special character set: `~!##$%^&*()_-+={}[]\|:;""'<>,.?/
Is this the right way to search for items within that special character set?
Regex.IsMatch(Result.Text, "^[`-/]")
I always get an error.... Do I need to use ASCII codes? If so, how can I do this? Thank you in advance!
Regular expressions have a different escape syntax and rules than VB.NET. Since you're dealing with a regex string in your code, you have to make sure the string is escaped properly for regex and VB.NET.
In your example, the - needs to be escaped with a ...
Regex.IsMatch(Result.Text, "^[`\-/]")
To match any character in the provided string, try this...
Regex.IsMatch(Result.Text, "[`~!##\$%\^&\*\(\)_\-\+=\{\}\[\]\\\|:;""'<>,\.\?/]")
Try this:
[`~!##$%^&*()_+={}\[\]\\|:;""'<>,.?/-]
Working regex example:
http://regex101.com/r/aU3wO9
In a char set, only some of them need to be escaped. like ] and \ and if - is at the end it doesn't need to be escaped. I'm not sure about [ so I just escaped it anyway.
You can use an escape character in the string where there is a special character
Dim str() As Byte
Dim j, n As Long
Dim ContainsSpecialChar As Boolean
Dim TempVendorName As String
str = Trim(VendorName)
n = 1
For j = 0 To UBound(str) - 1 Step 2
If (str(j) > 32 And str(j) < 47) Or (str(j) > 57 And str(j) < 65) Or (str(j) > 90 And str(j) < 97) Or (str(j) > 122) Then
ContainsSpecialChar = True
TempVendorName = Left(VendorName, n - 1) + "\" + Mid(VendorName, n)
n = n + 1
End If
n = n + 1
Next
Related
I have a SQL sintax in the form of a string in which there are some parameters written in a standard way "<parameter1>, <parameter2>".
Then i have another string with the parameters values written in a standard way as well: "Parameter1=123; Parameter2=aaa".
I need to match the parameters in the first SQL with the values in the second one.
What I have so far:`
Dim BodySQL = "Blablabal WHERE X=<Parameter1> AND Y=<Parameter2>"
Dim vmp As RegularExpressions.MatchCollection = RegularExpressions.Regex.Matches("Parameter1=2555; Parameter2 = 12/02/2021", "([\w ]+)=([\w ]+)")
Dim vmc As RegularExpressions.MatchCollection = RegularExpressions.Regex.Matches(BodySQL, "(?<=\<).+?(?=\>)")
For Each vm As RegularExpressions.Match In vmc
Dim Vl As String = (From m As RegularExpressions.Match In vmp
Where m.Groups(1).Value.Trim = vm.Value.ToString).Select(Of String)(Function(f) f.Groups(2).Value).ElementAt(0).Trim
BodySQL = BodySQL.Replace(vm.Value, Vl)
Next
It works for the first parameter, but then i get
"System.ArgumentOutOfRangeException: 'Specified argument was out of the range of valid values.
Parameter name: index'"
Can I please ask why?
You can extract the keys and values with one regex from the param=value strings, create a dictionary out of them, and then use Regex.Replace to replace the matches with the dictionary values:
Imports System.Text.RegularExpressions
' ...
Dim BodySQL = "Blablabal WHERE X=<Parameter1> AND Y=<Parameter2>"
Dim args As New Dictionary(Of String, String)(StringComparer.InvariantCultureIgnoreCase)
' (StringComparer.InvariantCultureIgnoreCase) makes the dictionary keys case insensitive
For Each match As Match In Regex.Matches("PARAMETER1=2555; Parameter2 = 12/02/2021", "(\S+)\s*=\s*([^;]*[^;\s])")
args.Add(match.Groups(1).Value, match.Groups(2).Value)
Next
Console.WriteLine(Regex.Replace(BodySQL, "<([^<>]*)>",
Function(match)
Return If(args.ContainsKey(match.Groups(1).Value), args(match.Groups(1).Value), match.Value)
End Function))
Output:
Blablabal WHERE X=2555 AND Y=12/02/2021
The (\S+)\s*=\s*([^;]*[^;\s]) pattern matches
(\S+) - captures into Group 1 any one or more non-whitespace chars (the key value)
\s*=\s* - a = char enclosed with zero or more whitespace chars
([^;]*[^;\s]) - Group 2: any zero or more chars other than ; and then one char other than ; and whitespace (the value, it cannot be empty with this pattern. If you want it to be possible to match empty values, you will need to remove [^;\s] and then use Trim() on the match.Groups(2).Value in the code.)
The <([^<>]*)> regex matches
< - a < char (do not escape this char in any regex flavor please, it is never a special regex metachar)
([^<>]*) - Group 1: any zero or more chars other than < and >
> - a literal > char.
Since the key is in Group 1 and < and > on both ends are consumed, the < and > are removed when replacing with the found value.
zero or more chars other than > and < between < and >.
The error is self explanatory. You are trying to access an array or List and specifying an index value that is either negative or larger than the largest index available.
.ElementAt(0) / m.Groups(1) / f.Groups(2)
My guess is that one of them might go out of bounds. Try to debug it with a breakpoint and check the values of your variables.
This is what i did with your code:
Dim vmc = "\<(.*?)\>"
i changed this regex so that it could also give me the "<>"
Dim BodySQL = "Blablabal WHERE X=<Parameter1> AND Y=<Parameter2>"
Dim args As New Dictionary(Of String, String)
For Each match As Match In Regex.Matches("Parameter1=2555; Parameter2 = 12/02/2021", "(\S+)\s*=\s*(\S[^;]+)")
i changed the regex expression to exclude the ";"
args.Add(match.Groups(1).Value, match.Groups(2).Value)
Next
Console.WriteLine(Regex.Replace(BodySQL, vmc,
Function(match As Match)
Return If(args.ContainsKey(match.Groups(1).Value), args(match.Groups(1).Value), match.Value)
End Function))
And now i have what i needed. Thank you a lot :)
Output:
WHERE X = 2555,Y = 12/02/2021
I am trying to match square brackets i.e. [] in regex VBA in excel. I am trying with the below code but it is not working.
Public Function IsSpecial(s As String) As Long
Dim L As Long, LL As Long
Dim sCh As String
IsSpecial = 0
For L = 1 To Len(s)
sCh = Mid(s, L, 1)
If sCh Like "[0-9a-zA-Z/;#%,'‚.+&/\(): ]" Or sCh = "_" Or sCh Like "[-]" Or sCh Like "\[" Then
Else
IsSpecial = 1
Exit Function
End If
Next L
End Function
According to Using the Like operator and wildcard characters in string comparisons:
You can use a group of one or more characters (charlist) enclosed in brackets ([ ]) to match any single character in expression, and charlist can include almost any characters in the ANSI character set, including digits. You can use the special characters opening bracket ([), question mark (?), number sign (#), and asterisk (*) to match themselves directly only if enclosed in brackets. You cannot use the closing bracket (]) within a group to match itself, but you can use it outside a group as an individual character.
So, you need to use
Ch Like "[[]"
However, the function you have is not following your logic, since it checks each char individually, and you want to make sure [] is checked as a char sequence.
With a regex, it will look like
Public Function IsSpecial(s As String) As Long
Dim L As Long, LL As Long
Dim rx As New regExp
rx.Pattern = "^(?:[0-9a-zA-Z/;#%,'‚.+&/\\(): _-]|\[])*$"
IsSpecial = 0
If Not rx.Test(s) Then IsSpecial = 1
End Function
I have thousands of lines of text that I need to work through and the lines I am interested with lines that look like the following:
01/04/2019 09:35:41 - Test user (Additional Comments)
I am currently using this code to filter out all the other rows:
If InStr(FullCell(i), " - ") <> 0 And InStr(FullCell(i), ":") <> 0 And InStr(FullCell(i), "(") <> 0 Then
FullCell is the array that I am working through.
which I know is not the best way to do it. Is there a way to check that there is a date at the beginning of the string in the format dd/mm/yyyy and then extract the user name inbetween the '-' and the '(' symbol.
I had a play with regex to see if that could help but i'm limited in skills to be able to pull off both VBA and regex in the same code.
Whats the best way to do this.
Assuming Fullcell(i) contains the string,
If Left(Fullcell(i), 10) Like "##/##/####"
Will return True if you have a date (note that it will not differentiate between dd/mm/yyyy and mm/dd/yyyy.
And
Mid(Fullcell(i), InStr(Fullcell(i), " - ") + 2, InStr(Fullcell(i), " (") - InStr(Fullcell(i), " - ") - 2)
Will return the username
I'm sure there is a more efficient way to do this, but I've used the following solution quite a few times:
This will select the date:
x = 1
Do While Mid(FullCell,1,x) <> " "
x = x + 1
Loop
strDate = Left(FullCell,x)
This will find the character number of the hyphen, the username starts 2 characters after.
x = 1
Do While Mid(FullCell,x,1) <> "-"
x = x + 1
Loop
Then we will find the end of the username
y = x + 2
Do While Mid(FullCell,y,1) <> " "
y = y + 1
Loop
The username should now be characters (x+2 to y-1)
strUsername = Mid(FullCell, x + 2, y - (x + 2) - 1)
Here's how I would do it
Dim your variables
Dim ring as Range
Dim dat as variant
Dim FullCell() as string
Dim User as string
Dim I as long
Set your range
Set rng = ` any way you choose
Dat = rng.value2
Loop dat
For i = 1 to UBound(dat, 1)
Split the data
FullCell = Trim(Split(FullCell, "-"))
Test if it split
If UBound(FullCell) > 0 Then
Test if it matches
If IsDate(FullCell(0)) Then
i = Instr(FullCell(1), "(")-1)
If i then
User = left$(FullCell(1), i)
' Found a user
End If
End If
End If
Next
Abstraction is your friend, it's always helpful to break these into their own private functions whenever you can. You could put your code in a function and call it something like ExtractUsername.
Below I did an example of this, and I decided to go with the RegExp approach (late binding), but you could use string functions like the examples above as well.
This function returns the username if it finds the pattern you mentioned above, otherwise, it returns an empty string.
Private Function ExtractUsername(ByVal SourceString As String) As String
Dim RegEx As Object
Set RegEx = CreateObject("vbscript.regexp")
'(FIRST GROUP FINDS THE DATE FORMATTED AS DD/MM/YYY, AS WELL AS THE FORWARD SLASH)
'(SECOND GROUP FINDS THE USERNAME) THIS WILL BE SUBMATCH 1
With RegEx
.Pattern = "(^\d{2}\/\d{2}\/\d{4}.*-)(.+)(\()"
.Global = True
End With
Dim Match As Object
Set Match = RegEx.Execute(SourceString)
'ONLY RETURN IF A MATCH WAS FOUND
If Match.Count > 0 Then
ExtractUsername = Trim(Match(0).SubMatches(1))
End If
Set RegEx = Nothing
End Function
The regex pattern is grouped into three parts, the date (and slash), username, and opening parentheses. What you are interested in is the username, which in the SubMatch would be number 1.
Regexr is a helpful site for practicing regular expressions and can show you a bit more of what the pattern I went with is doing.
Please note that using regular expressions might give you performance issues and you should test it against regular string functions to see what works best for your situation.
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 am trying to extract n 3-tuples (Si, Pi, Vi) from a string.
The string contains at least one such 3-tuple.
Pi and Vi are not mandatory.
SomeTextxyz#S1((property(P1)val(V1))#S2((property(P2)val(V2))#S3
|----------1-------------|----------2-------------|-- n
The desired output would be:
Si,Pi,Vi.
So for n occurrences in the string the output should look like this:
[S1,P1,V1] [S2,P2,V2] ... [Sn-1,Pn-1,Vn-1] (without the brackets)
Example
The input string could be something like this:
MyCarGarage#Mustang((property(PS)val(500))#Porsche((property(PS)val(425)).
Once processed the output should be:
Mustang,PS,500 Porsche,PS,425
Is there an efficient way to extract those 3-tuples using a regular expression
(e.g. using C++ and std::regex) and what would it look like?
#(.*?)\(\(property\((.*?)\)val\((.*?)\)\) should do the trick.
example at http://regex101.com/r/bD1rY2
# # Matches the # symbol
(.*?) # Captures everything until it encounters the next part (ungreedy wildcard)
\(\(property\( # Matches the string "((property(" the backslashes escape the parenthesis
(.*?) # Same as the one above
\)val\( # Matches the string ")val("
(.*?) # Same as the one above
\)\) # Matches the string "))"
How you should implement this in C++ i don't know but that is the easy part :)
http://ideone.com/S7UQpA
I used C's <regex.h> instead of std::regex because std::regex isn't implemented in g++ (which is what IDEONE uses). The regular expression I used:
" In C(++)? regexes are strings.
# Literal match
([^(#]+) As many non-#, non-( characters as possible. This is group 1
( Start another group (group 2)
\\(\\(property\\( Yet more literal matching
([^)]+) As many non-) characters as possible. Group 3.
\\)val\\( Literal again
([^)]+) As many non-) characters as possible. Group 4.
\\)\\) Literal parentheses
) Close group 2
? Group 2 optional
" Close Regex
And some c++:
int getMatches(char* haystack, item** items){
first, calculate the length of the string (we'll use that later) and the number of # found in the string (the maximum number of matches)
int l = -1, ats = 0;
while (haystack[++l])
if (haystack[l] == '#')
ats++;
malloc a large enough array.
*items = (item*) malloc(ats * sizeof(item));
item* arr = *items;
Make a regex needle to find. REGEX is #defined elsewhere.
regex_t needle;
regcomp(&needle, REGEX, REG_ICASE|REG_EXTENDED);
regmatch_t match[5];
ret will hold the return value (0 for "found a match", but there are other errors you may want to be catching here). x will be used to count the found matches.
int ret;
int x = -1;
Loop over matches (ret will be zero if a match is found).
while (!(ret = regexec(&needle, haystack, 5, match,0))){
++x;
Get the name from match1
int bufsize = match[1].rm_eo-match[1].rm_so + 1;
arr[x].name = (char *) malloc(bufsize);
strncpy(arr[x].name, &(haystack[match[1].rm_so]), bufsize - 1);
arr[x].name[bufsize-1]=0x0;
Check to make sure the property (match[3]) and the value (match[4]) were found.
if (!(match[3].rm_so > l || match[3].rm_so<0 || match[3].rm_eo > l || match[3].rm_so< 0
|| match[4].rm_so > l || match[4].rm_so<0 || match[4].rm_eo > l || match[4].rm_so< 0)){
Get the property from match[3].
bufsize = match[3].rm_eo-match[3].rm_so + 1;
arr[x].property = (char *) malloc(bufsize);
strncpy(arr[x].property, &(haystack[match[3].rm_so]), bufsize - 1);
arr[x].property[bufsize-1]=0x0;
Get the value from match[4].
bufsize = match[4].rm_eo-match[4].rm_so + 1;
arr[x].value = (char *) malloc(bufsize);\
strncpy(arr[x].value, &(haystack[match[4].rm_so]), bufsize - 1);
arr[x].value[bufsize-1]=0x0;
} else {
Otherwise, set both property and value to NULL.
arr[x].property = NULL;
arr[x].value = NULL;
}
Move the haystack to past the match and decrement the known length.
haystack = &(haystack[match[0].rm_eo]);
l -= match[0].rm_eo;
}
Return the number of matches.
return x+1;
}
Hope this helps. Though it occurs to me now that you never answered kind of a vital question: What have you tried?