid.
I'm guessing based on your code (as many things are unclear), but this should give the number of days between two dates:
difday = DateDiff("d", StartDate, EndDate)
I've used StartDate and EndDate to signify the start and end of the lone period which are used to set L1 and L2, as you shouldn't be converting from strings to dates for calculations.
So I have a text layer in aftereffects with text source expression:
D = new Date(Date(0));
D.getDate() + "/" + (D.getMonth()+1)
which gives me result like this 29/12 and I need to add 3 days to it.
I tried:
D = new Date(Date(0));
(D.getDate()+3) + "/" + (D.getMonth()+1)
But that results to 32/12 .. How do I make it so the result would be (in this particular case) 1/1? In Java people suggest to use calendar class. But I'm not sure if After Effects has such thing.
Try this:
D = new Date(Date(0));
D = new Date(D.getTime()+3*24*60*60*1000);
D.getDate() + "/" + (D.getMonth()+1)
you can use things like this:
DateTime.Now.AddDays(12);
DateTime.Now.AddYears(2);
in javascript you can set with the help of setDate() function
var d = new Date();
d.setDate(35); // this will give you 4 jan 2013
d.setDate(25); // this will give you 25 Dec 2012