• Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 232
  • Last Modified:

Do I need to use DateDiff to change times?

I've got a webcontrol that is supposed to enter in a start time and and end time based on information loaded into the MSFlexGrid. The only problem is, I cannot seem to get the end time to give the correct time base on a few factors:

'If the start hour = 11AM, then end hour = 2PM
'If the start hour = 11PM, then end hour = 2AM of next day
'If the start hour = 9AM, then end hour = 12PM
Private Sub MSFlexGrid_EnterCell()
On Error Resume Next
 
 
Dim strMonth As String
Dim strDay As String
Dim strYear As String
Dim strHour As String
Dim strMin As String
Dim strAMPM As String
 
'START TIME
'month
strMonth = MSFlexGrid.TextMatrix(MSFlexGrid.Row, 0)
strMonth = Split(strMonth, "/")(0)
strMonth = Replace(strMonth, "0", "")
wb.Document.All("start_time_month").Value = strMonth
 
'day
strDay = MSFlexGrid.TextMatrix(MSFlexGrid.Row, 0)
strDay = Split(strDay, "/")(1)
If Left(strDay, 1) = 0 Then strDay = Replace(strDay, "0", "")
wb.Document.All("start_time_day").Value = strDay
 
'year
strYear = MSFlexGrid.TextMatrix(MSFlexGrid.Row, 0)
strYear = Split(strYear, "/")(2)
wb.Document.All("start_time_year").Value = strYear
 
'hour
strHour = MSFlexGrid.TextMatrix(MSFlexGrid.Row, 1)
strHour = Split(strHour, ":")(0)
wb.Document.All("start_time_hour").Value = strHour
 
'min
strMin = MSFlexGrid.TextMatrix(MSFlexGrid.Row, 1)
strMin = Split(strMin, ":")(1)
strMin = Left(strMin, 2)
wb.Document.All("start_time_min").Value = strMin
 
'AM_PM
strAMPM = MSFlexGrid.TextMatrix(MSFlexGrid.Row, 1)
strAMPM = Right(strAMPM, 2)
wb.Document.All("start_time_ampm").Value = LCase(strAMPM)
'END START TIME
 
'END TIME
'month
strMonth = MSFlexGrid.TextMatrix(MSFlexGrid.Row, 0)
strMonth = Split(strMonth, "/")(0)
strMonth = Replace(strMonth, "0", "")
wb.Document.All("end_time_month").Value = strMonth
 
'day
strDay = MSFlexGrid.TextMatrix(MSFlexGrid.Row, 0)
strDay = Split(strDay, "/")(1)
If Left(strDay, 1) = 0 Then strDay = Replace(strDay, "0", "")
wb.Document.All("end_time_day").Value = strDay
 
'year
strYear = MSFlexGrid.TextMatrix(MSFlexGrid.Row, 0)
strYear = Split(strYear, "/")(2)
wb.Document.All("end_time_year").Value = strYear
 
'hour
strHour = MSFlexGrid.TextMatrix(MSFlexGrid.Row, 1)
strHour = Split(strHour, ":")(0)
strHour = strHour + 3
If strHour > 12 Then strHour = 12
'wb.Document.All("end_time_hour").Value = strHour
 
'min
strMin = MSFlexGrid.TextMatrix(MSFlexGrid.Row, 1)
strMin = Split(strMin, ":")(1)
strMin = Left(strMin, 2)
wb.Document.All("end_time_min").Value = strMin
 
'AM_PM
strAMPM = MSFlexGrid.TextMatrix(MSFlexGrid.Row, 1)
strAMPM = Right(strAMPM, 2)
wb.Document.All("end_time_ampm").Value = LCase(strAMPM)
'END END TIME
 
 
 
 
 
End Sub

Open in new window

0
Taylor814
Asked:
Taylor814
  • 7
  • 3
1 Solution
 
Taylor814Author Commented:
Here's a picture of the MSFlexGrid
Clipboard01.jpg
0
 
JohnBPriceCommented:
You just want to add 3 hours to the date time, running into next day?  Yes, DateAdd would be far easier.


dim dtStart as date
dim dtEnd as date
dtStart = CDate(MSFlexGrid.TextMatrix(MSFlexGrid.Row, 0) & " " & MSFlexGrid.TextMatrix(MSFlexGrid.Row, 1)
dtEnd = DateAdd("h",3,dtStart)

Open in new window

0
 
JohnBPriceCommented:
missing a paren in there

dtStart = CDate(MSFlexGrid.TextMatrix(MSFlexGrid.Row, 0) & " " & MSFlexGrid.TextMatrix(MSFlexGrid.Row, 1))

Open in new window

0
Free Tool: SSL Checker

Scans your site and returns information about your SSL implementation and certificate. Helpful for debugging and validating your SSL configuration.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

 
Taylor814Author Commented:
Thank you so much. It was exactly what I was looking for. Didn't realize it was THAT simple. Thank you again!
0
 
GrahamSkanRetiredCommented:
You need to work with date variables, e.g.

Dim iMonth as Integer
iMonth = Month(MSFlexGrid.TextMatrix(MSFlexGrid.Row, 1))

Dim iHour as Integer
iHour = Hour(MSFlexGrid.TextMatrix(MSFlexGrid.Row, 2))

Also arithmetic is easier if the 24-hour clock is used,

so if iHour = 23 means, that means 11 o'clock in the evening.
0
 
Taylor814Author Commented:
Quick question though, if the time starts at 9PM, why doesn't dtEND show the next day with "12:00AM". It seems to leave that part off.

If it starts at 9:30PM, then dtEND is nextday+ 12:30AM
0
 
JohnBPriceCommented:
If the date is EXACTLY 12:00:00 AM, the default date display format is to leave off the time and just show the date.  You can force it with the format function, e.g.

dim dtEndDate as date
dim strEndDate as string
strEndDate = format(dtEndDate,"mm/dd/yyyy Hh:Nn AM/PM")

Open in new window

0
 
Taylor814Author Commented:
Hmm, when using that code, I got:

?dtstart
5/9/2009 9:00:00 PM
?dtend
5/10/2009
?strenddate
12/30/1899 12:00 AM
0
 
Taylor814Author Commented:
Ah, nevermind. Changed dtEndDate to dtEnd...works...thanks
0
 
Taylor814Author Commented:
Here's what I don't understand. In the immediate window I get:

?Format(dtEnd, "mm/dd/yyyy hh:nn AM/PM")
05/09/2009 12:00 AM
?strenddate
5/9/2009

strEndDate doesn't show 12:00AM even though the format says it should.
0
 
Taylor814Author Commented:
Nevermind. strEndDate was Dim Date instead of string.
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

Join & Write a Comment

Featured Post

Cloud Class® Course: Microsoft Azure 2017

Azure has a changed a lot since it was originally introduce by adding new services and features. Do you know everything you need to about Azure? This course will teach you about the Azure App Service, monitoring and application insights, DevOps, and Team Services.

  • 7
  • 3
Tackle projects and never again get stuck behind a technical roadblock.
Join Now