How do I count the number of times one string appears within another in Access VBA? For example, how would I count how many times "the" occurs in "The quick brown fox jumps over the lazy dog."?
As you are ok with substrings/case sensitivity
matches = (len(lookin) - len(replace$(lookin, find, ""))) / len(find)
You can simplify the function with far less variables and avoid the usage of For by the following function.
Public Function getOccuranceCount(Expression As String, Find As String) As Long
'*******************************************************************************
'Code Courtesy of
' Paul Eugin
'
' Input - Expression, the String to check
' Find, the String pattern to be checked for
' Output - The number of occurance of the Find String in the Expression String
' Usage - getOccuranceCount("The quick brown fox jumped over the lazy dog.","saw")
' 0
' getOccuranceCount("The quick brown fox jumped over the lazy dog.","the")
' 2
'*******************************************************************************
On Error GoTo errDisplay
Dim strArr() As String, i As Long
strArr = Split(Expression, Find)
getOccuranceCount = UBound(strArr)
errExit:
Exit Function
errDisplay:
MsgBox "The following error has occured while trying to get the count." & vbCrLf & vbCrLf & _
"Error (" & Err.Number & ") - " & Err.Description, vbCritical, "Contact the DB Admin."
Resume errExit
End Function
The function will Split it into an array then all you need is the Count of the Boundaries. Hope this helps.
Dim lookin As String
Dim tofind As String
lookin = "The quick brown fox jumps over the lazy dog."
tofind = "the "
Dim r As Object, matches
Set r = CreateObject("VBScript.RegExp")
r.Pattern = tofind
r.IgnoreCase = True
r.Multiline = False
r.Global = True
Set matches = r.Execute(lookin)
matches finds two hits. One at index = 0 and one at index = 31.
You can use this function, which uses InStr:
Function CountStringOccurances(strStringToCheck As String, strValueToCheck As String) As Integer
'Purpose: Counts the number of times a string appears in another string.
On Error GoTo ErrorMessage
Dim intStringPosition As Integer
Dim intCursorPosition As Integer
CountStringOccurances = 0
intCursorPosition = 1
For i = 0 To Len(strStringToCheck)
intStringPosition = InStr(intCursorPosition, strStringToCheck, strValueToCheck)
If intStringPosition = 0 Then
Exit Function
Else
CountStringOccurances = CountStringOccurances + 1
intCursorPosition = intStringPosition + Len(strValueToCheck)
End If
Next i
Exit Function
ErrorMessage:
MsgBox "The database has generated an error. Please contact the database administrator, quoting the following error message: '" & Err.Description & "'", vbCritical, "Database Error"
End
End Function
The function returns the position of your value within the string. If that's 0, it exits and returns the number. If it returns a positive value (and it will if the value is present), then it adds one to the number of occurances, and then checks again, moving its starting position to just after the previous occurance. This function is case insenstive.
Using the examples above:
CountStringOccurances("The quick brown fox jumped over the lazy dog.","the")
would return 2.
You can use Split in the same solution as this question has. It's the tidiest solution that I've seen so far.
Related
I want to find if a string contains a ","(comma) in it. Do we have any other option other than reading char-by-char?
Use the Instr function (old version of MSDN doc found here)
Dim pos As Integer
pos = InStr("find the comma, in the string", ",")
will return 15 in pos
If not found it will return 0
If you need to find the comma with an excel formula you can use the =FIND(",";A1) function.
Notice that if you want to use Instr to find the position of a string case-insensitive use the third parameter of Instr and give it the const vbTextCompare (or just 1 for die-hards).
Dim posOf_A As Integer
posOf_A = InStr(1, "find the comma, in the string", "A", vbTextCompare)
will give you a value of 14.
Note that you have to specify the start position in this case as stated in the specification I linked: The start argument is required if compare is specified.
You can also use the special word like:
Public Sub Search()
If "My Big String with, in the middle" Like "*,*" Then
Debug.Print ("Found ','")
End If
End Sub
There is also the InStrRev function which does the same type of thing, but starts searching from the end of the text to the beginning.
Per #rene's answer...
Dim pos As Integer
pos = InStrRev("find the comma, in the string", ",")
...would still return 15 to pos, but if the string has more than one of the search string, like the word "the", then:
Dim pos As Integer
pos = InStrRev("find the comma, in the string", "the")
...would return 20 to pos, instead of 6.
Building on Rene's answer, you could also write a function that returned either TRUE if the substring was present, or FALSE if it wasn't:
Public Function Contains(strBaseString As String, strSearchTerm As String) As Boolean
'Purpose: Returns TRUE if one string exists within another
On Error GoTo ErrorMessage
Contains = InStr(strBaseString, strSearchTerm)
Exit Function
ErrorMessage:
MsgBox "The database has generated an error. Please contact the database administrator, quoting the following error message: '" & Err.Description & "'", vbCritical, "Database Error"
End
End Function
You wouldn't really want to do this given the existing Instr/InstrRev functions but there are times when it is handy to use EVALUATE to return the result of Excel worksheet functions within VBA
Option Explicit
Public Sub test()
Debug.Print ContainsSubString("bc", "abc,d")
End Sub
Public Function ContainsSubString(ByVal substring As String, ByVal testString As String) As Boolean
'substring = string to test for; testString = string to search
ContainsSubString = Evaluate("=ISNUMBER(FIND(" & Chr$(34) & substring & Chr$(34) & ", " & Chr$(34) & testString & Chr$(34) & "))")
End Function
Translate formula quotation marks incl. replacements into VBA-readable formulae
I was inspired to write this post by the recent question of
formula substitution using a constant.
At the same time, the frequent problem emerged that quotation marks
within a formula string should be replaced by double quotation marks in order
to make them readable in VBA.
Practical use case
A practical use case is to copy a table formula directly from a SO website
and "translate" it into a string-readable format.
But how is this supposed to be done with VBA means, since the direct input of
such an incomplete formula string in a procedure code without manually
added double quotation marks would immediately lead to an error?
Another feature would be to make replacements at certain points within
a formula template, for example with a constant or even with several
numerically identifiable markers.
I found a quick & dirty solution (without error handling) by analyzing a FormulaContainer procedure containing
exclusively outcommented formulae as these would allow any prior direct code input.
In order to distinguish them from the usual commentaries,
I decided with a heavy heart to use the Rem prefix (i.e. Remark) as an alternative, which we may still be familiar with from ancient Basic times.
My intention is not to show a perfect solution, but to stimulate further solutions
by demonstrating a possible way.
Question
Are there other work arounds allowing to copy tabular formulae with quotation marks directly and as possible replacement pattern into VBA procedures?
///////////////////////////////////
Main function QuickFormula()
References a FormulaContainer procedure containing exclusively formulae with Rem prefixes, such as e.g.
Sub FormulaContainer()
Rem =....
Rem =....
End Sub
This allows formula inputs with quotation marks similar to tabular cell inputs;
furthermore these inputs may contain string identifiers facilitating wanted replacements.
Option Explicit
'Site: https://stackoverflow.com/questions/70399681/how-many-quotes-to-put-around-a-formula-that-is-sending-an-empty-string
'Auth: https://stackoverflow.com/users/6460297/t-m
Function QuickFormula(ByVal no As Long, ParamArray repl() As Variant) As String
'Purp: - change indicated code line in FormulaContainer to code readable string and
' - replace enumerated identifiers with given value(s)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
'1) get REMark code line indicated by ordinal argument no
QuickFormula = getCodeLine("modFormula", "FormulaContainer", no)
'2a)replace "#" identifyer(s) with constant repl value
If Not IsArray(repl(0)) Then
QuickFormula = Replace(QuickFormula, "{1}", "#")
QuickFormula = Replace(QuickFormula, "#", repl(0))
If Len(QuickFormula) = 0 Then QuickFormula = "Error NA!"
Debug.Print no & " ~~> " & Chr(34) & QuickFormula & Chr(34)
Exit Function
End If
'2b)replace 1-based "{i}" identifiers by ParamArray values
Dim i As Long
For i = LBound(repl(0)) To UBound(repl(0))
QuickFormula = Replace(QuickFormula, "{" & i + 1 & "}", repl(0)(i))
Next
'3) optional display in immediate window
Debug.Print no & " ~~> " & Chr(34) & QuickFormula & Chr(34)
End Function
Help function getCodeLine()
Gets a given code line of the indicated procedure
Function getCodeLine(ByVal ModuleName As String, ByVal ProcedureName As String, Optional ByVal no As Long = 1) As String
'Purp: return a code line in given procedure containing "Rem "
'Note: assumes no line breaks; needs a library reference to
' "Microsoft Visual Basic for Applications Extensibility 5.3"
Const SEARCH As String = "Rem =", QUOT As String = """"
'1) set project
Dim VBProj As Object
Set VBProj = ThisWorkbook.VBProject
If VBProj.Protection = vbext_pp_locked Then Exit Function ' escape locked projects
'2) set component
Dim VBComp As Object
Set VBComp = VBProj.VBComponents(ModuleName)
Dim pk As vbext_ProcKind
'3) get no + 3 top code line(s)
With VBComp.CodeModule
'a)count procedure header lines
Dim HeaderCount As Long: HeaderCount = .ProcBodyLine(ProcedureName, pk) - .ProcStartLine(ProcedureName, pk)
'b) get procedure code
Dim codelines
'codelines = Split(.Lines(.ProcBodyLine(ProcedureName, pk), .ProcCountLines(ProcedureName, pk) - HeaderCount), vbNewLine)
codelines = Split(.Lines(.ProcBodyLine(ProcedureName, pk), no + 1), vbNewLine)
'c) filter code lines containing "Rem" entries
codelines = Filter(codelines, SEARCH, True)
End With
'4) return (existing) codeline no
If no - 1 > UBound(codelines) Then Exit Function ' check existance
getCodeLine = Replace(Replace(codelines(no - 1), QUOT, String(2, QUOT)), "Rem =", "=")
End Function
Example call
References all three formulae in the FormulaContainer (including an example of a non-existing number):
Sub EnterFormula()
With Sheet1.Range("X1") ' << change to any wanted target range
.Offset(1).Formula2 = QuickFormula(1, 6)
.Offset(2).Formula2 = QuickFormula(2, Array(10, 20, 30))
'two single argument inputs with same result
.Offset(3).Formula2 = QuickFormula(3, Array(17))
.Offset(4).Formula2 = QuickFormula(3, 17)
'not existing formula number in Rem code container
.Offset(5).Formula2 = QuickFormula(333, 17)
End With
End Sub
Example FormulaContainer
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
'Purp: formula container to be adjusted to code readable strings
'Note: Insert only Formulae starting with "Rem "-prefix!
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
' # identifies constant replacement(s)
' {i} stands for enumerated replacements {1},{2}..{n}
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Sub FormulaContainer()
Rem =IF($V#>0,IF($G#>$S#,($S#-$H#)*$K#+$Y#,($G#-$H#)*$K#+$Y#),"")
Rem =A{1}*B{3}+C{2}
Rem =A{1}+100
End Sub
Example output in immediate window
1 ~~> "=IF($V6>0,IF($G6>$S6,($S6-$H6)*$K6+$Y6,($G6-$H6)*$K6+$Y6),"""")"
2 ~~> "=A10*B30+C20"
3 ~~> "=A17+100"
3 ~~> "=A17+100"
333 ~~> "Error NA!"
Keep it simple stupid
Assuming either the currently selected cell formula or a textbox input, a simple Userform might act as a formula translator into a line of VBA code:
Basic Userform code
Needed: TextBox1, TextBox2, CommandButton1
Option Explicit
Private Sub CommandButton1_Click()
'Purp: Redouble inside quotation marks
Const Quot As String = """"
Dim assignTo As String
assignTo = "ws.Range(""" & Selection.Address(False, False) & """).Formula2 = "
Me.TextBox2.Text = assignTo & Quot & Replace(Me.TextBox1.Text, Quot, String(2, Quot)) & Quot
End Sub
Private Sub UserForm_Initialize()
'Purp: assume active formula as wanted input
Me.TextBox1 = Selection.Formula2
End Sub
Private Sub UserForm_Layout()
'Purp: example layout textboxes
'a) define textboxes
Dim textboxes() As String
textboxes = Split("Textbox1,Textbox2", ",")
'b) format
Dim i As Long
For i = 0 To UBound(textboxes)
With Me.Controls(textboxes(i))
.Font.Name = "Courier New"
.Font.Size = 12
.MultiLine = True
.EnterKeyBehavior = True
End With
Next i
End Sub
Possible extensions
Of course you might add an insertion routine (inserting e.g. {} brackets) as well as some replacement procedures like in my workaround above.
Just for fun, a basic insertion routine here:
Private Sub CommandButton2_Click()
'Purp: Insert brackets {}
With Me.TextBox1
.SetFocus
If InsertAtCursor("{}", Me.TextBox1) Then
.SelStart = .SelStart - 1
End If
End With
End Sub
Public Function InsertAtCursor(s As String, ctrl As MSForms.Control, Optional ErrMsg As String) As Boolean
'Purpose: Insert the characters at the cursor in the active control.
'Site: http://allenbrowne.com/func-InsertChar.html
'Return: True if characters were inserted.
'Arguments: s = the character(s) you want inserted at the cursor.
' ErrMsg = string to append any error messages to.
'Note: Control must have focus.
On Error GoTo Err_Handler
Dim prior As String 'Text before the cursor.
Dim after As String 'Text after the cursor.
Dim cnt As Long 'Number of characters
Dim iSelStart As Long 'Where cursor is.
Dim txt As String 'text with LineFeeds only
If s <> vbNullString Then
With ctrl ' --> UserForm Control
txt = Replace(.Text, vbCrLf, vbLf) ' LineFeeds only (MultiLine)
If .Enabled And Not .Locked Then
cnt = Len(txt) ' Zählung ohne vbCr's !
'SelStart can't cope with more than 32k characters.
If cnt <= 32767& - Len(s) Then
'Remember characters before cursor.
iSelStart = .SelStart
If iSelStart > 1 Then
prior = Left$(txt, iSelStart)
End If
'Remember characters after selection.
If iSelStart + .SelLength < cnt Then
after = Mid$(txt, iSelStart + .SelLength + 1) ' OP:2
End If
'Assign prior characters, new ones, and later ones.
.value = prior & s & after
'Put the cursor back where it as, after the new ones.
.SelStart = iSelStart + Len(s)
'Return True on success
InsertAtCursor = True
End If
End If
End With
End If
Exit_Handler:
Exit Function
Err_Handler:
Debug.Print Err.Number, Err.Description
Select Case Err.Number
Case 438&, 2135&, 2144& 'Object doesn't support this property. Property is read-only. Wrong data type.
ErrMsg = ErrMsg & "You cannot insert text here." & vbCrLf
Case 2474&, 2185& 'No active control. Control doesn't have focus.
ErrMsg = ErrMsg & "Cannot determine which control to insert the characters into." & vbCrLf
Case Else
ErrMsg = ErrMsg & "Error " & Err.Number & ": " & Err.Description & vbCrLf
End Select
Resume Exit_Handler
End Function
I am trying to utilize Vlookup function, according to the Textbox1 value user put in in Userform Guntest, automatically looking for corresponding features of the gun.
However the program currently doesn't run as it reminds me
'Runtime error '1004', method 'Range of object' _Global' failed.
The error appears on Retrieve1=…
I will be appreciated if you could help me to check where the problem is as I have really limited knowledge and experience on using VBA.
Thanks in advance.
It looks like some objects is undefined but I can't figure out where.
The module 1 code is:
Public Guncode As String
Option Explicit
Sub Test()
Call Vlookup
End Sub
Sub Vlookup()
Dim Retrieve1 As String
Dim Retrieve2 As String
Dim FinalRow As Long
Dim FinalColumn As Long
Dim WholeRange As String
If GunTest.TextBox1 = "" Then
Exit Sub
If GunTest.TextBox1 <> "" Then
MsgBox Guncode
End If
End If
With Sheets(1)
FinalRow = Range("A65536").End(xlUp).Row
FinalColumn = Range("IV1").End(xlToLeft).Column
WholeRange = "A2:" & CStr(FinalColumn) & CStr(FinalRow)
Retrieve1 = Application.WorksheetFunction.Vlookup(Trim(Guncode), Range(WholeRange), 1, False) 'Locate specific tool according to QR code number
Retrieve2 = Application.WorksheetFunction.Vlookup(Trim(Guncode), Range(WholeRange), 5, False) 'Locate specific gun type according to QR code number
If Guncode = "" Then
MsgBox "This gun doesn't exist in database!"
Else
MsgBox "The tool number is:" & Retrieve1 & vbCrLf & "The gun type is:" & Retrieve2
End If
End With
End Sub
The userform code is:
Option Explicit
Private Sub Label1_Click()
End Sub
Private Sub CommandButton1_Click()
If TextBox1 = "" Then Exit Sub 'Set condition 1 of exiting the program
Guncode = GunTest.TextBox1
With Me
Call Module1.Test
End With
End Sub
Private Sub PartID_Click()
End Sub
Private Sub TextBox1_Change()
End Sub
Private Sub UserForm_Click()
End Sub
It should run properly but it doesn't. Any help would be appreciated, thanks!
First off, you were passing in a number as the column letter value. CSTR() doesnt magically transform it into the letter equivalent but I like your enthusiasm.
Second, your method will bomb if the value isnt found - so you'll need to write your own error handling for it.
Sub Vlookup()
Dim Retrieve1 As String
Dim Retrieve2 As String
Dim FinalRow As Long
Dim FinalColumn As Long
Dim WholeRange As String
Dim vArr
Dim col_Letter As String
If GunTest.TextBox1 = "" Then
Exit Sub
If GunTest.TextBox1 <> "" Then
MsgBox Guncode
End If
End If
With ThisWorkbook.Sheets("Sheet1")
FinalRow = .Range("A65536").End(xlUp).Row
FinalColumn = .Range("IV1").End(xlToLeft).Column
vArr = Split(Cells(1, FinalColumn).Address(True, False), "$")
col_Letter = vArr(0)
WholeRange = "A2:" & col_Letter & CStr(FinalRow) '<---- you were passing a number in as the column value
Retrieve1 = Application.WorksheetFunction.Vlookup(Trim(Guncode), .Range(WholeRange), 1, False) 'Locate specific tool according to QR code number
Retrieve2 = Application.WorksheetFunction.Vlookup(Trim(Guncode), .Range(WholeRange), 5, False) 'Locate specific gun type according to QR code number
If Guncode = "" Then
MsgBox "This gun doesn't exist in database!"
Else
MsgBox "The tool number is:" & Retrieve1 & vbCrLf & "The gun type is:" & Retrieve2
End If
End With
End Sub
1. I am not sure what is the reason using Address(True, False) for row number.
This comes from a combination of these two functions. The true/false setting is telling the funciton to use/not use absolute references in the address.
Split ( expression [,delimiter] [,limit] [,compare] )
https://www.techonthenet.com/excel/formulas/split.php
expression.Address (RowAbsolute, ColumnAbsolute, ReferenceStyle, External, RelativeTo)
https://learn.microsoft.com/en-us/office/vba/api/excel.range.address
Shouldn't Cell (1, FinalColumn) stands for the column number?
No, the cells fucntiosn basically returns an intersection/address of rows & column.
Try this for example: debug.Print; thisworkbook.Sheets("Sheet1").Cells(2,2)
You mentioned CSTR doesn't magically transform to letter equivalent so what would it transform to? Could you further elaborate?
This is a data type conversion function. CSTR(666) essentially does this: this 666 becomes this "666"
2. vArr(0). I am confused with what does the parameter 0 stands for in the bracket. Actually this is a general question I always have regarding to parameter specification.
This is an array position refence. The split function returns an array of strings. Since we're using to capture the column label value, we only need to reference the first position.
(3) I tried copy your code and run it but still reminds me error on the same row.
Works fine for me unless there is no returning value, which returns an error which is what I meant by "bomb."
I want to find if a string contains a ","(comma) in it. Do we have any other option other than reading char-by-char?
Use the Instr function (old version of MSDN doc found here)
Dim pos As Integer
pos = InStr("find the comma, in the string", ",")
will return 15 in pos
If not found it will return 0
If you need to find the comma with an excel formula you can use the =FIND(",";A1) function.
Notice that if you want to use Instr to find the position of a string case-insensitive use the third parameter of Instr and give it the const vbTextCompare (or just 1 for die-hards).
Dim posOf_A As Integer
posOf_A = InStr(1, "find the comma, in the string", "A", vbTextCompare)
will give you a value of 14.
Note that you have to specify the start position in this case as stated in the specification I linked: The start argument is required if compare is specified.
You can also use the special word like:
Public Sub Search()
If "My Big String with, in the middle" Like "*,*" Then
Debug.Print ("Found ','")
End If
End Sub
There is also the InStrRev function which does the same type of thing, but starts searching from the end of the text to the beginning.
Per #rene's answer...
Dim pos As Integer
pos = InStrRev("find the comma, in the string", ",")
...would still return 15 to pos, but if the string has more than one of the search string, like the word "the", then:
Dim pos As Integer
pos = InStrRev("find the comma, in the string", "the")
...would return 20 to pos, instead of 6.
Building on Rene's answer, you could also write a function that returned either TRUE if the substring was present, or FALSE if it wasn't:
Public Function Contains(strBaseString As String, strSearchTerm As String) As Boolean
'Purpose: Returns TRUE if one string exists within another
On Error GoTo ErrorMessage
Contains = InStr(strBaseString, strSearchTerm)
Exit Function
ErrorMessage:
MsgBox "The database has generated an error. Please contact the database administrator, quoting the following error message: '" & Err.Description & "'", vbCritical, "Database Error"
End
End Function
You wouldn't really want to do this given the existing Instr/InstrRev functions but there are times when it is handy to use EVALUATE to return the result of Excel worksheet functions within VBA
Option Explicit
Public Sub test()
Debug.Print ContainsSubString("bc", "abc,d")
End Sub
Public Function ContainsSubString(ByVal substring As String, ByVal testString As String) As Boolean
'substring = string to test for; testString = string to search
ContainsSubString = Evaluate("=ISNUMBER(FIND(" & Chr$(34) & substring & Chr$(34) & ", " & Chr$(34) & testString & Chr$(34) & "))")
End Function
I need to convert a string, obtained from excel, in VBA to an interger. To do so I'm using CInt() which works well. However there is a chance that the string could be something other than a number, in this case I need to set the integer to 0. Currently I have:
If oXLSheet2.Cells(4, 6).Value <> "example string" Then
currentLoad = CInt(oXLSheet2.Cells(4, 6).Value)
Else
currentLoad = 0
End If
The problem is that I cannot predict all possible non numeric strings which could be in this cell. Is there a way I can tell it to convert if it's an integer and set to 0 if not?
Use IsNumeric. It returns true if it's a number or false otherwise.
Public Sub NumTest()
On Error GoTo MyErrorHandler
Dim myVar As Variant
myVar = 11.2 'Or whatever
Dim finalNumber As Integer
If IsNumeric(myVar) Then
finalNumber = CInt(myVar)
Else
finalNumber = 0
End If
Exit Sub
MyErrorHandler:
MsgBox "NumTest" & vbCrLf & vbCrLf & "Err = " & Err.Number & _
vbCrLf & "Description: " & Err.Description
End Sub
Cast to long or cast to int, be aware of the following.
These functions are one of the view functions in Excel VBA that are depending on the system regional settings. So if you use a comma in your double like in some countries in Europe, you will experience an error in the US.
E.g., in european excel-version 0,5 will perform well with CDbl(), but in US-version it will result in 5.
So I recommend to use the following alternative:
Public Function CastLong(var As Variant)
' replace , by .
var = Replace(var, ",", ".")
Dim l As Long
On Error Resume Next
l = Round(Val(var))
' if error occurs, l will be 0
CastLong = l
End Function
' similar function for cast-int, you can add minimum and maximum value if you like
' to prevent that value is too high or too low.
Public Function CastInt(var As Variant)
' replace , by .
var = Replace(var, ",", ".")
Dim i As Integer
On Error Resume Next
i = Round(Val(var))
' if error occurs, i will be 0
CastInt = i
End Function
Of course you can also think of cases where people use commas and dots, e.g., three-thousand as 3,000.00. If you require functionality for these kind of cases, then you have to check for another solution.
Try this:
currentLoad = ConvertToLongInteger(oXLSheet2.Cells(4, 6).Value)
with this function:
Function ConvertToLongInteger(ByVal stValue As String) As Long
On Error GoTo ConversionFailureHandler
ConvertToLongInteger = CLng(stValue) 'TRY to convert to an Integer value
Exit Function 'If we reach this point, then we succeeded so exit
ConversionFailureHandler:
'IF we've reached this point, then we did not succeed in conversion
'If the error is type-mismatch, clear the error and return numeric 0 from the function
'Otherwise, disable the error handler, and re-run the code to allow the system to
'display the error
If Err.Number = 13 Then 'error # 13 is Type mismatch
Err.Clear
ConvertToLongInteger = 0
Exit Function
Else
On Error GoTo 0
Resume
End If
End Function
I chose Long (Integer) instead of simply Integer because the min/max size of an Integer in VBA is crummy (min: -32768, max:+32767). It's common to have an integer outside of that range in spreadsheet operations.
The above code can be modified to handle conversion from string to-Integers, to-Currency (using CCur() ), to-Decimal (using CDec() ), to-Double (using CDbl() ), etc. Just replace the conversion function itself (CLng). Change the function return type, and rename all occurrences of the function variable to make everything consistent.
Just use Val():
currentLoad = Int(Val([f4]))
Now currentLoad has a integer value, zero if [f4] is not numeric.
To put it on one line:
currentLoad = IIf(IsNumeric(oXLSheet2.Cells(4, 6).Value), CInt(oXLSheet2.Cells(4, 6).Value), 0)
Here are a three functions that might be useful. First checks the string for a proper numeric format, second and third function converts a string to Long or Double.
Function IsValidNumericEntry(MyString As String) As Boolean
'********************************************************************************
'This function checks the string entry to make sure that valid digits are in the string.
'It checks to make sure the + and - are the first character if entered and no duplicates.
'Valid charcters are 0 - 9, + - and the .
'********************************************************************************
Dim ValidEntry As Boolean
Dim CharCode As Integer
Dim ValidDigit As Boolean
Dim ValidPlus As Boolean
Dim ValidMinus As Boolean
Dim ValidDecimal As Boolean
Dim ErrMsg As String
ValidDigit = False
ValidPlus = False
ValidMinus = False
ValidDecimal = False
ValidEntry = True
For x = 1 To Len(MyString)
CharCode = Asc(Mid(MyString, x, 1))
Select Case CharCode
Case 48 To 57 ' Digits 0 - 9
ValidDigit = True
Case 43 ' Plus sign
If ValidPlus Then 'One has already been detected and this is a duplicate
ErrMsg = "Invalid entry....too many plus signs!"
ValidEntry = False
Exit For
ElseIf x = 1 Then 'if in the first positon it is valide
ValidPlus = True
Else 'Not in first position and it is invalid
ErrMsg = "Invalide entry....Plus sign not in the correct position! "
ValidEntry = False
Exit For
End If
Case 45 ' Minus sign
If ValidMinus Then 'One has already been detected and this is a duplicate
ErrMsg = "Invalide entry....too many minus signs! "
ValidEntry = False
Exit For
ElseIf x = 1 Then 'if in the first position it is valid
ValidMinus = True
Else 'Not in first position and it is invalid
ErrMsg = "Invalide entry....Minus sign not in the correct position! "
ValidEntry = False
Exit For
End If
Case 46 ' Period
If ValidDecimal Then 'One has already been detected and this is a duplicate
ErrMsg = "Invalide entry....too many decimals!"
ValidEntry = False
Exit For
Else
ValidDecimal = True
End If
Case Else
ErrMsg = "Invalid numerical entry....Only digits 0-9 and the . + - characters are valid!"
ValidEntry = False
Exit For
End Select
Next
If ValidEntry And ValidDigit Then
IsValidNumericEntry = True
Else
If ValidDigit = False Then
ErrMsg = "Text string contains an invalid numeric format." & vbCrLf _
& "Use only one of the following formats!" & vbCrLf _
& "(+dd.dd -dd.dd +dd -dd dd.d or dd)! "
End If
MsgBox (ErrMsg & vbCrLf & vbCrLf & "You Entered: " & MyString)
IsValidNumericEntry = False
End If
End Function
Function ConvertToLong(stringVal As String) As Long
'Assumes the user has verified the string contains a valide numeric entry.
'User should call the function IsValidNumericEntry first especially after any user input
'to verify that the user has entered a proper number.
ConvertToLong = CLng(stringVal)
End Function
Function ConvertToDouble(stringVal As String) As Double
'Assumes the user has verified the string contains a valide numeric entry.
'User should call the function IsValidNumericEntry first especially after any user input
'to verify that the user has entered a proper number.
ConvertToDouble = CDbl(stringVal)
End Function