VBA/Excel/Access/Word/Data Type/Integer
Содержание
Add integer together
<source lang="vb">
Sub Accumulate()
Dim n As Integer Dim t As Integer For n = 1 To 10 t = t + n Next n MsgBox " The total is " & t
End Sub
</source>
Add two numbers together
<source lang="vb">
Sub addNumbers()
"Declare the variables Dim intNumber1 As Integer Dim intNumber2 As Integer Dim intSum As Integer "Create InputBoxes to enter numbers intNumber1 = InputBox("Enter the first number") intNumber2 = InputBox("Enter the second number") "Add numbers intSum = intNumber1 + intNumber2
" Create an output
Debug.Print "The numbers entered were " & intNumber1 & " and " & intNumber2
End Sub
</source>
Input and Output
<source lang="vb">
Sub addNumbers()
Dim intNumber1 As Integer Dim intNumber2 As Integer Dim intSum As Integer "Create InputBoxes to enter numbers intNumber1 = InputBox("Enter the first number") intNumber2 = InputBox("Enter the second number")
End Sub
</source>
Integer number
<source lang="vb">
Sub MyNumber()
Dim intNum As Integer intNum = 23.11 MsgBox intNum
End Sub
</source>
Local Integer variable
<source lang="vb">
Public Sub TestLocal1()
Dim intVariable1 As Integer intVariable1 = intVariable1 + 1 Debug.Print intVariable1
End Sub
</source>
Overflow error
<source lang="vb">
Sub macro_overflow()
Dim l As Integer l = 255 * 256 "overflow error
End Sub Sub macro_no_overflow()
Dim l As Long l = 255& * 256 "now it works
End Sub
</source>
Select Case statement with Integer value
<source lang="vb">
Sub cmdCase_Click()
Dim intAge As Integer intAge = 12 Select Case intAge Case 0 MsgBox "You Must Enter a Number" Case 1 To 18 MsgBox "You Are Just a Kid" Case 19, 20, 21 MsgBox "You are Almost an Adult" Case 22 To 40 MsgBox "Good Deal" Case Is > 40 MsgBox "Getting Up There!" Case Else MsgBox "You Entered an Invalid Number" End Select
End Sub
</source>
==The Integer data type is the most efficient way of handling numbers from -o /td>
<source lang="vb">
Sub intDemo()
Dim intMyVar As Integer For intMyVar = 1 To 300 "repeat actions Next intMyVar
End Sub
</source>
Use If statement with Integer
<source lang="vb">
Sub ifTest()
Dim intNum As Integer Dim strMessage As String intNum = 12 If intNum > 10 Then strMessage = "The number is " & intNum End If Debug.Print strMessage
End Sub
</source>