I'm trying to get the correct date from calendar using expression. But with my code, it gives 32 days and I need 31 days. How can I use DateAdd with this?
="Date: " + cstr(left(Parameters!KP2Ky.Value, 4) + "." + Mid(Parameters!KP2Ky.Value, 5, 2)+ "." + Right(Parameters!KP2Ky.Value,2))
I tried but I couldn't figure out how you are getting 32 in your result.
I would make the parameter a DATE type.
For the last day of the month previous to the selected month, you could use
="Date: " & FORMAT(DATEADD("d", 0 - DAY(Parameters!KP2Ky.Value), Parameters!KP2Ky.Value), "yyyy.MM.dd")
The last day of the selected month is a little more complicated:
="Date: " & FORMAT(DATEADD("d", 0 - DAY(DATEADD("M", 1, DATEADD("d", 1 - DAY(Parameters!KP2Ky.Value), Parameters!KP2Ky.Value))), DATEADD("M", 1, DATEADD("d", 1 - DAY(Parameters!KP2Ky.Value), Parameters!KP2Ky.Value))), "yyyy.MM.dd")
Related
Can someone explain this condition, because I'm getting wrong Time data eg : I'm expecting sch departure time as 15.10 but I'm getting 15.01
[Sch Dep Time] = IIF(DATEPART(Hour,[Journey and Details.schdeptime]) < 10 AND DATEPART(Minute,[Journey and Details.schdeptime]) < 10,
('0' + DATEPART(Hour,[Journey and Details.schdeptime]) + ':0' + DATEPART(Minute,[Journey and Details.schdeptime])),
IIF(DATEPART(Hour,[Journey and Details.schdeptime]) < 10,
('0' + DATEPART(Hour,[Journey and Details.schdeptime]) + ':' + DATEPART(Minute,[Journey and Details.schdeptime])),
(DATEPART(Hour,[Journey and Details.schdeptime]) + ':0' + DATEPART(Minute,[Journey and Details.schdeptime]))))
Update
It suddenly hit me that [Sch Dep Time] should contain the time component of the dataTime value stored in [Journey and Details.schdeptime], in a minute resolution. For that, you don't need to mess around with specific date parts and string concatenation, all you have to do is use convert:
[Sch Dep Time] = CONVERT(char(5), [Journey and Details.schdeptime], 108)
The 108 style returns hh:mm:ss (24 hours), and by using char(5) you are just taking the first 5 chars of that string - hh:mm.
First version
You have overcomplicated things. Try using the old right('00' + val, 2) trick instead:
[Sch Dep Time] = RIGHT('00' + CAST(DATEPART(Hour,[Journey and Details.schdeptime]) AS VARCHAR(2)), 2) + ':' +
RIGHT('00' + CAST(DATEPART(Minute,[Journey and Details.schdeptime]) AS VARCHAR(2)), 2)
Exlpanation:
You start off by concatenating leading zeroes to the string you want.
Suppose you have a string representing a number that must always have 4 digits, but it might be 1234 or 0003 - so you start by doing '0000' + #YourNumber.
Then, you use RIGHT to trim off any unwanted zeros - suppose you now have 000023, but you want 0023 - you do RIGHT('000023', 4) to get the last 4 chars.
I am trying to use asp classic to find how many working days (mon - sat) are in the month and how many are left.
any help or pointers greatly appreciated!
Here's how you can find the number of Sundays in a month without iteration. Somebody posted a JavaScript solution a few months back and I ported it to VBScript:
Function GetSundaysInMonth(intMonth, intYear)
dtmStart = DateSerial(intYear, intMonth, 1)
intDays = Day(DateAdd("m", 1, dtmStart) - 1)
GetSundaysInMonth = Int((intDays + (Weekday(dtmStart) + 5) Mod 7) / 7)
End Function
So, your total work days would just be the number of days in the month minus the number of Sundays.
Edit:
As #Lankymart pointed out in the comments, the above function gives you the number of Sundays in the month but it doesn't tell you how many are left.
Here's another version that does just that. Pass in any date and it will tell you how many Sundays are left in the month starting with that date. If you want to know how many Sundays are in a full month, just pass in the first day of the month (e.g., DateSerial(2014, 8, 1)).
Function GetSundaysRemainingInMonth(dtmStart)
intDays = Day(DateSerial(Year(dtmStart), Month(dtmStart) + 1, 1) - 1)
intDays = intDays - Day(dtmStart) + 1
GetSundaysRemainingInMonth = Int((intDays + (Weekday(dtmStart) + 5) Mod 7) / 7)
End Function
Edit 2:
#Cheran Shunmugavel was interested in some specifics about how this works. First, I just want to restate that I didn't develop this method originally. I just ported it to VBScript and tailored it to the OP's requirement (Sundays).
Imagine a February during a leap year. We have 29 days during the month. We know from the start that we have four full weeks, so each weekday will be represented at least four times. But that still leaves one addition day that's unaccounted for (29 Mod 7 = 1). How do we know if we get an extra Sunday from that one day? Well, in this case, it's pretty simple. Only if our start date is a Sunday can we count an extra Sunday for the month.
What if the month has 30 days? Then we have two extra days to account for. In that case, the start date can be a Saturday or a Sunday and we can count an extra Sunday for the month. And so it goes. So we can see that if we're X additional days within an upcoming Sunday, we can count an extra Sunday.
Let's put this in tabular form:
Addl Days Needed
Day To Count Sunday
---------- ----------------
Sunday 1
Saturday 2
Friday 3
Thursday 4
Wednesday 5
Tuesday 6
Monday 7
So what we need is a formula that we can apply to these situations so that they all result in the same value. We'll need to assign some value to each day and combine that value with the number of addition days needed for Sunday to count. Seems reasonable that if we assign an inverse value to the weekdays and add that to the number of additional days, we can get the same result.
Addl Days Needed Value Assigned
Day To Count Sunday To Weekday Sum
---------- ---------------- -------------- ---
Sunday 1 6 7
Saturday 2 5 7
Friday 3 4 7
Thursday 4 3 7
Wednesday 5 2 7
Tuesday 6 1 7
Monday 7 0 7
So, if weekday_value + addl_days = 7 then we count an extra Sunday. (We'll divide this by 7 later to give us 1 additional Sunday). But how do we assign the values we want to the weekdays? Well, VBScript's Weekday() function already does this but, unfortunately, it doesn't use the values we need by default (it uses 1 for Sunday through 7 for Saturday). We could change the way Weekday() works by using the second param, or we could just use a Mod(). This is where the + 5 Mod 7 comes in. If we take the Weekday() value and add 5, then mod that by 7, we get the values we need.
Day Weekday() +5 Mod 7
---------- --------- -- -----
Sunday 1 6 6
Saturday 7 12 5
Friday 6 11 4
Thursday 5 10 3
Wednesday 4 9 2
Tuesday 3 8 1
Monday 2 7 0
That's how the + 5 Mod 7 was determined. And, with that solved, the rest is easy(er)!
#Zam is on the right track you need to use WeekDay() function, here is a basic idea of how to script it;
<%
Dim month_start, month_end, currentdate, dayofmonth
Dim num_weekdays, num_past, num_future
Dim msg
'This can be configured how you like even use Date().
month_start = CDate("01/08/2014")
month_end = DateAdd("d", -1, DateAdd("m", 1, month_start))
msgbox(Day(month_end))
For dayofmonth = 1 To Day(month_end)
currentdate = CDate(DateAdd("d", dayofmonth, month_start))
'Only ignore Sundays
If WeekDay(currentdate) <> vbSunday Then
num_weekdays = num_weekdays + 1
If currentdate <= Date() Then
num_past = num_past + 1
Else
num_future = num_future + 1
End If
End If
Next
msg = ""
msg = msg & "Start: " & month_start & "<br />"
msg = msg & "End: " & month_end & "<br />"
msg = msg & "Number of Weekdays: " & num_weekdays & "<br />"
msg = msg & "Weekdays Past: " & num_past & "<br />"
msg = msg & "Weekdays Future: " & num_future & "<br />"
Response.Write msg
%>
How about using "The Weekday function returns a number between 1 and 7, that represents the day of the week." ?
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 am new in SSRS world.
I have set the X-axis label interval '1' and Interval Type 'day'. It is working properly. Now I want to hide the all label Except the 1st, 15th,31th Date of every month. Is it possible to set the expression in hide properties for X-axis?
Can anyone help me regarding that?
EDIT #2
Open the Category Group properties - there you will have the option to define expression to the labal. Then use the following expression:
Switch(Day(Fields!SAMPLE_DATE.Value) = 31 , "31", Day(Fields!SAMPLE_DATE.Value)> 1 AND Day(Fields!SAMPLE_DATE.Value) < 14 , " ", Day(Fields!SAMPLE_DATE.Value) = 15 , "15", Day(Fields!SAMPLE_DATE.Value) > 15 AND Day(Fields!SAMPLE_DATE.Value) < 30, " ", Day(Fields!SAMPLE_DATE.Value) = 30 , "30")
I want to be able to present "today" and "yesterday" for recent dates in my application. I've got a date formatter in use currently to show dates (retrieved from data records) and will keep using this for anything more than a couple of days old. I just really like the way the SMS app in the iPhone shows dates for recent messages and would like to emulate this.
The time-stamps that I have to work with are generated on a server that the phone downloads the data records from. All times are therefore generated at UTC (i.e. GMT) time.
I've been fiddling about with this for a while the solutions I've devised just seem horribly long-winded.
Can anyone suggest how to implement a method that could do this?
Cheers - Steve.
If this is a web app, you might find PrettyDate useful. I made a vb.net implementation that could easily be converted to another language:
Public Function formatDate(ByVal time As DateTime) As String
Dim datediff As TimeSpan = Now.Subtract(time)
Dim days As Integer = datediff.TotalDays
If days < 1 Then
Dim seconds As Integer = datediff.TotalSeconds
Select Case seconds
Case 0 To 60
Return "just now"
Case 61 To 120
Return "1 minute ago"
Case 121 To 3600
Return Math.Floor(seconds / 60) & " minutes ago"
Case 3601 To 7200
Return "1 hour ago"
Case 7201 To 86400
Return Math.Floor(seconds / 3600) & " hours ago"
End Select
ElseIf days < 31 Then
Select Case days
Case 1
Return "yesterday"
Case 2 To 7
Return days & " days ago"
Case Is > 7
Return Math.Ceiling(days / 7) & " weeks ago"
End Select
Else : Return time.ToString("MM/dd/yyyy")
End If
End Function