I'm trying to delete string content before a certain word contained within the string. For example
master_of_desaster#live.de
I'd like to use VBA in order to replace that with
master_of_desaster
Everything after the "word" (#) should be removed, including the "word" itself.
I found a similar topic here, but he asks the opposite.
email = "master_of_desaster#live.de"
ret = Left(email, InStr(1, email, "#") - 1)
Result: master_of_desaster
Thanks to Shai Rado
=split("master_of_desaster#live.de","#")(0)
Just for fun - a regex approach.
Public Sub reg()
Dim re_pattern As String
Dim re As RegExp
Dim email As String
Dim match As Object
Set re = New RegExp
email = "master_of_desaster#live.de"
re_pattern = "(.*)#.*"
With re
.Global = True
.MultiLine = True
.IgnoreCase = False
.Pattern = re_pattern
End With
Set match = re.Execute(email)
Debug.Print match.Item(0).SubMatches(0)
End Sub
A bit hacky but fast ( most Windows API accept zero terminated strings )
ret = Replace("master_of_disaster#live.de", "#", vbNullChar, , 1) ' Chr(0)
I usually use the Split method but with Limit:
ret = Split("master_of_disaster#live.de", "#", 2)(0)
ret = evaluate("left(" & string & ", search(""#"", " & string & ") - 1)")
Related
I have a list of strings defined as
Dim replyFormat(0 To 999) As String
and a list of answers as
Dim answers(0 to 999) As String
and throughout the code certain strings get added to replyFormat that look similar to this:
Name: {1} {3}
When everything is done, I define a string called sendBack and start looping through each line in replyFormat. I want to set sendBack equal to itself plus what replyFormat is, evaluating answers for the numbers in the curly brackets and finally adding vbCrLf to the end. For exmaple if answers contains { Yes, John, H, Doe } and replyFormat is "Name: {1} {3}" it would ouput "Name: John Doe"
It sounds like you're referring to reflection which isn't supported in VBA. You can however achieve the desired result by using Regular Expressions (RegEx):
Function FormattedString(stringToFormat As String, replacements() As String) As String
Dim placeholder As Variant
Dim index As Long
With CreateObject("VBScript.RegExp")
.Pattern = "\{([\d]{1,3})\}"
.Global = True
.MultiLine = False
.IgnoreCase = True
If .Test(stringToFormat) Then
For Each placeholder In .Execute(stringToFormat)
index = CLng(placeholder.SubMatches(0))
stringToFormat = Replace$(stringToFormat, placeholder, replacements(index))
Next
End If
End With
FormattedString = stringToFormat
End Function
Example use:
Sub FooBar()
Dim answers(0 To 3) As String
Const testString = "Name: {1} {3}"
answers(0) = "Test"
answers(1) = "John"
answers(2) = "Testing"
answers(3) = "Doe"
Debug.Print FormattedString(testString, answers) '// "Name: John Doe"
End Sub
If this is your object:
Ob = { Yes, John, H, Doe},
You could select object item like this:
Ob(1), Ob(3)
For more information, Please refer to this link:
Retrieve the index of an object stored in a collection using its key (VBA)
I'm trying to do something simple and I don't understand why it's not working. I'm really new to MS Access VBA.
I have a string in a textbox :
\\p9990cdc\C$\Temp
I want to turn it into :
C:\Temp
I'm trying :
strSelectedFile = Replace(strSelectedFile, "\\*\C$", "C:")
and it's not working.
Not sure why RegEx doesn't work either :
strSelectedFile = Replace(strSelectedFile, "\\[\w]\C$", "C:")
Everything is set properly so the problem lies exactly in that replace code, because if I try for example :
strSelectedFile = Replace(strSelectedFile, "C$", "C:")
It works and sucessfully replaces the C$ with C:
\p9990cdc\C:\Temp
How can I make this work?
Thanks a lot for your time!
Replace doesn't do wildcards. You can implement your own function that does, or use regex using VBScript.RegEx.
I've written a small function that does this for you. Performance is suboptimal, however, and I've only done a little testing. It works for your sample input.
Public Function LikeReplace(strInput As String, strPattern As String, strReplace As String, Optional start As Long = 1)
Dim LenCompare As Long
Do While start <= Len(strInput)
For LenCompare = Len(strInput) - start + 1 To 1 Step -1
If Mid(strInput, start, LenCompare) Like strPattern Then
strInput = Left(strInput, start - 1) & strReplace & Right(strInput, Len(strInput) - (start + LenCompare - 1))
End If
Next
start = start + 1
Loop
LikeReplace = strInput
End Function
Using your inputs and swapping Replace with this LikeReplace should just work.
You can use just VBScript.RegEx and the correct pattern for this.
Public Function ReplacePattern(ByRef iString As String, iSearchPattern As String, iReplaceWith As Variant)
'---------------------------------------------------------------------------------------
' Procedure : ReplacePattern
' Purpose : Searches a string by pattern and replaces the text with given value
' Returns : Replaced string.
'---------------------------------------------------------------------------------------
'
Dim RE As Object
Set RE = CreateObject("VBScript.RegExp")
RE.ignorecase = True
RE.Global = True
RE.Pattern = iSearchPattern
iString = RE.Replace(iString, iReplaceWith)
ReplacePattern = iString
On Error Resume Next
Set RE = Nothing
End Function
Read more about patterns Here
pattern: "^\\\\.*C\$" => Begins with \\ + any number of any character except linebreak + C$
usage
??replacepattern("\\p9990cdc\C$\Temp","^\\\\.*C\$","C:") => C:\Temp
You could instead use Mid(Instr()) to find the index of $ and grab the string from there (minus 1 to keep the directory letter) onwards.
strSelectedFile = Replace(Mid(strSelectedFile, InStr(strSelectedFile, "$") - 1, Len(strSelectedFile)), "$", ":")
I am trying to split data using VBA within word.
I have got the data using the following method
d = ActiveDocument.Tables(1).Cell(1, 1).Range.Text
This works and gets the correct data. Data for this example is
This
is
a
test
However, when I need to split the string into a list of strings using the delimiter as \n
Here is an example of the desired output
This,is,a,test
I am currently using
Dim dataTesting() As String
dataTesting() = Split(d, vbLf)
Debug.Print dataTesting(0)
However, this returns all the data and not just the first line.
Here is what I have tried within the Split function
\n
\n\r
\r
vbNewLine
vbLf
vbCr
vbCrLf
Word uses vbCr (ANSI 13) to write a "new" paragraph (created when you press ENTER) - represented in the Word UI by ¶ if the display of non-printing characters is activated.
In this case, the table cell content you show would look like this
This¶
is¶
a¶
test¶
The correct way to split an array delimited by a pilcro in Word is:
Dim d as String
d = ActiveDocument.Tables(1).Cell(1, 1).Range.Text
Dim dataTesting() As String
dataTesting() = Split(d, vbCr)
Debug.Print dataTesting(0) 'result is "This"
You can try this (regex splitter from this thread)
Sub fff()
Dim d As String
Dim dataTesting() As String
d = ActiveDocument.Tables(1).Cell(1, 1).Range.Text
dataTesting() = SplitRe(d, "\s+")
Debug.Print "1:" & dataTesting(0)
Debug.Print "2:" & dataTesting(1)
Debug.Print "3:" & dataTesting(2)
Debug.Print "4:" & dataTesting(3)
End Sub
Public Function SplitRe(Text As String, Pattern As String, Optional IgnoreCase As Boolean) As String()
Static re As Object
If re Is Nothing Then
Set re = CreateObject("VBScript.RegExp")
re.Global = True
re.MultiLine = True
End If
re.IgnoreCase = IgnoreCase
re.Pattern = Pattern
SplitRe = Strings.Split(re.Replace(Text, ChrW(-1)), ChrW(-1))
End Function
If this doesn't work, there may be strange unicode/Wprd characters in your Word doc. It may be soft breaks, for instance. You could try to not split with "\W+" in stead of "\s+". I cannot test this without your document.
Dim dataTesting() As String
dataTesting() = Split(d, vbLf)
Debug.Print dataTesting(0)
works fine and thank you very much for your example,
for why it have returned a whole array is because you have used 0 as index, in many programming languages 0 is the whole array, so the first element is ,
so in my case counting from 1 this perfectly split a string that I had troubles with.
To be more exact this is how it was used in my case
Dim dataTesting() As String
dataTesting() = Split(Document.LatheMachineSetup.Heads.Item(1).Comment, vbCrLf)
MsgBox (dataTesting(1))
And that comment is a multiline string.
Image
So this msg box returned exactly first line.
I am trying to remove words appearing in one string from a different string using a custom function. For instance:
A1:
the was why blue hat
A2:
the stranger wanted to know why his blue hat was turning orange
The ideal outcome in this example would be:
A3:
stranger wanted to know his turning orange
I need to have the cells in reference open to change so that they can be used in different situations.
The function will be used in a cell as:
=WORDREMOVE("cell with words needing remove", "cell with list of words being removed")
I have a list of 20,000 rows and have managed to find a custom function that can remove duplicate words (below) and thought there may be a way to manipulate it to accomplish this task.
Function REMOVEDUPEWORDS(txt As String, Optional delim As String = " ") As String
Dim x
'Updateby20140924
With CreateObject("Scripting.Dictionary")
.CompareMode = vbTextCompare
For Each x In Split(txt, delim)
If Trim(x) <> "" And Not .exists(Trim(x)) Then .Add Trim(x), Nothing
Next
If .Count > 0 Then REMOVEDUPEWORDS = Join(.keys, delim)
End With
End Function
If you can guarantee that your words in both strings will be separated by spaces (no comma, ellipses, etc), you could just Split() both strings then Filter() out the words:
Function WORDREMOVE(ByVal strText As String, strRemove As String) As String
Dim a, w
a = Split(strText) ' Start with all words in an array
For Each w In Split(strRemove)
a = Filter(a, w, False, vbTextCompare) ' Remove every word found
Next
WORDREMOVE = Join(a, " ") ' Recreate the string
End Function
You can also do this using Regular Expressions in VBA. The version below is case insensitive and assumes all words are separated only by space. If there is other punctuation, more examples would aid in crafting an appropriate solution:
Option Explicit
Function WordRemove(Str As String, RemoveWords As String) As String
Dim RE As Object
Set RE = CreateObject("vbscript.regexp")
With RE
.ignorecase = True
.Global = True
.Pattern = "(?:" & Join(Split(WorksheetFunction.Trim(RemoveWords)), "|") & ")\s*"
WordRemove = .Replace(Str, "")
End With
End Function
My example is certainly not the best code, but it should work
Function WORDREMOVE(FirstCell As String, SecondCell As String)
Dim FirstArgument As Variant, SecondArgument As Variant
Dim FirstArgumentCounter As Integer, SecondArgumentCounter As Integer
Dim Checker As Boolean
WORDREMOVE = ""
FirstArgument = Split(FirstCell, " ")
SecondArgument = Split(SecondCell, " ")
For SecondArgumentCounter = 0 To UBound(SecondArgument)
Checker = False
For FirstArgumentCounter = 0 To UBound(FirstArgument)
If SecondArgument(SecondArgumentCounter) = FirstArgument(FirstArgumentCounter) Then
Checker = True
End If
Next FirstArgumentCounter
If Checker = False Then WORDREMOVE = WORDREMOVE & SecondArgument(SecondArgumentCounter) & " "
Next SecondArgumentCounter
WORDREMOVE = Left(WORDREMOVE, Len(WORDREMOVE) - 1)
End Function
Let's assume that we have one module with only one Sub in it, and there are no comments. How to identify all variable names ? Is it possible to identify names of variables which are not defined using Dim ? I would like to identify them and replace each with some random name to obfuscate my code (O0011011010100101 for example), replace part is much easier.
List of characters which could be use in names of macros, functions and variables :
ABCDEFGHIJKLMNOPQRSTUVWXYZdefghijklmnopqrstuvwxyzg€‚„…†‡‰Š‹ŚŤŽŹ‘’“”•–—™š›śťžź ˇ˘Ł¤Ą¦§¨©Ş«¬®Ż°±˛ł´µ¶·¸ąş»Ľ˝ľżŔÁÂĂÄĹĆÇČÉĘËĚÍÎĎĐŃŇÓÔŐÖ×ŘŮÚŰÜÝŢßŕáâăäĺćçčéęëěíîďđńňóôőö÷řůúűüýţ˙ÉĘËĚÍÎĎĐŃŇÓÔŐÖ×ŘŮÚŰÜÝŢßŕáâăäĺćçčéęëěíîďđńňóôőö÷řůúűüýţ˙
Below are my function I've wrote recenlty :
Function randomName(n as integer) as string
y="O"
For i = 2 To n:
If Rnd() > 0.5 Then
y = y & "0"
Else
y = y & "1"
End If
Next i
randomName=y
End Function
In goal to replace given strings in another string which represent the code of module I use below sub :
Sub substituteNames()
'count lines in "Module1" which is part of current workbook
linesCount = ActiveWorkbook.VBProject.VBComponents("Module1").CodeModule.CountOfLines
'read code from module
code = ActiveWorkbook.VBProject.VBComponents("Module1").CodeModule.Lines(StartLine:=1, Count:=linesCount)
inputStr = Array("name1", "name2", "name2") 'some hardwritten array with string to replace
namesLength = 20 'length of new variables names
For i = LBound(inputStr) To UBound(inputStr)
outputString = randomName(namesLength-1)
code = Replace(code, inputStr(i), outputString)
Next i
Debug.Print code 'view code
End Sub
then we simply substitute old code with new one, but how to identify strings with names of variables ?
Edition
Using **Option Explicit ** decrease safety of my simple method of obfuscation, because to reverse changes you only have to follow Dim statements and replace ugly names with something normal. Except that to make such substitution harder, I think it's good idea to break the line in the middle of variable name :
O0O000O0OO0O0000 _
0O00000O0OO0
the simple method is also replacing some strings with chains based on chr functions chr(104)&chr(101)&chr(108)&chr(108)&chr(111) :
Sub stringIntoChrChain()
strInput = "hello"
strOutput = ""
For i = 1 To Len(strInput)
strOutput = strOutput & "chr(" & Asc(Mid(strInput, i, 1)) & ")&"
Next i
Debug.Print Mid(strOutput, 1, Len(strOutput) - 1)
End Sub
comments like below could make impression on user and make him think that he does not poses right tool to deal with macro etc.:
'(k=Äó¬)w}ż^¦ů‡ÜOyúm=ěËnóÚŽb W™ÄQó’ (—*-ĹTIäb
'R“ąNPÔKZMţ†üÍQ‡
'y6ű˛Š˛ŁŽ¬=iýQ|˛^˙ ‡ńb ¬ĂÇr'ń‡e˘źäžŇ/âéç;1qýěĂj$&E!V?¶ßšÍ´cĆ$Âű׺Ůî’ﲦŔ?TáÄu[nG¦•¸î»éüĽ˙xVPĚ.|
'ÖĚ/łó®Üă9Ę]ż/ĹÍT¶Mµę¶mÍ
'q[—qëýY~Pc©=jÍ8˘‡,Ú+ń8ŐűŻEüńWü1ďëDZ†ć}ęńwŠbŢ,>ó’Űçµ™Š_…qÝăt±+‡ĽČgřÍ!·eŠP âńđ:ŶOážű?őë®ÁšńýĎáËTbž}|Ö…ăË[®™
You can use a regular expression to find variable assignments by looking for the equals sign. You'll need to add a reference to the Microsoft VBScript Regular Expressions 5.5 and Microsoft Visual Basic for Applications Extensibility 5.3 libraries as I've used early binding.
Please be sure to back up your work and test this before using it. I could have gotten the regex wrong.
UPDATE:
I've refined the regular expressions so that it no longer catches datatypes of strongly typed constants (Const ImAConstant As String = "Oh Noes!" previously returned String). I've also added another regex to return those constants as well. The last version of the regex also mistakenly caught things like .Global = true. That was corrected. The code below should return all variable and constant names for a given code module. The regular expressions still aren't perfect, as you'll note that I was unable to stop false positives on double quotes. Also, my array handling could be done better.
Sub printVars()
Dim linesCount As Long
Dim code As String
Dim vbPrj As VBIDE.VBProject
Dim codeMod As VBIDE.CodeModule
Dim regex As VBScript_RegExp_55.RegExp
Dim m As VBScript_RegExp_55.match
Dim matches As VBScript_RegExp_55.MatchCollection
Dim i As Long
Dim j As Long
Dim isInDatatypes As Boolean
Dim isInVariables As Boolean
Dim datatypes() As String
Dim variables() As String
Set vbPrj = VBE.ActiveVBProject
Set codeMod = vbPrj.VBComponents("Module1").CodeModule
code = codeMod.Lines(1, codeMod.CountOfLines)
Set regex = New RegExp
With regex
.Global = True ' match all instances
.IgnoreCase = True
.MultiLine = True ' "code" var contains multiple lines
.Pattern = "(\sAs\s)([\w]*)(?=\s)" ' get list of datatypes we've used
' match any whole word after the word " As "
Set matches = .Execute(code)
End With
ReDim datatypes(matches.count - 1)
For i = 0 To matches.count - 1
datatypes(i) = matches(i).SubMatches(1) ' return second submatch so we don't get the word " As " in our array
Next i
With regex
.Pattern = "(\s)([^\.\s][\w]*)(?=\s\=)" ' list of variables
' begins with a space; next character is not a period (handles "with" assignments) or space; any alphanumeric character; repeat until... space
Set matches = .Execute(code)
End With
ReDim variables(matches.count - 1)
For i = 0 To matches.count - 1
isInDatatypes = False
isInVariables = False
' check to see if current match is a datatype
For j = LBound(datatypes) To UBound(datatypes)
If matches(i).SubMatches(1) = datatypes(j) Then
isInDatatypes = True
Exit For
End If
'Debug.Print matches(i).SubMatches(1)
Next j
' check to see if we already have this variable
For j = LBound(variables) To i
If matches(i).SubMatches(1) = variables(j) Then
isInVariables = True
Exit For
End If
Next j
' add to variables array
If Not isInDatatypes And Not isInVariables Then
variables(i) = matches(i).SubMatches(1)
End If
Next i
With regex
.Pattern = "(\sConst\s)(.*)(?=\sAs\s)" 'strongly typed constants
' match anything between the words " Const " and " As "
Set matches = .Execute(code)
End With
For i = 0 To matches.count - 1
'add one slot to end of array
j = UBound(variables) + 1
ReDim Preserve variables(j)
variables(j) = matches(i).SubMatches(1) ' again, return the second submatch
Next i
' print variables to immediate window
For i = LBound(variables) To UBound(variables)
If variables(i) <> "" And variables(i) <> Chr(34) Then ' for the life of me I just can't get the regex to not match doublequotes
Debug.Print variables(i)
End If
Next i
End Sub