I want to display the weeks dates based on a week number that I get from my db.
So if I get week=43 then it would display all 7 dates for that week.
Like this.
M=23
T=24
W=25
T=26
F=27
S=28
S=29
Have tested with a lot of date formating but I can't get it working.
So any input really appreciated, thanks!
I ended up with this, works perfect!
currentDate = Date
weekNumber=DatePart("ww", currentDate, vbMonday, vbFirstFourDays)
y = Year(Date)
Public Function FirstDayOfWeek(Year, Week)
Dim TempDate
TempDate = DateSerial(Year, 1, 1)
Do Until DatePart("ww", TempDate, vbMonday, vbFirstFourDays) = 1
TempDate = TempDate + 7
Loop
TempDate = TempDate + (7 * (Week - 1))
FirstDayOfWeek = TempDate - Weekday(TempDate, vbMonday) + 1
End Function
Dim startDatum
Dim slutDatum
startDatum = FirstDayOfWeek(y, weekNumber)
slutDatum = startDatum + 6
mon=DatePart("d", startDatum)
tus=DatePart("d", startDatum+1)
wen=DatePart("d", startDatum+2)
tur=DatePart("d", startDatum+3)
fri=DatePart("d", startDatum+4)
sat=DatePart("d", startDatum+5)
sun=DatePart("d", startDatum+6)
Well, there isn't really an inverse DatePart() function, so you have to make your own calculations.
dim w, wd, y, m, i
y = Year(Date) '- year of the week in question; I'm using today's date
w = 43
wd = DateAdd("d",w*7,CDate("1/1/" & y)) '- adjust as needed for 1st week of year
m = DateAdd("d",2-Weekday(wd),wd) '- find Monday of week
Hopefully, you can go from there.
Related
I am (still) programming in VB6, but I think this problem is not program language related.
I need to show one Picture out of 7 Pictures (named 1.jpg to 7.jpg) every calendar week, beginning with Calendar week 3, which shows picture nr.1
Example:
(year and calendar week together as 2022/03)
202203=1.jpg
202204=2.jpg
202205=3.jpg
202206=4.jpg
202207=5.jpg
202208=6.jpg
202209=7.jpg
202210=1.jpg
202211=2.jpg
etc.
How can I do this??
Also, Years can have 52 or 53 weeks, but this has to continue like above...
Appreciate any help, thank you in advance.
Here is some code that gives the results you desire. It should be fairly self-explanatory:
Option Explicit
Private Sub Test()
Dim dt As Date
Dim i As Integer
dt = DateSerial(2022, 1, 17) 'start at 3rd calendar week
For i = 1 To 100
Debug.Print Format(dt, "mm/dd/yyyy") & vbTab & Year(dt) & _
Format(ISOWeekNumber(dt), "00") & vbTab & GetPicture(dt)
dt = DateAdd("d", 1, dt)
Next
End Sub
Private Function GetPicture(ByVal dt As Date) As String
Dim dp As Integer
dp = ISOWeekNumber(dt) Mod 7 - 2
If dp <= 0 Then dp = dp + 7
GetPicture = dp & ".jpg"
End Function
Private Function ISOWeekNumber(ByVal dt As Date) As Integer
ISOWeekNumber = DatePart("ww", dt, vbMonday, vbFirstFourDays)
End Function
This produces the following results:
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
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
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.