I would like to remove all comments from a PHP source file from within a VB.NET application. Another stackoverflow question showed how to do this in C# code
I came up with this conversion, but it does not work unfortunately:
Dim blockComments As String = "/\*(.*?)\*/"
Dim lineComments As String = "//(.*?)\r?\n"
Dim strings As String = """((\\[^\n]|[^""\n])*)"""
Dim verbatimStrings As String = "#(""[^""]*"")+"
regex = New Regex(blockComments & "|" & lineComments)
srcT = regex.Replace(srcT, "")
You need to pass the flag RegexOptions.Singleline when constructing the Regex object. Otherwise, the block-comments can't span multiple lines.
regex = New Regex(blockComments & "|" & lineComments, RegexOptions.Singleline)
The . normally matches any character except newline (\n). The RegexOptions.Singleline flag makes it match any character, including newline.
Related
I am trying to use regex to 'extract' paragraphs in a document. Each paragraph is preceded and followed by a '-' on separate line and each paragraph starts with a number.
For example
-
1. This is a paragraph
It may go over multiple lines
-
Ideally, I would like to not include the '-', but it doesn't really matter as I will be placing it in a string and running another regex against it (One that I know works)
The code I am trying to use is basically as follows
Dim matchPara as Object
Dim regex as Object
Dim theMatch as Object
Dim matches as Object
Dim fileName as String
Dim fileNo as Integer
Dim document as String
matchPara = "-?(\d.*?)?-"
Set regex = CreateObject("VBScript.RegExp")
regex.Pattern = matchPara
regex.Global = True
regex.Multiline = True
fileName = "C:\file.txt"
fileNo = FreeFile
Open fileName For Input As #fileNo
document = Input$(LOF(fileNo), fileNo)
set matches = regex.Execute(document)
For Each theMatch in matches
MsgBox(theMatch.Value)
Next theMatch
Close #fileNo
I have tested this regex on regex101 and it appeared to do what I wanted. I have also tested it without the grouping
-?\d.*?-
However when I run the code the theMatch.Value only ever contains a single '-'. After some messing around with the regex I got it to display the first line of text but never any more than the first line.
I have checked the length of theMatch.Value with:
MsgBox(len(theMatch.Value))
and placed the contents of theMatch.Value in a cell on the worksheet to see if It was cutting off in the message box, but both theories were proved wrong.
I am at a complete loss now and I am beginning to suspect it is possibly a VBA thing and not a regex thing. There is no requirement to use regex, I just assumed it would be the easiest thing to do.
The paragraphs contain data that I am trying to extract. So the idea was to regex each paragraph out place that in a string then run other regex to get the information that I need. Some paragraphs wont contain the data that I need so the idea was to loop through each individual paragraph and then error handle better if the data I need wasn't in that paragraph (ie get what I can and drop the rest with an error message)
Here is a screenshot:
This simple approach does not use Regex. It assumes the data is in column A and the paragraphs are placed in column B:
Sub paragraph_no_regex()
Dim s As String
Dim ary
With Application.WorksheetFunction
s = .TextJoin(" ", False, Columns(1).SpecialCells(2))
End With
ary = Split(s, "-")
i = 1
For Each a In ary
Cells(i, 2) = a
i = i + 1
Next a
End Sub
Sub F()
Dim re As New RegExp
Dim sMatch As String
Dim document As String
re.Pattern = "-\n((.|\n)+?)\n-"
'Getting document
document = ...
sMatch = re.Execute(document)(0).SubMatches(0)
End Sub
If you need dashes -, then just include them into capture group (the outer parenthesis).
This RegEx matches your description and successfully extracts paragraphs (as tested on regex101.com):
matchPara = "-\n\d+\.\s*((?:.|\n)+?)\s*\n-"
It needs the 'global' flag but not the 'multiline' flag. Instead, the end-of-line token is matched in the regex. The main point is that the innermost matching group will match any character including the end-of-line (given as an alternative) but does so in a non-greedy way ("+?"). It doesn't care about word boundaries as this is not necessary here. Also, "-" is not a special character where used in the regex so it doesn't have to be escaped.
As added benefit leading and trailing whitespace is cut off ("\s*" outside the group).
script file content is
//{input: x(width),y(height);
//output: z(area);}
function(x,y)
z=x*y
I have to read only these lines. What will be the regular expression for data that is in curly braces
//{input: x(width),y(height);
//output: z(area);}
I tried the following
Dim sr As StreamReader = New StreamReader(scriptpath)
' Dim textToParse As String
Dim scriptText As String
scriptText = sr.ReadToEnd
Dim extractCommentRegex As New Regex("\/\/\{(.*?)\}")
Dim textToParse As Match = extractCommentRegex.Match(scriptText)
Try this
^\/\/\{.*\}
with /m option to make dot match newlines.
Hi, sorry haven't been writing vb for quite a while so did not make the answer clear enough. I've created a console project to test following code:
Dim sr As StreamReader = New StreamReader("d:\script1.txt")
' Dim textToParse As String
Dim scriptText As String
scriptText = sr.ReadToEnd
Dim match = Regex.Match(scriptText, "^\/\/\{.*\}", RegexOptions.Singleline Or RegexOptions.Multiline Or RegexOptions.IgnorePatternWhitespace)
Console.WriteLine(match.Success)
Dim sw As StreamWriter = New StreamWriter("d:\output.txt")
sw.Write(match.Value)
sw.Flush()
sw.Close()
Console.ReadLine()
And I'll get following for output.txt.
//{input: x(width),y(height);
//output: z(area);}
I think you need to provide RegexOptions If you have Windows format LF for the input file. for detail of the issue please see this thread:
.NET Regex dot character matches carriage return?
I am doing this task as part of a larger sub in order to massively reduce the workload for a different team.
I am trying to read in a string and use Regular Expressions to replace one-to-many spaces with a single space (or another character). At the moment I am using a local string, however in the main sub this data will come from an external .txt file. The number of spaces between elements in this .txt can vary depeneding on the row.
I am using the below code, and replacing the spaces with a dash. I have tried different variations and different logic on the below code, but always get "Run-time error '91': Object Variable or with clock variable not set" on line "c = re.Replace(s, replacement)"
After using breakpoints, I have found out that my RegularExpression (re) is empty, but I can't quite figure out how to progress from here. How do I replace my spaces with dashes? I have been at this problem for hours and spent most of that time on Google to see if someone has had a similar issue.
Sub testWC()
Dim s As String
Dim c As String
Dim re As RegExp
s = "hello World"
Dim pattern As String
pattern = "\s+"
Dim replacement As String
replacement = "-"
c = re.Replace(s, replacement)
Debug.Print (c)
End Sub
Extra information: Using Excel 2010. Have successfully linked all my references (Microsoft VBScript Regular Expressions 5.5". I was sucessfully able to replace the spaces using the vanilla "Replace" function, however as the number of spaces between elements vary I am unable to use that to solve my issue.
Ed: My .txt file is not fixed either, there are a number of rows that are different lengths so I am unable to use the MID function in excel to dissect the string either
Please help
Thanks,
J.H.
You're not setting up the RegExp object correctly.
Dim pattern As String
pattern = "\s+" ' pattern is just a local string, not bound to the RegExp object!
You need to do this:
Dim re As RegExp
Set re = New RegExp
re.Pattern = "\s+" ' Now the pattern is bound to the RegExp object
re.Global = True ' Assuming you want to replace *all* matches
s = "hello World"
Dim replacement As String
replacement = "-"
c = re.Replace(s, replacement)
Try setting the pattern inside your Regex object. Right now, re is just a regex with no real pattern assigned to it. Try adding in re.Pattern = pattern after you initialize your pattern string.
You initialized the pattern but didn't actually hook it into the Regex. When you ended up calling replace it didn't know what it was looking for pattern wise, and threw the error.
Try also setting the re as a New RegExp.
Sub testWC()
Dim s As String
Dim c As String
Dim re As RegExp
Set re = New RegExp
s = "hello World"
Dim pattern As String
pattern = "\s+"
re.Pattern = pattern
Dim replacement As String
replacement = "-"
c = re.Replace(s, replacement)
Debug.Print (c)
End Sub
I have a list of original words and replace with words which I want to replace occurrence of the original words in some sentences to the replace words.
For example my list:
theabove the above
myaddress my address
So the sentence "This is theabove." will become "This is the above."
I am using Regular Expression in VB like this:
Dim strPattern As String
Dim regex As New RegExp
regex.Global = True
If Not IsEmpty(myReplacementList) Then
For intRow = 0 To UBound(myReplacementList, 2)
strReplaceWith = IIf(IsNull(myReplacementList(COL_REPLACEMENTWORD, intRow)), " ", varReplacements(COL_REPLACEMENTWORD, intRow))
strPattern = "\b" & myReplacementList(COL_ORIGINALWORD, intRow) & "\b"
regex.Pattern = strPattern
TextToCleanUp = regex.Replace(TextToReplace, strReplaceWith)
Next
End If
I loop all entries in my list myReplacementList against the text TextToReplace I want to process, and the replacement have to be whole word so I used the "\b" token around the original word.
It works well but I have a problem when the original words contain some special characters for example
overla) overlay
I try to escape the ) in the pattern but it does not work:
\boverla\)\\b
I can't replace the sentence "This word is overla) with that word." to "This word is overlay with that word."
Not sure what is missing? Is regular expression the way to the above scenario?
I'd use string.replace().
That way you don't have to escape special chars .. only these: ""!
See here for examples: http://www.dotnetperls.com/replace-vbnet
Regex is good if your looking for patterns. Or renaming your mp3 collection ;-) and much, much more. But in your case, I'd use string.replace().
I'm trying to parse csv file with VB.NET.
csv files contains value like 0,"1,2,3",4 which splits in 5 instead of 3. There are many examples with other languages in Stockoverflow but I can't implement it in VB.NET.
Here is my code so far but it doesn't work...
Dim t As String() = Regex.Split(str(i), ",(?=([^\""]*\""[^\""]*\"")*[^\""]*$)")
Assuming your csv is well-formed (ie no " besides those used to delimit string fields, or besides ones escaped like \"), you can split on a comma that's followed by an even number of non-escaped "-marks. (If you're inside a set of "" there's only an odd number left in the line).
Your regex you've tried looks like you're almost there.
The following looks for a comma followed by an even number of any sort of quote marks:
,(?=([^"]*"[^"]*")*[^"]*$)
To modify it to look for an even number of non-escaped quote marks (assuming quote marks are escaped with backslash like \"), I replace each [^"] with ([^"\\]|\\.). This means "match a character that isn't a " and isn't a blackslash, OR match a backslash and the character immediately following it".
,(?=(([^"\\]|\\.)*"([^"\\]|\\.)*")*([^"\\]|\\.)*$)
See it in action here.
(The reason the backslash is doubled is I want to match a literal backslash).
Now to get it into vb.net you just need to double all your quote marks:
splitRegex = ",(?=(([^""\\]|\\.)*""([^""\\]|\\.)*"")*([^""\\]|\\.)*$)"
Instead of a regular expression, try using the TextFieldParser class for reading .csv files. It handles your situation exactly.
TextFieldParserClass
Especially look at the HasFieldsEnclosedInQuotes property.
Example:
Note: I used a string instead of a file, but the result would be the same.
Dim theString As String = "1,""2,3,4"",5"
Using rdr As New StringReader(theString)
Using parser As New TextFieldParser(rdr)
parser.TextFieldType = FieldType.Delimited
parser.Delimiters = New String() {","}
parser.HasFieldsEnclosedInQuotes = True
Dim fields() As String = parser.ReadFields()
For i As Integer = 0 To fields.Length - 1
Console.WriteLine("Field {0}: {1}", i, fields(i))
Next
End Using
End Using
Output:
Field 0: 1
Field 1: 2,3,4
Field 2: 5
This worked great for parsing a Shipping Notice .csv file we receive. Thanks for keeping this solution here.
This is my version of the code:
Try
Using rdr As New IO.StringReader(Row.FlatFile)
Using parser As New FileIO.TextFieldParser(rdr)
parser.TextFieldType = FileIO.FieldType.Delimited
parser.Delimiters = New String() {","}
parser.HasFieldsEnclosedInQuotes = True
Dim fields() As String = parser.ReadFields()
Row.Account = fields(0).ToString().Trim()
Row.AccountName = fields.GetValue(1).ToString().Trim()
Row.Status = fields.GetValue(2).ToString().Trim()
Row.PONumber = fields.GetValue(3).ToString().Trim()
Row.ErrorMessage = ""
End Using
End Using
Catch ex As Exception
Row.ErrorMessage = ex.Message
End Try
It's possible to do it with regex VB.NET in the following way:
,(?=(?:[^"]*"[^"]*")*[^"]*$)
The positive lookahead ((?= ... )) ensures that there is an even number of quotes ahead of the comma to split on (i.e. either they occur in pairs, or there are none).
[^"]* matches non-quote characters.
Given below is a VB.NET example to apply the regex.
Imports System
Imports System.Text.RegularExpressions
Public Class Test
Public Shared Sub Main()
Dim theString As String = "1,""2,3,4"",5"
Dim theStringArray As String() = Regex.Split(theString, ",(?=(?:[^""\\]*""[^""\\]*"")*[^""\\]*$)")
For i As Integer = 0 To theStringArray.Length - 1
Console.WriteLine("theStringArray {0}: {1}", i, theStringArray(i))
Next
End Sub
End Class
'Output:
'theStringArray 0: 1
'theStringArray 1: "2,3,4"
'theStringArray 2: 5