I continue trying to perform string format matching using RegExp in VBScript & VB6. I am now trying to match a short, single-line string formatted as:
Seven characters:
a. Six alphanumeric plus one "-" OR
b. Five alphanumeric plus two "-"
Three numbers
Two letters
Literal "65"
A two-digit hex number.
Examples include 123456-789LM65F2, 4EF789-012XY65A5, A2345--789AB65D0 & 23456--890JK65D0.
The RegExp pattern ([A-Z0-9\-]{12})([65][A-F0-9]{2}) lumps (1) - (3) together and finds these OK.
However, if I try to:
c) Break (3) out w/ pattern ([A-Z0-9\-]{10})([A-Z]{2})([65][A-F0-9]{2}),
d) Break out both (2) & (3) w/ pattern ([A-Z0-9\-]{7})([0-9]{3})([A-Z]{2})([65][A-F0-9]{2}), or
e) Tighten up (1) with alternation pattern ([A-Z0-9]{5}[-]{2}|[A-Z0-9]{6}[-]{1})([0-9]{3})([A-Z]{2})([65][A-F0-9]{2})
it refuses to find any of them.
What am I doing wrong? Following is a VBScript that runs and checks these.
' VB Script
Main()
Function Main() ' RegEx_Format_sample.vbs
'Uses two paterns, TestPttn for full format accuracy check & SplitPttn
'to separate the two desired pieces
Dim reSet, EtchTemp, arrSplit, sTemp
Dim sBoule, sSlice, idx, TestPttn, SplitPttn, arrMatch
Dim arrPttn(3), arrItems(3), idxItem, idxPttn, Msgtemp
Set reSet = New RegExp
' reSet.IgnoreCase = True ' Not using
' reSet.Global = True ' Not using
' load test case formats to check & split
arrItems(0) = "0,6 nums + 1 '-',123456-789LM65F2"
arrItems(1) = "1,6 chars + 1 '-',4EF789-012XY65A5"
arrItems(2) = "2,5 chars + 2 '-',A2345--789AB65D0"
arrItems(3) = "3,5 nums + 2 '-',23456--890JK65D0"
SplitPttn = "([A-Z0-9]{5,6})[-]{1,2}([A-Z0-9]{9})" ' split pattern has never failed to work
' load the patterns to try
arrPttn(0) = "([A-Z0-9\-]{12})([65][A-F0-9]{2})"
arrPttn(1) = "([A-Z0-9\-]{10}[A-Z]{2})([65][A-F0-9]{2})"
arrPttn(2) = "([A-Z0-9\-]{7})([0-9]{3})([A-Z]{2})([65][A-F0-9]{2})"
arrPttn(3) = "([A-Z0-9]{5}[-]{2}|[A-Z0-9]{6}[-]{1})([0-9]{3})([A-Z]{2})([65][A-F0-9]{2})"
For idxPttn = 0 To 3 ' select Test pattern
TestPttn = arrPttn(idxPttn)
TestPttn = TestPttn & "[%]" ' append % "ender" char
SplitPttn = SplitPttn & "[%]" ' append % "ender" char
For idxItem = 0 To 3
reSet.Pattern = TestPttn ' set to Test pattern
sTemp = arrItems(idxItem )
arrSplit = Split(sTemp, ",") ' arrSplit is Split array
EtchTemp = arrSplit(2) & "%" ' append % "ender" char to Item sub (2) as the "phrase" under test
If reSet.Test(EtchTemp) = False Then
MsgBox("RegEx " & TestPttn & " false for " & EtchTemp & " as " & arrSplit(1) )
Else ' test OK; now switch to SplitPttn
reSet.Pattern = SplitPttn
Set arrMatch = reSet.Execute(EtchTemp) ' run Pttn as Exec this time
If arrMatch.Count > 0 then ' If test OK then Count s/b > 0
Msgtemp = ""
Msgtemp = "RegEx " & TestPttn & " TRUE for " & EtchTemp & " as " & arrSplit(1)
For idx = 0 To arrMatch.Item(0).Submatches.Count - 1
Msgtemp = Msgtemp & Chr(13) & Chr(10) & "Split segment " & idx & " as " & arrMatch.Item(0).submatches.Item(idx)
Next
MsgBox(Msgtemp)
End If ' Count OK
End If ' test OK
Next ' idxItem
Next ' idxPttn
End Function
Try this Regex:
(?:[A-Z0-9]{6}-|[A-Z0-9]{5}--)[0-9]{3}[A-Z]{2}65[0-9A-F]{2}
Click for Demo
Explanation:
(?:[A-Z0-9]{6}-|[A-Z0-9]{5}--) - matches either 6 Alphanumeric characters followed by a - or 5 Alphanumeric characters followed by a --
[0-9]{3} - matches 3 Digits
[A-Z]{2} - matches 2 Letters
65 - matches 65 literally
[0-9A-F]{2} - matches 2 HEX symbols
You can get some idea from the following code:
VBScript Code:
Option Explicit
Dim objReg, strTest
strTest = "123456-789LM65F2" 'Change the value as per your requirements. You can also store a list of values in an array and run the code in loop
set objReg = new RegExp
objReg.Global = True
objReg.IgnoreCase = True
objReg.Pattern = "(?:[A-Z0-9]{6}-|[A-Z0-9]{5}--)[0-9]{3}[A-Z]{2}65[0-9A-F]{2}"
if objReg.test(strTest) then
msgbox strTest&" matches with the Pattern"
else
msgbox strTest&" does not match with the Pattern"
end if
set objReg = Nothing
Your patterns do not work because:
([A-Z0-9\-]{12})([65][A-F0-9]{2}) - matches 12 occurrences of either an AlphaNumeric character or - followed by either 6 or 5 followed by 2 HEX characters
([A-Z0-9\-]{10}[A-Z]{2})([65][A-F0-9]{2}) - matches 10 occurrences of either an AlphaNumeric character or - followed by 2 Letters followed by either 6 or 5 followed by 2 HEX characters
([A-Z0-9\-]{7})([0-9]{3})([A-Z]{2})([65][A-F0-9]{2}) - matches 7 occurrences of either an AlphaNumeric character or - followed by 3 digits followed by 2 Letters followed by either 6 or 5 followed by 2 HEX characters
([A-Z0-9]{5}[-]{2}|[A-Z0-9]{6}[-]{1})([0-9]{3})([A-Z]{2})([65][A-F0-9]{2}) - matches either 5 occurrences of an AlphaNumeric character followed by -- or 6 occurrences of an Alphanumeric followed by a -. This is then followed by 3 digits followed by 2 Letters followed by either 6 or 5 followed by 2 HEX characters
Try this pattern :
(([A-Z0-9]{5}--)|([A-Z0-9]{6}-))[0-9]{3}[A-Z]{2}65[0-9A-F]{2}
Or, if the last part doesn't like the [A-F]
(([A-Z0-9]{5}--)|([A-Z0-9]{6}-))[0-9]{3}[A-Z]{2}65[0-9ABCDEF]{2}
All, tanx again for your help!!
trincot, everything in each arrItems() between the commas, incl the the "plus", is merely part of a shorthand description of each item's characteristics, such as "5 characters plus 2 dashes".
Gurman, your pttn breakdowns were helpful, but, if I read it right, the addition of the ? prefix is a "Match zero or one occurrences" and this must match exactly one occurrence. Also, my 1st pattern (matches 12) actually DID work for all my test cases.
jNevill, & JMichelB your suggestions are very close to what I ended up with.
I was "over-classing". After some tinkering, I was able to get the Test Pttn to successfully recognize these test cases by taking the [65] out of the [] in my original Alternation pattern. That is I went from ([65]) to (65) and Zammo! it worked.
Orig pattern:
([A-Z0-9]{5}[-]{2}|[A-Z0-9]{6}[-]{1})([0-9]{3})([A-Z]{2})([65][A-F0-9]{2})
Wkg pattern:
([A-Z0-9]{5}[-]{2}|[A-Z0-9]{6}[-]{1})([0-9]{3})([A-Z]{2})(65)([A-F0-9]{2})
Oh, and I moved the
SplitPttn = SplitPttn & "[%]" ' append % "ender" char
stmt up out of the For...Next loop. That helped w/ the splitting.
T-Bone
Related
I'm trying to write a Ruby method that will return true only if the input is a valid phone number, which means, among other rules, it can have spaces and/or dashes between the digits, but not before or after the digits.
In a sense, I need a method that does the opposite of String#strip! (remove all spaces except leading and trailing spaces), plus the same for dashes.
I've tried using String#gsub!, but when I try to match a space or a dash between digits, then it replaces the digits as well as the space/dash.
Here's an example of the code I'm using to remove spaces. I figure once I know how to do that, it will be the same story with the dashes.
def valid_phone_number?(number)
phone_number_pattern = /^0[^0]\d{8}$/
# remove spaces
number.gsub!(/\d\s+\d/, "")
return number.match?(phone_number_pattern)
end
What happens is if I call the method with the following input:
valid_phone_number?(" 09 777 55 888 ")
I get false because line 5 transforms the number into " 0788 ", i.e. it gets rid of the digits around the spaces as well as the spaces. What I want it to do is just to get rid of the inner spaces, so as to produce " 0977755888 ".
I've tried
number.gsub!(/\d(\s+)\d/, "") and number.gsub!(/\d(\s+)\d/) { |match| "" } to no avail.
Thank you!!
If you want to return a boolean, you might for example use a pattern that accepts leading and trailing spaces, and matches 10 digits (as in your example data) where there can be optional spaces or hyphens in between.
^ *\d(?:[ -]?\d){9} *$
For example
def valid_phone_number?(number)
phone_number_pattern = /^ *\d(?:[ -]*\d){9} *$/
return number.match?(phone_number_pattern)
end
See a Ruby demo and a regex demo.
To remove spaces & hyphen inbetween digits, try:
(?:\d+|\G(?!^)\d+)\K[- ]+(?=\d)
See an online regex demo
(?: - Open non-capture group;
d+ - Match 1+ digits;
| - Or;
\G(?!^)\d+ - Assert position at end of previous match but (negate start-line) with following 1+ digits;
)\K - Close non-capture group and reset matching point;
[- ]+ - Match 1+ space/hyphen;
(?=\d) - Assert position is followed by digits.
p " 09 777 55 888 ".gsub(/(?:\d+|\G(?!^)\d+)\K[- ]+(?=\d)/, '')
Prints: " 0977755888 "
Using a very simple regex (/\d/ tests for a digit):
str = " 09 777 55 888 "
r = str.index(/\d/)..str.rindex(/\d/)
str[r] = str[r].delete(" -")
p str # => " 0977755888 "
Passing a block to gsub is an option, capture groups available as globals:
>> str = " 09 777 55 888 "
# simple, easy to understand
>> str.gsub(/(^\s+)([\d\s-]+?)(\s+$)/){ "#$1#{$2.delete('- ')}#$3" }
=> " 0977755888 "
# a different take on #steenslag's answer, to avoid using range.
>> s = str.dup; s[/^\s+([\d\s-]+?)\s+$/, 1] = s.delete("- "); s
=> " 0977755888 "
Benchmark, not that it matters that much:
n = 1_000_000
puts(Benchmark.bmbm do |x|
# just a match
x.report("match") { n.times {str.match(/^ *\d(?:[ -]*\d){9} *$/) } }
# use regex in []=
x.report("[//]=") { n.times {s = str.dup; s[/^\s+([\d\s-]+?)\s+$/, 1] = s.delete("- "); s } }
# use range in []=
x.report("[..]=") { n.times {s = str.dup; r = s.index(/\d/)..s.rindex(/\d/); s[r] = s[r].delete(" -"); s } }
# block in gsub
x.report("block") { n.times {str.gsub(/(^\s+)([\d\s-]+?)(\s+$)/){ "#$1#{$2.delete('- ')}#$3" }} }
# long regex
x.report("regex") { n.times {str.gsub(/(?:\d+|\G(?!^)\d+)\K[- ]+(?=\d)/, "")} }
end)
Rehearsal -----------------------------------------
match 0.997458 0.000004 0.997462 ( 0.998003)
[//]= 1.822698 0.003983 1.826681 ( 1.827574)
[..]= 3.095630 0.007955 3.103585 ( 3.105489)
block 3.515401 0.003982 3.519383 ( 3.521392)
regex 4.761748 0.007967 4.769715 ( 4.772972)
------------------------------- total: 14.216826sec
user system total real
match 1.031670 0.000000 1.031670 ( 1.032347)
[//]= 1.859028 0.000000 1.859028 ( 1.860013)
[..]= 3.074159 0.003978 3.078137 ( 3.079825)
block 3.751532 0.011982 3.763514 ( 3.765673)
regex 4.634857 0.003972 4.638829 ( 4.641259)
I have been trying to change a string like this, {X=5, Y=9} to a string like this (5, 9), as it would be used as an on-screen coordinate.
I finally came up with this code:
Dim str As String = String.Empty
Dim regex As Regex = New Regex("\d+")
Dim m As Match = regex.Match("{X=9")
If m.Success Then str = m.Value
Dim s As Match = regex.Match("Y=5}")
If s.Success Then str = "(" & str & ", " & s.Value & ")"
MsgBox(str)
which does work, but surely there must be a better way to do this (I not familiar with Regex).
I have many to convert in my program, and doing it like above would be torturous.
You may use
Dim result As String = Regex.Replace(input, ".*?=(\d+).*?=(\d+).*", "($1, $2)")
The regex means
.*? - any 0+ chars other than newline chars as few as possible
= - an equals sign
(\d+) - Group 1: one or more digits
.*?= - any 0+ chars other than newline chars as few as possible and then a = char
(\d+) - Group 2: one or more digits
.* - any 0+ chars other than newline chars as many as possible
The $1 and $2 in the replacement pattern are replacement backreferences that point to the values stored in Group 1 and 2 memory buffer.
I need a Regular Expression to check whether a value contains any other characters than digits between 0 and 9.
I also want to check the length of the value.
The RegEx I´ve made: ^([0-9]\d{6})$
My test value is: 123Z45 and 123456
The ABAP code:
FIND ALL OCCURENCES OF REGEX '^([0-9]\d{6})$' IN L_VALUE RESULTS DATA(LT_RESULTS).
I´m expecting a result in LT_RESULTS, when I´m testing the first test value '123Z45', because there is a non-digit character.
But LT_RESULTS is in nearly every test case empty.
Your expression ^([0-9]\d{6})$ translates to:
^ - start of input
( - begin capture group
[0-9] - a character between 0 and 9
\d{6} - six digits (digit = character between 0 and 9)
) - end capture group
$ - end of input
So it will only match 1234567 (7 digit strings), not 123456, or 123Z45.
If you just need to find a string that contains non digits you could use the following instead: ^\d*[^\d]+\d*$
* - previous element may occur zero, one or more times
[^\d] - ^ right after [ means "NOT", i.e. any character which is not a digit
+ - previous element may occur one or more times
Example:
const expression = /^\d*[^\d]+\d*$/;
const inputs = ['123Z45', '123456', 'abc', 'a21345', '1234f', '142345'];
console.log(inputs.filter(i => expression.test(i)));
You can also use this character class if you want to extract non-digit group:
DATA(l_guid) = '0074162D8EAA549794A4EF38D9553990680B89A1'.
DATA(regx) = '[[:alpha:]]+'.
DATA(substr) = match( val = l_guid
regex = regx
occ = 1 ).
It finds a first occured non-digit group of characters and shows it.
If you want to just check if they are exists or how much of them reside in your string, count built-in function is your friend:
DATA(how_many) = count( val = l_guid regex = regx ).
DATA(yes) = boolc( count( val = l_guid regex = regx ) > 0 ).
Match and count exist since ABAP 7.50.
If you don't need a Regular Expression for something more complex, ABAP has some nice comparison operators CO (Contains Only), CA, NA etc for you. Something like:
IF L_VALUE CO '0123456789' AND STRLEN( L_VALUE ) = 6.
I'm trying to create a pattern for finding placeholders within a string to be able to replace them with variables later. I'm stuck on a problem to find all these placeholders within a string according to my requirement.
I already found this post, but it only helped a little:
Regex match ; but not \;
Placeholders will look like this
{&var} --> Variable stored in a dictionary --> dict("var")
{$prop} --> Property of a class cls.prop read by CallByName and PropGet
{#const} --> Some constant values by name from a function
Generally I have this pattern and it works well
Dim RegEx As Object
Set RegEx = CreateObject("VBScript.RegExp")
RegEx.pattern = "\{([#\$&])([\w\.]+)\}"
For example I have this string:
"Value of foo is '{&var}' and bar is '{$prop}'"
I get 2 matches as expected
(&)(var)
($)(prop)
I also want to add a formating part like in .Net to this expression.
String.Format("This is a date: {0:dd.mm.yyyy}", DateTime.Now());
// This is a date: 05.07.2019
String.Format("This is a date, too: {0:dd.(mm).yyyy}", DateTime.Now());
// This is a date, too: 05.(07).2019
I extended the RegEx to get that optional formatting string
Dim RegEx As Object
Set RegEx = CreateObject("VBScript.RegExp")
RegEx.pattern = "\{([#\$&])([\w\.]+):{0,1}([^\}]*)\}"
RegEx.Execute("Value of foo is '{&var:DD.MM.YYYY}' and bar is '{$prop}'")
I get 2 matches as expected
(&)(var)(DD.MM.YYYY)
($)(prop)()
At this point I noticed I have to take care for escapet "{" and "}", because maybe I want to have some brackets within the formattet result.
This does not work properly, because my pattern stops after "...{MM"
RegEx.Execute("Value of foo is '{&var:DD.{MM}.YYYY}' and bar is '{$prop}'")
It would be okay to add escape signs to the text before checking the regex:
RegEx.Execute("Value of foo is '{&var:DD.\{MM\}.YYYY}' and bar is '{$prop}'")
But how can I correctly add the negative lookbehind?
And second: How does this also works for variables, that should not be resolved, even if they have the correct syntax bus the outer bracket is escaped?
RegEx.Execute("This should not match '\{&var:DD.\{MM\}.YYYY\}' but this one '{&var:DD.\{MM\}.YYYY}'")
I hope my question is not confusing and someone can help me
Update 05.07.19 at 12:50
After the great help of #wiktor-stribiżew the result is completed.
As requested i provide some example code:
Sub testRegEx()
Debug.Print FillVariablesInText(Nothing, "Date\\\\{$var01:DD.\{MM\}.YYYY}\\\\ Var:\{$nomatch\}{$var02} Double: {#const}{$var01} rest of string")
End Sub
Function FillVariablesInText(ByRef dict As Dictionary, ByVal txt As String) As String
Const c_varPattern As String = "(?:(?:^|[^\\\n])(?:\\{2})*)\{([#&\$])([\w.]+)(?:\:([^}\\]*(?:\\.[^\}\\]*)*))?(?=\})"
Dim part As String
Dim snippets As New Collection
Dim allMatches, m
Dim i As Long, j As Long, x As Long, n As Long
' Create a RegEx object and execute pattern
Dim RegEx As Object
Set RegEx = CreateObject("VBScript.RegExp")
RegEx.pattern = c_varPattern
RegEx.MultiLine = True
RegEx.Global = True
Set allMatches = RegEx.Execute(txt)
' Start at position 1 of txt
j = 1
n = 0
For Each m In allMatches
n = n + 1
Debug.Print "(" & n & "):" & m.value
Debug.Print " [0] = " & m.SubMatches(0) ' Type [&$#]
Debug.Print " [1] = " & m.SubMatches(1) ' Name
Debug.Print " [2] = " & m.SubMatches(2) ' Format
part = "{" & m.SubMatches(0)
' Get offset for pre-match-string
x = 1 ' Index to Postion at least +1
Do While Mid(m.value, x, 2) <> part
x = x + 1
Loop
' Postition in txt
i = m.FirstIndex + x
' Anything to add to result?
If i <> j Then
snippets.Add Mid(txt, j, i - j)
End If
' Next start postition (not Index!) + 1 for lookahead-positive "}"
j = m.FirstIndex + m.Length + 2
' Here comes a function get a actual value
' e.g.: snippets.Add dict(m.SubMatches(1))
' or : snippets.Add Format(dict(m.SubMatches(1)), m.SubMatches(2))
snippets.Add "<<" & m.SubMatches(0) & m.SubMatches(1) & ">>"
Next m
' Any text at the end?
If j < Len(txt) Then
snippets.Add Mid(txt, j)
End If
' Join snippets
For i = 1 To snippets.Count
FillVariablesInText = FillVariablesInText & snippets(i)
Next
End Function
The function testRegEx gives me this result and debug print:
(1):e\\\\{$var01:DD.\{MM\}.YYYY(2):}{$var02
[0] = $
[1] = var02
[2] =
(1):e\\\\{$var01:DD.\{MM\}.YYYY
[0] = $
[1] = var01
[2] = DD.\{MM\}.YYYY
(2):}{$var02
[0] = $
[1] = var02
[2] =
(3): {#const
[0] = #
[1] = const
[2] =
(4):}{$var01
[0] = $
[1] = var01
[2] =
Date\\\\<<$var01>>\\\\ Var:\{$nomatch\}<<$var02>> Double: <<#const>><<$var01>> rest of string
You may use
((?:^|[^\\])(?:\\{2})*)\{([#$&])([\w.]+)(?::([^}\\]*(?:\\.[^}\\]*)*))?}
To make sure the consecutive matches are found, too, turn the last } into a lookahead, and when extracting matches just append it to the result, or if you need the indices increment the match length by 1:
((?:^|[^\\])(?:\\{2})*)\{([#$&])([\w.]+)(?::([^}\\]*(?:\\.[^}\\]*)*))?(?=})
^^^^^
See the regex demo and regex demo #2.
Details
((?:^|[^\\])(?:\\{2})*) - Group 1 (makes sure the { that comes next is not escaped): start of string or any char but \ followed with 0 or more double backslashes
\{ - a { char
([#$&]) - Group 2: any of the three chars
([\w.]+) - Group 3: 1 or more word or dot chars
(?::([^}\\]*(?:\\.[^}\\]*)*))? - an optional sequence of : and then Group 4:
[^}\\]* - 0 or more chars other than } and \
(?:\\.[^}\\]*)* - zero or more reptitions of a \-escaped char and then 0 or more chars other than } and \
} - a } char
Welcome to the site! If you need to only match balanced escapes, you will need something more powerful. If not --- I haven't tested this, but you could try replacing [^\}]* with [^\{\}]|\\\{|\\\}. That is, match non-braces and escaped brace sequences separately. You may need to change this depending on how you want to handle backslashes in your formatting string.
VBScript's Trim function only trims spaces. Sometimes I want to trim TABs as well. For this I've been using this custom trimSpTab function that is based on a regular expression.
Today I ran into a performance problem. The input consisted of rather long lines (several 1000 chars).
As it turns out
- the function is slow, only if the string is long AND contains many spaces
- the right-hand part of the regular expression is reponsible for the poor performance
- the run time seems quadratic to the line length (O(n^2))
So why is this line trimmed fast
" aaa xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx bbb " '10000 x's
and this one trimmed slowly
" aaa bbb " '10000 spaces
Both contain only 6 characters to be trimmed.
Can you propose a modification to my trimSpTab function?
Dim regex
Set regex = new regexp
' TEST 1 - executes in no time
' " aaa XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX bbb "
t1 = Timer
character = "X"
trimTest character
MsgBox Timer-t1 & " sec",, "with '" & character & "' in the center of the string"
' TEST 2 - executes in 1 second on my machine
' " aaa bbb "
t1 = Timer
character = " "
trimTest character
MsgBox Timer-t1 & " sec",, "with '" & character & "' in the center of the string"
Sub trimTest (character)
sInput = " aaa " & String (10000, character) & " bbb "
trimmed = trimSpTab (sInput)
End Sub
Function trimSpTab (byval s)
'trims spaces & tabs
regex.Global = True
regex.Pattern = "^[ \t]+|[ \t]+$" 'trim left+right
trimSpTab = regex.Replace (s, "")
End Function
I have tried this (with regex.Global = false) but to no avail
regex.Pattern = "^[ \t]+" 'trim left
s = regex.Replace (s, "")
regex.Pattern = "[ \t]+$" 'trim right
trimSpTab = regex.Replace (s, "")
UPDATE
I've come up with this alternative in the mean time. It processes a 100 million character string is less than a second.
Function trimSpTab (byval s)
'trims spaces & tabs
regex.Pattern = "^[ \t]+"
s = strReverse (s)
s = regex.Replace (s, "")
s = strReverse (s)
s = regex.Replace (s, "")
trimSpTab = s
End Function
Solution
As mentioned in the question, your current solution is to reverse the string. However, this is not necessary, since .NET regex supports RightToLeft matching option. For the same regex, the engine will start matching from right to left instead of default behavior of matching from left to right.
Below is sample code in C#, which I hope you can adapt to VB solution (I don't know VB enough to write sample code):
input = new Regex("^[ \t]+").Replace(input, "", 1)
input = new Regex("[ \t]+$", RegexOptions.RightToLeft).Replace(input, "", 1)
Explanation
The long run time is due to the engine just trying to match [ \t]+ indiscriminately in the middle of the string and end up failing when it is not an trailing blank sequence.
The observation that the complexity is quadratic is correct.
We know that the regex engine starts matching from index 0. If there is a match, then the next attempt starts at the end of the last match. Otherwise, the next attempt starts at the (current index + 1). (Well, to simplify things, I don't mention the case where a zero-length match is found).
Below shall illustrate the farthest attempt (some is a match, some are not) of the engine matching the regex ^[ \t]+|[ \t]+$. _ is used to denote space (or tab character) for clarity.
_____ab_______________g________
^----
^
^
^--------------
^-------------
^------------
...
^
^
^-------
When there is a long sequence of spaces & tabs in the middle of the string (which will not produce a match), the engine attempts matching at every index in the long sequence of spaces & tabs. As the result, the engine ends up going through O(k2) characters on a non-matching sequence of spaces & tabs of length k.
Your evidence proves that VBScript's RegExp implementation does not optimize for the $ anchor: It spends time (backtracking?) for each of the spaces in the middle of your test string. Without doubt, that's a fact good to know.
If this causes you real world problems, you'll have to find/write a better (R)Trim function. I came up with:
Function trimString(s, p)
Dim l : l = Len(s)
If 0 = l Then
trimString = s
Exit Function
End If
Dim ps, pe
For ps = 1 To l
If 0 = Instr(p, Mid(s, ps, 1)) Then
Exit For
End If
Next
For pe = l To ps Step -1
If 0 = Instr(p, Mid(s, pe, 1)) Then
Exit For
End If
Next
trimString = Mid(s, ps, pe - ps + 1)
End Function
It surely needs testing and benchmarks for long heads or tails of white space, but I hope it gets you started.