I have some trouble with {}. When i get max value like this {1,8} it not work and i don't now why. Min vale is valid well
Private Sub Highlvl_Expression()
Dim strPattern As String: strPattern = "[a-zA-Z0-9_]{1,8}"
Dim strReplace As String: strReplace = ""
Dim regEx As New RegExp
Dim Test As Boolean
With regEx
.Global = True
.MultiLine = True
.IgnoreCase = False
.Pattern = strPattern
End With
Test = regEx.Test(Highlvl.Value)
If regEx.Test(Highlvl.Value) Then
MsgBox ("Validate")
Else
MsgBox ("Not Validate")
End If
End Sub
You specified the pattern that looks for 1 to 8 alphanumeric characters inside a string. If you run the regex against a 9-character string "ABCDE6789" (regEx.Execute("ABCDE6789")), you will have 2 matches: ABCDE678 and 9.
If you want to validate a string that should have a minimum or a maximum number of characters, you need to use anchors, i.e. start and end of string assertions ^ and $. So, use
Dim strPattern As String: strPattern = "^[a-zA-Z0-9_]{1,8}$"
And
.Global = False
The global flag is not necessary since we are not looking for multiple matches, but for a single true or false result with test.
Related
I'm trying to use Regex to match any character (This is just a piece of code from a larger project). I got the below to work, but seems like it is wrong, is there a proper way to search for any character via RegEx?
strPattern = "([!##$%^&*()]?[a-z]?[0-9]?)"
Eg: MCVE
Public Sub RegExSearch()
Dim regexp As Object
Dim rng As Range, rcell As Range
Dim strInput As String, strPattern As String
Set regexp = CreateObject("vbscript.regexp")
Set rng = ActiveSheet.Range("A1:A1")
With regexp
.Global = False
.MultiLine = False
.ignoreCase = True
.Pattern = strPattern
End With
For Each rcell In rng.Cells
strPattern = "([!##$%^&*()]?[a-z]?[0-9]?)" ' This matches everything, but seems improper
If strPattern <> "" Then
strInput = rcell.Value
If regexp.test(strInput) Then
MsgBox rcell & " Matched in Cell" & rcell.Address
End If
End If
Next
End Sub
. "Wildcard." The unescaped period matches any character, except a new line.
strPattern = "."
Or as #RonRosenfeld pointed out, if you need to match everything INCLUDING a "new line" then this would work.
strPattern = "[/S/s]*"
https://wellsr.com/vba/2018/excel/vba-regex-regular-expressions-guide/
The below code matches and replaces, but the digit next to the capture group is consumed. Where am I going wrong?
Sub test()
Dim regex As Object 'Regexp object.
Set regex = CreateObject("VBScript.RegExp") 'Regexp object.
Dim strPattern As String: strPattern = "\d(AM|PM)" 'Declare regex pattern.
Dim strReplace As String 'Placeholder string for replace operation.
Dim target As String
target = "1:05PM"
strReplace = " $1"
With regex
.Global = True
.IgnoreCase = False
.Pattern = strPattern
End With
If regex.test(target) Then
Debug.Print regex.Replace(target, strReplace)
End If
End Sub
Output:
1:0 PM
It's because you have an un-captured \d in your regex. Try putting () around the \d i.e. (\d)(AM|PM).
You also need to change strReplace to "$1 $2"
I want to return 5 consecutive digits from a string (working in VBA).
Based on this post Regex I'm using the pattern [^\d]\d{5}[^\d], but this picks up the single letters immediately before and after the targeted 5 digits and returns h92345W(from "....South92345West").
How can I modify to return only the 5 consecutive digits: 92345
Sub RegexTest()
Dim strInput As String
Dim strPattern As String
strInput = "9129 Nor22 999123456 South92345West"
'strPattern = "^\d{5}$" 'No match
strPattern = "[^\d]\d{5}[^\d]" 'Returns additional letter before and after digits
'In this case returns: "h12345W"
MsgBox RegxFunc(strInput, strPattern)
End Sub
Function RegxFunc(strInput As String, regexPattern As String) As String
Dim regEx As New RegExp
With regEx
.Global = True
.MultiLine = True
.IgnoreCase = False
.Pattern = regexPattern
End With
If regEx.test(strInput) Then
Set matches = regEx.Execute(strInput)
RegxFunc = matches(0).Value
Else
RegxFunc = "not matched"
End If
End Function
([^\d])(\d{5})([^\d])
You can use this regex, the matched terms should be in the 2nd group
You need to use a group:
"[^\d](\d{5})[^\d]"
And then the number will be in the first group. Not sure about the VBA syntax for grouping.
I'm trying to use a regex to find cells in a range that have a comma, but no space after that comma. Then, I want to simply add a space between the comma and the next character. For example, a cell has Wayne,Bruce text inside, but I want to turn it to Wayne, Bruce.
I have a regex pattern that can find cells with characters and commas without spaces, but when I replace this, it cuts off some characters.
Private Sub simpleRegexSearch()
' adapted from http://stackoverflow.com/questions/22542834/how-to-use-regular-expressions-regex-in-microsoft-excel-both-in-cell-and-loops
Dim strPattern As String: strPattern = "[a-zA-Z]\,[a-zA-Z]"
Dim strReplace As String: strReplace = ", "
Dim regEx As New RegExp
Dim strInput As String
Dim Myrange As Range
Set Myrange = ActiveSheet.Range("P1:P5")
For Each cell In Myrange
If strPattern <> "" Then
strInput = cell.Value
With regEx
.Global = True
.MultiLine = True
.IgnoreCase = False
.Pattern = strPattern
End With
If regEx.TEST(strInput) Then
Debug.Print (regEx.Replace(strInput, strReplace))
Else
Debug.Print ("No Regex Not matched in " & cell.address)
End If
End If
Next
Set regEx = Nothing
End Sub
If I run that against "Wayne,Bruce" I get "Wayn, ruce". How do I keep the letters, but separate them?
Change the code the following way:
Dim strPattern As String: strPattern = "([a-zA-Z]),(?=[a-zA-Z])"
Dim strReplace As String: strReplace = "$1, "
Output will be Bruce, Wayne.
The problem is that you cannot use a look-behind in VBScript, so we need a workaround in the form of a capturing group for the letter before the comma.
For the letter after the comma, we can use a look-ahead, it is available in this regex flavor.
So, we just capture with ([a-zA-Z]) and restore it in the replacing call with a back-reference $1. Look-ahead does not consume characters, so we are covered.
(EDIT) REGEX EXPLANATION
([a-zA-Z]) - A captured group that includes a character class matching just 1 English character
, - Matching a literal , (you actually do not have to escape it as it is not a special character)
(?=[a-zA-Z]) - A positive look-ahead that only checks (does not match, or consume) if the immediate character following the comma is and English letter.
If we replace all commas with comma+space and then replace comma+space+space with comma+space, we can meet your requirement:
Sub NoRegex()
Dim r As Range
Set r = Range("P1:P5")
r.Replace What:=",", Replacement:=", "
r.Replace What:=", ", Replacement:=", "
End Sub
Uses the same RegExp as in the solution from stribizhev but with two optimisations for speed
Your current code sets the RegExp details for every cell tested, these only need setting once.
Looping through a varinat array is much faster than a cell range
code
Private Sub simpleRegexSearch()
' adapted from http://stackoverflow.com/questions/22542834/how-to-use-regular-expressions-regex-in-microsoft-excel-both-in-cell-and-loops
Dim strPattern As String:
Dim strReplace As String:
Dim regEx As Object
Dim strInput As String
Dim X, X1
Dim lngnct
Set regEx = CreateObject("vbscript.regexp")
strPattern = "([a-zA-Z])\,(?=[a-zA-Z])"
strReplace = "$1, "
With regEx
.Global = True
.MultiLine = True
.IgnoreCase = False
.Pattern = strPattern
X = ActiveSheet.Range("P1:P5").Value2
For X1 = 1 To UBound(X)
If .TEST(X(X1, 1)) Then
Debug.Print .Replace(X(X1, 1), strReplace)
Else
Debug.Print ("No Regex Not matched in " & [p1].Offset(X1 - 1).Address(0, 0))
End If
Next
End With
Set regEx = Nothing
End Sub
What you are doing via Regex is to find a pattern
(any Alphabet),(any Alphabet)
and then replace such pattern to
,_
where _ implies a space.
So if you have Wayne,Bruce then the pattern matches where e,B. Therefore the result becomes Wayn, ruce.
Try
Dim strPattern As String: strPattern = "([a-zA-Z]),([a-zA-Z])"
Dim strReplace As String: strReplace = "$1, $2"
.
This is my sample record in a Text format with comma delimited
901,BLL,,,BQ,ARCTICA,,,,
i need to replace ,,, to ,,
The Regular expression that i tried
With regex
.MultiLine = False
.Global = True
.IgnoreCase = False
.Pattern="^(?=[A-Z]{3})\\,{3,}",",,"))$ -- error
Now i want to pass Line from file to Regex to correct the record, can some body guide me to fix this i am very new to VBA
I want to read the file line by line pass it to Regex
Looking at your original pattern I tried using .Pattern = "^\d{3},\D{3},,," which works on the sample record as with the 3 number characters , 3 letters,,,
In the answer I have used a more generalised pattern .Pattern = "^\w*,\w*,\w*,," This also works on the sample and mathces 3 commas each preceded with 0 or more alphanumeric characters followed directly by a fourth comma. Both patterns require a match to be from the begining of the string.
Pattern .Pattern = "^\d+,[a-zA-Z]+,\w*,," also works on the sample record. It would specify that before the first comma there should be 1 or greater numeric characters (and only numeric characters) and before the second comma ther should be 1 or more letters (and only letters). Before the 3rd comma there could be 0 or more alphanumeric characters.
The left function removes the rightmost character in the match ie. the last comma to generate the string used by the Regex.Replace.
Sub Test()
Dim str As String
str = "901,BLL,,,BQ,ARCTICA,,,,"
Debug.Print
Debug.Print str
str = strConv(str)
Debug.Print str
End Sub
Function strConv(ByVal str As String) As String
Dim objRegEx As Object
Dim oMatches As Object
Dim oMatch As Object
Set objRegEx = CreateObject("VBScript.RegExp")
With objRegEx
.MultiLine = False
.IgnoreCase = False
.Global = True
.Pattern = "^\w*,\w*,\w*,,"
End With
Set oMatches = objRegEx.Execute(str)
If oMatches.Count > 0 Then
For Each oMatch In oMatches
str = objRegEx.Replace(str, Left(oMatch.Value, oMatch.Length - 1))
Next oMatch
End If
strConv = str
End Function
Try this
Sub test()
Dim str As String
str = "901,BLL,,,BQ,ARCTICA,,,,"
str = strConv(str)
MsgBox str
End Sub
Function strConv(ByVal str As String) As String
Dim objRegEx As Object, allMatches As Object
Set objRegEx = CreateObject("VBScript.RegExp")
With objRegEx
.MultiLine = False
.IgnoreCase = False
.Global = True
.Pattern = ",,,"
End With
strConv = objRegEx.Replace(str, ",,")
End Function