RegEx for replacing part of a string including single quotes (')

Question

I am trying to replace a portion of a text that is between apostrophes, not all, just a part. For example, I need to replace the characters /* and */ that are only within text between quotes by null text but not outside them.

My input text, for example:

A = 'THIS IS AN ALPHABETIC /* CONSTANT' || WS_CON1 /* */ || 'TEST STRING */';

Expected output:

A = 'THIS IS AN ALPHABETIC  CONSTANT' || WS_CON1 /* */ || 'TEST STRING ';

I extracted the texts in quotes but I do not know how to replace the /* and */ with null text.

Sub ReplaceWithRegex()
    Dim strPattern As String
    Dim strReplace As String
    Dim regEx As Variant
    Dim strtxt As String

    Set regEx = CreateObject("vbscript.regexp")
    strtxt = "A = 'THIS IS AN ALPHABETIC /* CONSTANT' || WS_CON1 /* */ || ' TEST STRING */';"
    strPattern = "\'([^\']*)\'"
    strReplace = ""

    With regEx
        .Global = True
        .MultiLine = True
        .IgnoreCase = False
        .Pattern = strPattern
    End With

    If regEx.Test(strtxt) Then
        Debug.Print regEx.Replace(strtxt, strReplace)
    Else
        MsgBox ("Not matched")
    End If
End Sub

Obviously, this replace all text between quotes to null string.

How do I solve this problem?

Answer 1

This expression might help you to replace those undesired /*:

[A-Z]\s\/\*\s[A-Z]

We can simply wrap that in a capturing groups (), similar to:

([A-Z])\s(\/\*)\s([A-Z])

Then, we can replace it with $1 $3 and ignore the second undesired capturing group:

This tool helps you to modify/change/edit your expressions, as you wish.

RegEx Descriptive Graph

This link helps you to visualize your expressions:

Edit

If you may have more patterns in these capturing groups, you can simply add them using an |, such as:

([A-Z])\s(\/\*|\*\/)(\s[A-Z]|\x27)

You might also want to use \x27 instead of ' such that your code would become easy to read.

JavaScript Demo

const regex = /([A-Z])\s(\/\*|\*\/)(\s[A-Z]|\x27)/gm;
const str = `A = 'THIS IS AN ALPHABETIC /* CONSTANT\' || WS_CON1 /* */ || \'TEST STRING */';`;
const subst = `$1$3`;

// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);

console.log('Substitution result: ', result);

Performance Test

This snippet returns the runtime of a 1-million times for loop.

const repeat = 1000000;
const start = Date.now();

for (var i = repeat; i >= 0; i--) {
	const regex = /([A-Z])\s(\/\*|\*\/)(\s[A-Z]|\x27)/gm;
	const str = `A = 'THIS IS AN ALPHABETIC /* CONSTANT\' || WS_CON1 /* */ || \'TEST STRING */';`;
	const subst = `$1$3`;

	var match = str.replace(regex, subst);
}

const end = Date.now() - start;
console.log("YAAAY! \"" + match + "\" is a match 💚💚💚 ");
console.log(end / 1000 + " is the runtime of " + repeat + " times benchmark test. 😳 ");

Answer 2

Here is another VBA method using Regular Expressions.

Option Explicit
'Set Reference to Microsoft VBScript Regular Expressions 5.5
Function reReplaceComment(S As String) As String
  Dim RE As RegExp
  Dim I As Long, J As Long

Set RE = New RegExp
With RE
    .Global = True
    .Pattern = "('[^']*?)(?:(?:/\*)|(?:\*/))([^']*?')"
    reReplaceComment = .Replace(S, "$1$2")
End With
End Function

Remove tokens within single quotes

('[^']*?)(?:(?:/\*)|(?:\*/))([^']*?')

• Match the regex below and capture its match into backreference number 1 ('[^']*?)
  • Match the character “'” literally '
  • Match any character that is NOT a “'” [^']*?
    • Between zero and unlimited times, as few times as possible, expanding as needed (lazy) *?
• Match the regular expression below (?:(?:/\*)|(?:\*/))
  • Match this alternative (attempting the next alternative only if this one fails) (?:/\*)
    • Match the regular expression below (?:/\*)
      • Match the character “/” literally /
      • Match the character “*” literally \*
  • Or match this alternative (the entire group fails if this one fails to match) (?:\*/)
    • Match the regular expression below (?:\*/)
      • Match the character “*” literally \*
      • Match the character “/” literally /
• Match the regex below and capture its match into backreference number 2 ([^']*?')
  • Match any character that is NOT a “'” [^']*?
    • Between zero and unlimited times, as few times as possible, expanding as needed (lazy) *?
  • Match the character “'” literally '

$1$2

• Insert the text that was last matched by capturing group number 1 $1
• Insert the text that was last matched by capturing group number 2 $2

Created with RegexBuddy