Related
From ActiveWorkbook.name, I would like to extract the strings that are before (left side of ) the numbers. Since I want to use the same code in multiple workbooks, the file names would be variable, but every file name has date info in the middle (yyyymmdd).
In case of excel file, I can use the below formula, but can I apply the same kind of method in VBA?
=LEFT(A1,MIN(FIND({0,1,2,3,4,5,6,7,8,9},ASC(A1)&1234567890))-1)
Example: MyExcelWorkbook_Management_20200602_MyName.xlsm
In above case, I want to extract "MyExcelWorkbook_Management_".
The most basic thing you could do is to replicate something that worked for you in Excel through Evaluate:
Sub Test()
Dim str As String: str = "MyExcelWorkbook_Management_20200602_MyName.xlsm"
Debug.Print Evaluate(Replace("=LEFT(""X"",MIN(FIND({0,1,2,3,4,5,6,7,8,9},ASC(""X"")&1234567890))-1)", "X", str))
End Sub
Pretty? Not really, but it does the job and got it's limitations.
You could use Regular Expressions to extract any letters / underscores before the number as well
Dim str As String
str = "MyExcelWorkbook_Management_20200602_MyName.xlsm"
With CreateObject("vbscript.regexp")
.Pattern = "^\D*"
.Global = True
MsgBox .Execute(str)(0)
End With
Gives:
MyExcelWorkbook_Management_
So basically you want to use the Midfunction to look for the first numerical character in your input string, and then cut your input string to that position.
That means we need to loop through the string from left to right, look at one character at a time and see if it is a digit or not.
This code does exactly that:
Option Explicit
Sub extratLeftText()
Dim someString As String
Dim result As String
someString = "Hello World1234"
Dim i As Long
Dim c As String 'one character of your string
For i = 1 To Len(someString)
c = Mid(someString, i, 1)
If IsNumeric(c) = True Then 'should write "If IsNumeric(c) = True AND i>1 Then" to avoid an "out of bounds" error
result = Left(someString, i - 1)
Exit For
End If
Next i
MsgBox result
End Sub
Last thing you need to do is to load in some workbook name into your VBA function. Generally this is done with the .Name method of the workbookobject:
Sub workbookName()
Dim wb As Workbook
Set wb = ActiveWorkbook
MsgBox wb.Name
End Sub
Of course you would need to find some way to replace the Set wb = ActiveWorkbook line with code that suits your purpose.
I need to find numbers from a string. How does one find numbers from a string in VBA Excel?
Assuming you mean you want the non-numbers stripped out, you should be able to use something like:
Function onlyDigits(s As String) As String
' Variables needed (remember to use "option explicit"). '
Dim retval As String ' This is the return string. '
Dim i As Integer ' Counter for character position. '
' Initialise return string to empty '
retval = ""
' For every character in input string, copy digits to '
' return string. '
For i = 1 To Len(s)
If Mid(s, i, 1) >= "0" And Mid(s, i, 1) <= "9" Then
retval = retval + Mid(s, i, 1)
End If
Next
' Then return the return string. '
onlyDigits = retval
End Function
Calling this with:
Dim myStr as String
myStr = onlyDigits ("3d1fgd4g1dg5d9gdg")
MsgBox (myStr)
will give you a dialog box containing:
314159
and those first two lines show how you can store it into an arbitrary string variable, to do with as you wish.
Regular expressions are built to parse. While the syntax can take a while to pick up on this approach is very efficient, and is very flexible for handling more complex string extractions/replacements
Sub Tester()
MsgBox CleanString("3d1fgd4g1dg5d9gdg")
End Sub
Function CleanString(strIn As String) As String
Dim objRegex
Set objRegex = CreateObject("vbscript.regexp")
With objRegex
.Global = True
.Pattern = "[^\d]+"
CleanString = .Replace(strIn, vbNullString)
End With
End Function
Expanding on brettdj's answer, in order to parse disjoint embedded digits into separate numbers:
Sub TestNumList()
Dim NumList As Variant 'Array
NumList = GetNums("34d1fgd43g1 dg5d999gdg2076")
Dim i As Integer
For i = LBound(NumList) To UBound(NumList)
MsgBox i + 1 & ": " & NumList(i)
Next i
End Sub
Function GetNums(ByVal strIn As String) As Variant 'Array of numeric strings
Dim RegExpObj As Object
Dim NumStr As String
Set RegExpObj = CreateObject("vbscript.regexp")
With RegExpObj
.Global = True
.Pattern = "[^\d]+"
NumStr = .Replace(strIn, " ")
End With
GetNums = Split(Trim(NumStr), " ")
End Function
Use the built-in VBA function Val, if the numbers are at the front end of the string:
Dim str as String
Dim lng as Long
str = "1 149 xyz"
lng = Val(str)
lng = 1149
Val Function, on MSDN
I was looking for the answer of the same question but for a while I found my own solution and I wanted to share it for other people who will need those codes in the future. Here is another solution without function.
Dim control As Boolean
Dim controlval As String
Dim resultval As String
Dim i as Integer
controlval = "A1B2C3D4"
For i = 1 To Len(controlval)
control = IsNumeric(Mid(controlval, i, 1))
If control = True Then resultval = resultval & Mid(controlval, i, 1)
Next i
resultval = 1234
This a variant of brettdj's & pstraton post.
This will return a true Value and not give you the #NUM! error. And \D is shorthand for anything but digits. The rest is much like the others only with this minor fix.
Function StripChar(Txt As String) As Variant
With CreateObject("VBScript.RegExp")
.Global = True
.Pattern = "\D"
StripChar = Val(.Replace(Txt, " "))
End With
End Function
This is based on another answer, but is just reformated:
Assuming you mean you want the non-numbers stripped out, you should be able to use something like:
'
' Skips all characters in the input string except digits
'
Function GetDigits(ByVal s As String) As String
Dim char As String
Dim i As Integer
GetDigits = ""
For i = 1 To Len(s)
char = Mid(s, i, 1)
If char >= "0" And char <= "9" Then
GetDigits = GetDigits + char
End If
Next i
End Function
Calling this with:
Dim myStr as String
myStr = GetDigits("3d1fgd4g1dg5d9gdg")
Call MsgBox(myStr)
will give you a dialog box containing:
314159
and those first two lines show how you can store it into an arbitrary string variable, to do with as you wish.
Alternative via Byte Array
If you assign a string to a Byte array you typically get the number equivalents of each character in pairs of the array elements. Use a loop for numeric check via the Like operator and return the joined array as string:
Function Nums(s$)
Dim by() As Byte, i&, ii&
by = s: ReDim tmp(UBound(by)) ' assign string to byte array; prepare temp array
For i = 0 To UBound(by) - 1 Step 2 ' check num value in byte array (0, 2, 4 ... n-1)
If Chr(by(i)) Like "#" Then tmp(ii) = Chr(by(i)): ii = ii + 1
Next i
Nums = Trim(Join(tmp, vbNullString)) ' return string with numbers only
End Function
Example call
Sub testByteApproach()
Dim s$: s = "a12bx99y /\:3,14159" ' [1] define original string
Debug.Print s & " => " & Nums(s) ' [2] display original string and result
End Sub
would display the original string and the result string in the immediate window:
a12bx99y /\:3,14159 => 1299314159
Based on #brettdj's answer using a VBScript regex ojbect with two modifications:
The function handles variants and returns a variant. That is, to take care of a null case; and
Uses explicit object creation, with a reference to the "Microsoft VBScript Regular Expressions 5.5" library
Function GetDigitsInVariant(inputVariant As Variant) As Variant
' Returns:
' Only the digits found in a varaint.
' Examples:
' GetDigitsInVariant(Null) => Null
' GetDigitsInVariant("") => ""
' GetDigitsInVariant(2021-/05-May/-18, Tue) => 20210518
' GetDigitsInVariant(2021-05-18) => 20210518
' Notes:
' If the inputVariant is null, null will be returned.
' If the inputVariant is "", "" will be returned.
' Usage:
' VBA IDE Menu > Tools > References ...
' > "Microsoft VBScript Regular Expressions 5.5" > [OK]
' With an explicit object reference to RegExp we can get intellisense
' and review the object heirarchy with the object browser
' (VBA IDE Menu > View > Object Browser).
Dim regex As VBScript_RegExp_55.RegExp
Set regex = New VBScript_RegExp_55.RegExp
Dim result As Variant
result = Null
If IsNull(inputVariant) Then
result = Null
Else
With regex
.Global = True
.Pattern = "[^\d]+"
result = .Replace(inputVariant, vbNullString)
End With
End If
GetDigitsInVariant = result
End Function
Testing:
Private Sub TestGetDigitsInVariant()
Dim dateVariants As Variant
dateVariants = Array(Null, "", "2021-/05-May/-18, Tue", _
"2021-05-18", "18/05/2021", "3434 ..,sdf,sfd 444")
Dim dateVariant As Variant
For Each dateVariant In dateVariants
Debug.Print dateVariant & ": ", , GetDigitsInVariant(dateVariant)
Next dateVariant
Debug.Print
End Sub
Public Function ExtractChars(strRef$) As String
'Extract characters from a string according to a range of charactors e.g'+.-1234567890'
Dim strA$, e%, strExt$, strCnd$: strExt = "": strCnd = "+.-1234567890"
For e = 1 To Len(strRef): strA = Mid(strRef, e, 1)
If InStr(1, strCnd, strA) > 0 Then strExt = strExt & strA
Next e: ExtractChars = strExt
End Function
In the immediate debug dialog:
? ExtractChars("a-5d31.78K")
-531.78
i want to get only letters from string.
eg.
Lets say the string is this :123abc456d
I want to get: abcd
Looking for something like this but for letters in a string:
Dim mytext As String = "123a123"
Dim myChars() As Char = mytext.ToCharArray()
For Each ch As Char In myChars
If Char.IsDigit(ch) Then
MessageBox.Show(ch)
End If
Next
Thanks
You can do it like this :
Dim mytext As String = "123a123"
Dim RemoveChars As String = "0123456789" 'These are the chars that you want to remove from your mytext string
Dim FinalResult As String
Dim myChars() As Char = mytext.ToCharArray()
For Each ch As Char In myChars
If Not RemoveChars.Contains(ch) Then
FinalResult &= ch
End If
Next
MsgBox(FinalResult)
OR :
Dim mytext As String = "1d23ad123d"
Dim myChars() As Char = mytext.ToCharArray()
Dim FinalResult As String
For Each ch As Char In myChars
If Not Char.IsDigit(ch) Then
FinalResult &= ch
End If
Next
MsgBox(FinalResult)
Both will give you the same result.
Hope that helped you :)
You can use Regex to solve this problem. This regex basically says anything that is not in this class, the class being letters in the alphabet then remove by replacing it with nothing.
Dim mytext As String = "123a123"
Dim Result as String = Regex.Replace(myText, "[^a-zA-Z]", "")
Dim myChars() As Char = Result.ToCharArray()
For Each ch As Char In myChars
If Char.IsDigit(ch) Then
MessageBox.Show(ch)
End If
Next
Make sure you have this at the top of your code Imports System.Text.RegularExpressions
Here is a LINQ one liner:
Debug.Print(String.Concat("123abc456d".Where(AddressOf Char.IsLetter)))
Result: abcd.
Here, .Where(AddressOf Char.IsLetter) treats the string as a list of chars, and only keeps letters in the list. Then, String.Concat re-builds the string out of the char list by concatenating the chars.
I have a column in a spreadsheet.
The format of the data in each cell is aa-0001-xx.
I need to examine the whole column to find the highest value of the sequence number. this would be the substring from column4 thru column7.
I can find the sequence number using Mid(ActiveWorkbook.Sheets("Sheet1").Range("B2:B2"), 4, 4)
But I need to find the max sequence in the whole column.
I am doing this in VBA.
Any help would be appreciated.
Here is my code so far:
Private Sub CommandButton1_Click()
Dim sQuoteNumber As String
Dim sFileName As String
Dim sPathName As String
Dim checkit As String
'Log the Quote
'First, open the log file and determine the next sequential log number.
sPathName = "C:\Users\Rich\Documents\Bryan\BigProject\"
sFileName = "QuoteLog2016.xlsx"
ControlFile = ActiveWorkbook.Name
Workbooks.Open Filename:=sPathName & sFileName
'Create the new Quote Number
checkit = Mid(ActiveWorkbook.Sheets("Sheet1").Range("B2:B2"), 4, 4) ' This is a temp test line
If Mid(ActiveWorkbook.Sheets("Sheet1").Range("B2:B2"), 4, 4) = "" Then
sQuoteNumber = "16-0001"
Else
'find the biggest number
'Here I was looking to like pass the mid function to a Max function of some sort.
sQuoteNumber = "16-0002"
End If
MsgBox ("The new Quote Number is: " + sQuoteNumber)
'Save the log entry
Workbooks(sFileName).Close
All of the comments made to your answer would work well for you. It's also true that there's no evidence in your code at having attempted something, however rudimentary, and this is why answers to a rather trivial task are not forthcoming for you. Perhaps, in future, have a go at some kind of solution ( even if it feels more guesswork than anything) and people on this site will be much more supportive of you.
To set you on your way, you could make use of the Split() function which converts a String into a String array, separated by a nominated value - in the case of your quotations, you could use "-" as your separator. This might be easier than your Mid function and will deal with the case of different sized quotations.
The code below will get you started but you'd want some error handling in there to test, for example, that each cell splits appropriately or that any cells aren't blank. I'll leave all of that to you.
Option Explicit
Private mLastQuote As Long
Public Sub Test()
Initialise 'call this routine just once at the start of your project
MsgBox GetNextQuote(16) 'use the GetNextQuote() function to get next number
MsgBox GetNextQuote(16)
MsgBox GetNextQuote(16)
End Sub
Private Function GetNextQuote(prefix As Integer) As String
mLastQuote = mLastQuote + 1
GetNextQuote = CStr(prefix) & "-" & _
Format(mLastQuote, "000#")
End Function
Private Sub Initialise()
Const PATH_NAME As String = "C:\Users\Rich\Documents\Bryan\BigProject\"
Const FILE_NAME As String = "QuoteLog2016.xlsx"
Const QUOTE_COL As String = "B"
Dim wb As Workbook
Dim ws As Worksheet
Dim v As Variant
Dim r As Long
Dim parts() As String
Dim num As Long
Application.ScreenUpdating = False
Set wb = Workbooks.Open(PATH_NAME & FILE_NAME, True, True)
Set ws = wb.Worksheets("Sheet1")
'Read quote values into variant array
With ws
v = .Range(.Cells(2, QUOTE_COL), _
.Cells(.Rows.Count, QUOTE_COL).End(xlUp)) _
.Value2
End With
'Find max quote
For r = 1 To UBound(v, 1)
parts = Split(v(r, 1), "-") 'splits quote into 3 parts
num = CLng(parts(1)) 'index (1) is the middle part
If num > mLastQuote Then mLastQuote = num
Next
wb.Close False
Application.ScreenUpdating = True
End Sub
I need to find numbers from a string. How does one find numbers from a string in VBA Excel?
Assuming you mean you want the non-numbers stripped out, you should be able to use something like:
Function onlyDigits(s As String) As String
' Variables needed (remember to use "option explicit"). '
Dim retval As String ' This is the return string. '
Dim i As Integer ' Counter for character position. '
' Initialise return string to empty '
retval = ""
' For every character in input string, copy digits to '
' return string. '
For i = 1 To Len(s)
If Mid(s, i, 1) >= "0" And Mid(s, i, 1) <= "9" Then
retval = retval + Mid(s, i, 1)
End If
Next
' Then return the return string. '
onlyDigits = retval
End Function
Calling this with:
Dim myStr as String
myStr = onlyDigits ("3d1fgd4g1dg5d9gdg")
MsgBox (myStr)
will give you a dialog box containing:
314159
and those first two lines show how you can store it into an arbitrary string variable, to do with as you wish.
Regular expressions are built to parse. While the syntax can take a while to pick up on this approach is very efficient, and is very flexible for handling more complex string extractions/replacements
Sub Tester()
MsgBox CleanString("3d1fgd4g1dg5d9gdg")
End Sub
Function CleanString(strIn As String) As String
Dim objRegex
Set objRegex = CreateObject("vbscript.regexp")
With objRegex
.Global = True
.Pattern = "[^\d]+"
CleanString = .Replace(strIn, vbNullString)
End With
End Function
Expanding on brettdj's answer, in order to parse disjoint embedded digits into separate numbers:
Sub TestNumList()
Dim NumList As Variant 'Array
NumList = GetNums("34d1fgd43g1 dg5d999gdg2076")
Dim i As Integer
For i = LBound(NumList) To UBound(NumList)
MsgBox i + 1 & ": " & NumList(i)
Next i
End Sub
Function GetNums(ByVal strIn As String) As Variant 'Array of numeric strings
Dim RegExpObj As Object
Dim NumStr As String
Set RegExpObj = CreateObject("vbscript.regexp")
With RegExpObj
.Global = True
.Pattern = "[^\d]+"
NumStr = .Replace(strIn, " ")
End With
GetNums = Split(Trim(NumStr), " ")
End Function
Use the built-in VBA function Val, if the numbers are at the front end of the string:
Dim str as String
Dim lng as Long
str = "1 149 xyz"
lng = Val(str)
lng = 1149
Val Function, on MSDN
I was looking for the answer of the same question but for a while I found my own solution and I wanted to share it for other people who will need those codes in the future. Here is another solution without function.
Dim control As Boolean
Dim controlval As String
Dim resultval As String
Dim i as Integer
controlval = "A1B2C3D4"
For i = 1 To Len(controlval)
control = IsNumeric(Mid(controlval, i, 1))
If control = True Then resultval = resultval & Mid(controlval, i, 1)
Next i
resultval = 1234
This a variant of brettdj's & pstraton post.
This will return a true Value and not give you the #NUM! error. And \D is shorthand for anything but digits. The rest is much like the others only with this minor fix.
Function StripChar(Txt As String) As Variant
With CreateObject("VBScript.RegExp")
.Global = True
.Pattern = "\D"
StripChar = Val(.Replace(Txt, " "))
End With
End Function
This is based on another answer, but is just reformated:
Assuming you mean you want the non-numbers stripped out, you should be able to use something like:
'
' Skips all characters in the input string except digits
'
Function GetDigits(ByVal s As String) As String
Dim char As String
Dim i As Integer
GetDigits = ""
For i = 1 To Len(s)
char = Mid(s, i, 1)
If char >= "0" And char <= "9" Then
GetDigits = GetDigits + char
End If
Next i
End Function
Calling this with:
Dim myStr as String
myStr = GetDigits("3d1fgd4g1dg5d9gdg")
Call MsgBox(myStr)
will give you a dialog box containing:
314159
and those first two lines show how you can store it into an arbitrary string variable, to do with as you wish.
Alternative via Byte Array
If you assign a string to a Byte array you typically get the number equivalents of each character in pairs of the array elements. Use a loop for numeric check via the Like operator and return the joined array as string:
Function Nums(s$)
Dim by() As Byte, i&, ii&
by = s: ReDim tmp(UBound(by)) ' assign string to byte array; prepare temp array
For i = 0 To UBound(by) - 1 Step 2 ' check num value in byte array (0, 2, 4 ... n-1)
If Chr(by(i)) Like "#" Then tmp(ii) = Chr(by(i)): ii = ii + 1
Next i
Nums = Trim(Join(tmp, vbNullString)) ' return string with numbers only
End Function
Example call
Sub testByteApproach()
Dim s$: s = "a12bx99y /\:3,14159" ' [1] define original string
Debug.Print s & " => " & Nums(s) ' [2] display original string and result
End Sub
would display the original string and the result string in the immediate window:
a12bx99y /\:3,14159 => 1299314159
Based on #brettdj's answer using a VBScript regex ojbect with two modifications:
The function handles variants and returns a variant. That is, to take care of a null case; and
Uses explicit object creation, with a reference to the "Microsoft VBScript Regular Expressions 5.5" library
Function GetDigitsInVariant(inputVariant As Variant) As Variant
' Returns:
' Only the digits found in a varaint.
' Examples:
' GetDigitsInVariant(Null) => Null
' GetDigitsInVariant("") => ""
' GetDigitsInVariant(2021-/05-May/-18, Tue) => 20210518
' GetDigitsInVariant(2021-05-18) => 20210518
' Notes:
' If the inputVariant is null, null will be returned.
' If the inputVariant is "", "" will be returned.
' Usage:
' VBA IDE Menu > Tools > References ...
' > "Microsoft VBScript Regular Expressions 5.5" > [OK]
' With an explicit object reference to RegExp we can get intellisense
' and review the object heirarchy with the object browser
' (VBA IDE Menu > View > Object Browser).
Dim regex As VBScript_RegExp_55.RegExp
Set regex = New VBScript_RegExp_55.RegExp
Dim result As Variant
result = Null
If IsNull(inputVariant) Then
result = Null
Else
With regex
.Global = True
.Pattern = "[^\d]+"
result = .Replace(inputVariant, vbNullString)
End With
End If
GetDigitsInVariant = result
End Function
Testing:
Private Sub TestGetDigitsInVariant()
Dim dateVariants As Variant
dateVariants = Array(Null, "", "2021-/05-May/-18, Tue", _
"2021-05-18", "18/05/2021", "3434 ..,sdf,sfd 444")
Dim dateVariant As Variant
For Each dateVariant In dateVariants
Debug.Print dateVariant & ": ", , GetDigitsInVariant(dateVariant)
Next dateVariant
Debug.Print
End Sub
Public Function ExtractChars(strRef$) As String
'Extract characters from a string according to a range of charactors e.g'+.-1234567890'
Dim strA$, e%, strExt$, strCnd$: strExt = "": strCnd = "+.-1234567890"
For e = 1 To Len(strRef): strA = Mid(strRef, e, 1)
If InStr(1, strCnd, strA) > 0 Then strExt = strExt & strA
Next e: ExtractChars = strExt
End Function
In the immediate debug dialog:
? ExtractChars("a-5d31.78K")
-531.78