I have these requirements to follow:
Windows computer name cannot be more than 15 characters long, be
entirely numeric, or contain the following characters: ` ~ ! # # $ % ^
& * ( ) = + _ [ ] { } \ | ; : . ' " , < > / ?.
I want to create a RegEx to validate a given computer name.
I can see that the only permitted character is - and so far I have this:
/^[a-zA-Z0-9-]{1,15}$/
which matches almost all constraints except the "not entirely numeric" part.
How to add last constraints to my RegEx?
You could use a negative lookahead:
^(?![0-9]{1,15}$)[a-zA-Z0-9-]{1,15}$
Or simply use two regular expressions:
^[a-zA-Z0-9-]{1,15}$
AND NOT
^[0-9]{1,15}$;
Here is a live example:
var regex1 = /^(?![0-9]{1,15}$)[a-zA-Z0-9-]{1,15}$/;
var regex2 = /^[a-zA-Z0-9-]{1,15}$/;
var regex3 = /^[0-9]{1,15}$/;
var text1 = "lklndlsdsvlk323";
var text2 = "4214124";
console.log(text1 + ":", !!text1.match(regex1));
console.log(text1 + ":", text1.match(regex2) && !text1.match(regex3));
console.log(text2 + ":", !!text2.match(regex1));
console.log(text2 + ":", text2.match(regex2) && !text2.match(regex3));
Related
I'm trying to create a pattern for finding placeholders within a string to be able to replace them with variables later. I'm stuck on a problem to find all these placeholders within a string according to my requirement.
I already found this post, but it only helped a little:
Regex match ; but not \;
Placeholders will look like this
{&var} --> Variable stored in a dictionary --> dict("var")
{$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.
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
I have a project with Delphi and in that project I have to get a version info from a PHP file. In the PHP file there is a line that like below:
$app_version = 'X.X';
And while I am trying to get X.X by using
RegularExpression.Create('app_version.*''(?<appVersion>.*)''',[roIgnoreCase,roMultiline]);
I am getting
app_version = 'X.X'
without dollar $ and semicolon ; signs - that's okay. But I actually only need need the
X.X
part. So what regex rule could I use to only get version number part?
Just use the following regular expression:
\$app_version\s*=\s*["']([^"']*)["'];
This will match the whole expression on the line. The first group $1 will contain the version number.
\$ → escapes the dollar sign, otherwise it would be "end of string"
\s* → 0 or more whitespace characters
["'] → either " or '
[^"']* → 0 or more characters which are unequal to " and '
( anything ) → matches a group to extract from a query (referred to as $1, $2, etc.)
Here is an example in JavaScript:
var regex = /\$app_version\s*=\s*["']([^"']*)["']/g;
var text = document.getElementById('main').innerHTML;
var match = regex.exec(text);
while(match !== null) {
console.log(match[1]);
match = regex.exec(text);
}
#main {
white-space: pre;
font-family: monospace;
}
<div id="main">
$app_version = '10.2';
$app_version = "11.3";
$ap_version = "10.30;
$app_version = 9.30;
</div>
In Delphi (source):
regexpr := TRegEx.Create('\$app_version\s*=\s*["\']([^"\']*)["\'];',[roIgnoreCase,roMultiline]);
match := regexpr.Match(searchMe);
if not match.Success then
begin
WriteLn('No Match Found');
exit;
end;
while match.Success do
begin
WriteLn('Match : [' + match.Value + ']';
//group 0 is the entire match, so count will always be at least 1 for a match
if match.Groups.Count > 1 then
begin
for i := 1 to match.Groups.Count -1 do
WriteLn(' Group[' + IntToStr(i) + '] : [' + match.Value + ']';
end;
match := match.NextMatch;
end;
The code in Delphi is untested.
I need to use regexp for matching and the code below works fine. However, I need to KEEP the dollar sign ($) as a true dollar sign and not a special character.
I've tried excluding but nothing is working.
IE: [^$]
Here's the code. It works as expected except when the text contains a $ or IS the $.
textNode = "$19,000";
regex = RegExp("$19,000",'ig');
text = '$';
textReplacerFunc: function (textNode, regex, text) {
var sTag = '<span class="highlight">';
var eTag = '</span>';
var re = '(?![^<>]*>)(' + text + '(?!#8212;))';
var regExp = new RegExp(re, 'ig');
textNode.data = textNode.data.replace(regExp, sTag + '$1' + eTag);
},
RESULT: $ not highlighted. desired results:
$19,000
Make sure to double escape the $ as in :
text = '\\$';
Since you are using construction of RegExp instance using a string here.
I have a bunch of strings with punctuation in them that I'd like to convert to spaces:
"This is a string. In addition, this is a string (with one more)."
would become:
"This is a string In addition this is a string with one more "
I can go thru and do this manually with the stringr package (str_replace_all()) one punctuation symbol at a time (, / . / ! / ( / ) / etc. ), but I'm curious if there's a faster way I'd assume using regex's.
Any suggestions?
x <- "This is a string. In addition, this is a string (with one more)."
gsub("[[:punct:]]", " ", x)
[1] "This is a string In addition this is a string with one more "
See ?gsub for doing quick substitutions like this, and ?regex for details on the [[:punct:]] class, i.e.
‘[:punct:]’ Punctuation characters:
‘! " # $ % & ' ( ) * + , - . / : ; < = > ? # [ \ ] ^ _ ` { |
} ~’.
have a look at ?regex
library(stringr)
str_replace_all(x, '[[:punct:]]',' ')
"This is a string In addition this is a string with one more "