I am looking to extract initials from a single textbox using Left, mid and other similar functions. The user enters their first, middle initial and last name; assuming spaces and a period after the middle initial. I need to ensure the initials extracted are uppercase, even if the text is entered in lowercase. Any assistance in a code that can accomplish this in VBA for Access would be appreciated. I am able to use the functions individually but am new to coding and am unsure how to string them together correctly.
Private Sub cmdGreeting_Click()
strOutput = Left(txtInput.Value, 1) & Mid(txtinput.value, 1) & Right(txtinput.value, 1)
lblOutput.Caption = strOutput
End Sub
This is as far as I have gotten and I know it's incorrect because I have no idea how to account for the 3 separate names.
Here's a function based on regex. I'm sure someone will chip in to improve it, my VBA regex is rusty. It's based on the regex here where you can see examples of it matching. If you're not familiar with regex at all, they're frightening initially and it's beyond the scope of an answer to explain them.
However, it works by breaking any input into 5 strings:
Initial character of first name
remainder of name
Initial and . if present
Initial letter of last name
remainder of last name
Then, with some simple UCase and LCase, you can compile the require, formatted name. You may want to change the logic - you did imply there would be a middle initial but this assumes it won't always be there, and the dot after the initial may or may not be there.
Note: you need to enable regex in Excel - instructions
Sub normalise()
Debug.Print (proper("Reginald D. Hunter"))
Debug.Print (proper("reginald D. hunter"))
Debug.Print (proper("rEGINALD d. Hunter"))
Debug.Print (proper("Reginald D Hunter"))
Debug.Print (proper("Reginald Hunter"))
Debug.Print (proper("Reginald D. Hunter"))
End Sub
Function proper(text) As String
Dim regexMatch As Object
Dim matches As Object
With New RegExp
.Global = False
.MultiLine = False
.IgnoreCase = False
.Pattern = "([a-zA-Z])([^ ]*)\s*([a-zA-Z]?[. ])?\s*([a-zA-Z])([^ ]*)"
If .test(text) Then
For Each regexMatch In .Execute(text)
Set matches = regexMatch.SubMatches
Next
End If
End With
proper = UCase(matches(0)) + LCase(matches(1))
If Trim(matches(2)) <> "" Then
If InStr(matches(2), ".") Then
proper = proper + " " + Trim(UCase(matches(2))) + " "
Else
proper = proper + " " + Trim(UCase(matches(2))) + ". "
End If
Else
proper = proper + " "
End If
proper = proper + UCase(matches(3)) + LCase(matches(4))
End Function
Results in
Reginald D. Hunter
Reginald D. Hunter
Reginald D. Hunter
Reginald D. Hunter
Reginald Hunter
Reginald D. Hunter
Edit: I misread the question and if you just want initials then replace the last part of the function like so:
proper = UCase(matches(0))
If Trim(matches(2)) <> "" Then
If InStr(matches(2), ".") Then
proper = proper + Replace(Trim(UCase(matches(2))), ".", "")
Else
proper = proper + Trim(UCase(matches(2)))
End If
End If
proper = proper + UCase(matches(3))
gives:
RDH
RDH
RDH
RDH
RH
RDH
This is the code I've been using for a while. It will include the the initials of double-barreled names as well.
?GetInitials("Darren Bartrup-Cook") will return DBC.
?GetInitials("The quick brown fox jumps over the lazy dog") will return TQBFJOTLD.
Public Function GetInitials(FullName As String) As String
Dim RegEx As Object
Dim Ret As Object
Dim RetItem As Object
On Error GoTo ERR_HANDLE
Set RegEx = CreateObject("VBScript.RegExp")
With RegEx
.IgnoreCase = True
.Global = True
.Pattern = "(\b[a-zA-Z])[a-zA-Z]* ?"
Set Ret = .Execute(FullName)
For Each RetItem In Ret
GetInitials = GetInitials & UCase(RetItem.Submatches(0))
Next RetItem
End With
EXIT_PROC:
On Error GoTo 0
Exit Function
ERR_HANDLE:
'Add your own error handling here.
'DisplayError Err.Number, Err.Description, "mdl_GetInitials.GetInitials()"
Resume EXIT_PROC
End Function
Related
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'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)")
So I'm completely new to VBA. I have a java-fetish so I'm not new to programming, however manipulating office documents just seemed easier with VBA.
Anyway, on topic:
I'm currently automating things in the company (This example is creating a contract). However, using Java, I always learned to make robust code and although the VBA code now works, I'm not happy with it because it requires a lot of 'friendliness' of the user. So my question is (I hope you don't mind), could you give me a nudge in the right direction to make my code way more robust?
Here's the code:
Function spaties(Name As String) As String
' Function used to ensure the length of a String (Working with Range)
Dim index As Integer
While (Len(Name) < 30)
Name = Name + " "
Wend
spaties = Name
End Function
Sub Macro3()
'
' Macro3 Macro
'
'
'ActiveDocument.Range(26101, 26102).Text = "d"
StartUndoSaver
Dim firma As String
firma = InputBox("Voor welke onderaannemer? (Zonder hoofdletters)" + Chr(10) + "(nicu, sorin of marius)")
Dim werf As String
werf = InputBox("Over welke Werf gaat het?")
Dim datum As String
datum = InputBox("Op welke datum spreekt het contract? (dd/mm/yyyy)")
With ActiveDocument
.Range(25882, 25899).Text = datum
ActiveDocument.Range(575, 605).Text = spaties(werf)
ActiveDocument.Range(1279, 1309).Text = spaties(werf)
End With
Select Case Len(firma)
Case 4
With ActiveDocument
.Range(26168, 26181).Text = "Nicu Dinita"
.Range(26062, 26088).Text = "Badi Woodconstruct SRL"
.Range(11359, 11371).Text = "Nicu Dinita"
End With
Case 5
With ActiveDocument
.Range(26168, 26181).Text = "Asavei Sorin"
.Range(26062, 26088).Text = "BELRO INTERIOR DESIGN SRL"
.Range(11359, 11371).Text = "Asavei Sorin"
End With
Case 6
With ActiveDocument
.Range(26168, 26181).Text = "Ivan Maricel"
.Range(26062, 26088).Text = "Solomon & Aaron Construct"
.Range(11359, 11371).Text = "Ivan Maricel"
End With
End Select
Dim prijs As String
Dim besch As String
Dim eenh As String
Dim hoev As Integer
hoev = InputBox("Hoeveel artikels zijn er?")
Dim index As Integer
index = 1
While (index <= hoev)
besch = InputBox("Beschrijving van het artikel (engels)")
prijs = InputBox("prijs van het artikel")
eenh = InputBox("Eenheid van het artikel")
With ActiveDocument
.Range(5701, 5702).Text = "" + vbTab + spaties2(besch, prijs, eenh) + Chr(10) + vbTab
End With
index = index + 1
Wend
With ActiveDocument.Sections(1)
.Headers(wdHeaderFooterPrimary).Range.Text = "Raes G. Schrijnwerken BVBA" + vbTab + vbTab + datum + Chr(10) + "Robert Klingstraat 5" + Chr(10) + "8940 Wervik"
.Footers(wdHeaderFooterPrimary).Range.Text = "Overeenkomst tot onderaanneming" + Chr(10) + "met betrekking tot:" + werf
.Footers(wdHeaderFooterPrimary).PageNumbers.Add PageNumberAlignment:=wdAlignPageNumberRight
End With
If firma = "sorin" Then
ActiveDocument.Range(254, 255).ImportFragment "Z:\Raes Netwerk DATA\professioneel\004 Sjablonen\belro.docx", False
Else
If firma = "nicu" Then
With ActiveDocument
.Range(254, 255).ImportFragment "Z:\Raes Netwerk DATA\professioneel\004 Sjablonen\Nicu.docx", False
End With
Else
If firma = "marius" Then
ActiveDocument.Range(254, 255).ImportFragment "Z:\Raes Netwerk DATA\professioneel\004 Sjablonen\Marius.docx", False
End If
End If
End If
ActiveDocument.PrintOut
ActiveDocument.PrintOut
End Sub
Function spaties2(artikel As String, prijs As String, eenh As String) As String
'Another function to ensure length of String
Dim index As Integer
Dim eind As String
eind = "" + artikel + vbTab + vbTab + prijs + "€/" + eenh
While (Len(eind) < 100)
eind = eind + " "
Wend
spaties2 = eind
End Function
As you can see, the code is very basic. And although it works, it's no good to deliver.
The two defined Functions are simply formatting the String of the user because obviously the name of something is not always the same length.
I'd like to cut out the Range properties, because in my opinion, that's what makes the program so sensitive to changes.
Any and all suggestions are welcome.
note: For the moment, the contract can have three different 'target parties' so that's why the Select Case statement is there. It's going to be completely useless if it should grow but for now it works.
Here's one:
sName = Left(sName & Space(30), 30)
And I think it's better to use bookmarks as placeholders instead of using Range(start, end)
How to change programmatically the text of a Word Bookmark
I think that your code needs some Trim's, in order to avoid mistaken spaces before and after the names (when you use some inputboxes, I mean).
And you need to verify input dates, too.
For string concatenation, use the ampersand (&) better than the plus sign (+), in order to avoid mistaken sums.
Instead of Chr(10) I have some recommendations in order to make your code more readable:
Chr(13) = vbCr
Chr(10) = vbLf
Chr(13) & Chr(10) = vbCrLf
Verify that the files you are indicating exist.
Using Range with numerical values is definitely not reliable. Bookmarks, as Tim suggests or content controls if this is Word 2007 or later. Content Controls are Microsoft's recommendation, going forward, but I don't see any particular advantage one way or the other for your purpose.
Looking at all the InputBox calls I have to wonder whether displaying a VBA UserForm for the input might not be better? All the input fields in one place, rather than flashing multiple prompts. You can validate for correct input before the UserForm is removed from the screen, etc.
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