I am trying to perform this task in Microstation using VBA. I want to use it to find and replace multiple numbers in the drawing.
I know the program generally, however I am having trouble putting this together. I have 2 variables.
Thank you in advance and sorry for the badly written code. Just getting used to VBA!
Sub main()
Dim Find_text () As string = split ("150 160 170 180 190 200 210 220")
Dim Replace_text () As string = split ("15 16 17 18 19 20 21 22")
For i As Integer = 0 To Find_text.length - 1
'I will write my find and replace code here
Next
End Sub
You can use function InStr() to find substring in string and function Replace() to replace found string with replace_string
Sub main()
Dim Find_text() As String
Dim Replace_text() As String
Dim str As String
str = "test 150 test 160 test 170 test 200 test 220"
Find_text = Split("150 160 170 180 190 200 210 220")
Replace_text = Split("15 16 17 18 19 20 21 22")
For i = 0 To UBound(Find_text)
For j = 0 To UBound(Replace_text)
If InStr(str, Find_text(j)) > 0 Then
str = Replace(str, Find_text(j), Replace_text(j))
End If
Next
Next
MsgBox str
End Sub
The trick is to use the Microstation API to get a list of elements to operate on. You can use .GetSelectedElements or .Scan depending on how you want your tool to work. As you said you are just getting used to vba I have heavily commented the code below.
Other gotchas include text elements and text nodes (multi line text) need to be handled differently, and don't forget to .Rewrite the element once you have modified it.
Option Explicit
Sub replaceText()
Dim findText() As String
Dim replaceText() As String
'set find and replace
findText = Split("10 20 30 40 50")
replaceText = Split("a b c d e")
'guard against unequal length searches
If UBound(findText) <> UBound(replaceText) Then
MsgBox "Find and replace are not equal lengths"
Exit Sub
End If
' Scan Criteria are needed when looking for elements in a model
' Set up scan criteria to only include text types
Dim eSC As New ElementScanCriteria
eSC.ExcludeAllTypes
eSC.IncludeType msdElementTypeText 'text element
eSC.IncludeType msdElementTypeTextNode 'multiple line text element
Dim model As ModelReference
Set model = ActiveModelReference
' if you need to loop through multiple models you could use this
' Set model = ActiveDesignFile.Models(i)
' Element Enumerator is a list of elements
Dim elements As ElementEnumerator
'scan active model for text
Set elements = model.Scan(eSC)
' the elements could also be retrieved using
' Set elements = model.GetSelectedElements
' iterate through element set
' If there is another element in the list then
' MoveNext sets elements.Current to the next element and returns true
' otherwise it returns false and the loop exits.
Do While elements.MoveNext
Dim i As Integer
Dim textNodeI As Integer
Dim tempText As String
' elements.Current is a generic element
' we need to check its type to handle it correctly
If elements.Current.IsTextElement Then
' access the generic element using the text element interface
With elements.Current.AsTextElement
'.text here refers to elements.Current.AsTextElement.text as specified by the With statement
tempText = .text
'split is 0 indexed
For i = 0 To UBound(findText)
tempText = Replace(tempText, findText(i), replaceText(i))
Next
'set the elements text to the replaced text
.text = tempText
'rewrite the text element to the model
.Rewrite
End With
ElseIf elements.Current.IsTextNodeElement Then
With elements.Current.AsTextNodeElement
' TextNodes have an array of TextLines (1 indexed)
For textNodeI = 1 To .TextLinesCount
'same as for text but for each line of node
tempText = .TextLine(textNodeI)
For i = 0 To UBound(findText)
tempText = Replace(tempText, findText(i), replaceText(i))
Next
.TextLine(textNodeI) = tempText
Next
' Rewrite the text node after you have replaced each line.
.Rewrite
End With
End If
Loop
End Sub
Related
I am creating a third party add in for my CAD program that has a sub in it that goes through a drawing and finds all the parts lists (BOMS), if any items in the parts list are shared between the BOM (1 part being used in 2 weldments for example) then it changes the item number of the second instance to be that of the first instance. It does this by comparing full file names between the two values. When they match change the number to that of the matcher. I have got this to work but it runs a little slow because for a 100 item BOM each item is compared to 100 and thus that takes a little longer then I would like (about 60seconds to run). After thinking about it I realized I did not need to compare each item to all the items, I just needed to compare until it found a duplicate and then exit the search loop and go to the next value. Example being Item 1 does not need to compare to the rest of the 99 values because even if it does have a match in position 100 I do not want to change item 1s number to that of item 100. I want to change item 100 to that of 1(ie change the duplpicate to that of the first encountered double). For my code however I am having trouble exiting the comparison for loops which is causing me trouble. An example of the trouble is this:
I have 3 BOMs, each one shares Part X, and is numbered 1 in BOM 1, 4 in BOM 2, and 7 in BOM 3. when I run my button because I cannot get it to leave the comparison loop once it finds it first match all the Part X's ended up getting item number 7 from BOM 3 because it is the last instance. (I can get this to do what I want by stepping through my for loops backwards and thus everything ends up as the top most occurrence, but I would like to get my exit fors working because it saves me on unnecessary comparisons)
How do I go about breaking out of the nested for loops using an if conditional?
Here is my current code:
Public Sub MatchingNumberR1()
Debug.Print ThisApplication.Caption
'define active document as drawing doc. Will produce an error if its not a drawing doc
Dim oDrawDoc As DrawingDocument
Set oDrawDoc = ThisApplication.ActiveDocument
'Store all the sheets of drawing
Dim oSheets As Sheets
Set oSheets = oDrawDoc.Sheets
Dim oSheet As Sheet
'Loop through all the sheets
For Each oSheet In oSheets
Dim oPartsLists As PartsLists
Set oPartsLists = oSheet.PartsLists
'Loop through all the part lists on that sheet
Dim oPartList As PartsList
'For every parts list on the sheet
For Each oPartList In oPartsLists
For i3 = 1 To oPartList.PartsListRows.Count
'Store the Item number and file referenced in that row to compare
oItem = FindItem(oPartList)
oDescription = FindDescription(oPartList)
oDescripCheck = oPartList.PartsListRows.Item(i3).Item(oDescription).Value
oNumCheck = oPartList.PartsListRows.Item(i3).Item(oItem).Value
'Check to see if the BOM item is a virtual component if it is do not try and get the reference part
If oPartList.PartsListRows.Item(i3).ReferencedFiles.Count = 0 Then
oRefPart = " "
End If
'Check to see if the BOM item is a virtual component if it is try and get the reference part
If oPartList.PartsListRows.Item(i3).ReferencedFiles.Count > 0 Then
oRefPart = oPartList.PartsListRows.Item(i3).ReferencedFiles.Item(1).FullFileName
End If
MsgBox (" We are comparing " & oRefPart)
'''''Create a comparison loop to go through the drawing that checks the oRefPart against other BOM items and see if there is a match.'''''
'Store all the sheets of drawing
Dim oSheets2 As Sheets
Set oSheets2 = oDrawDoc.Sheets
Dim oSheet2 As Sheet
'For every sheet in the drawing
For Each oSheet2 In oSheets2
'Get all the parts list on a single sheet
Dim oPartsLists2 As PartsLists
Set oPartsLists2 = oSheet2.PartsLists
Dim oPartList2 As PartsList
'For every parts list on the sheet
For Each oPartList2 In oPartsLists2
oItem2 = FindItem(oPartList2)
oDescription2 = FindDescription(oPartList2)
'Go through all the rows of the part list
For i6 = 1 To oPartList2.PartsListRows.Count
'Check to see if the part is a not a virtual component, if not, get the relevent comparison values
If oPartList2.PartsListRows.Item(i6).ReferencedFiles.Count > 0 Then
oNumCheck2 = oPartList2.PartsListRows.Item(i6).Item(oItem2).Value
oRefPart2 = oPartList2.PartsListRows.Item(i6).ReferencedFiles.Item(1).FullFileName
'Compare the file names, if they match change the part list item number for the original to that of the match
If oRefPart = oRefPart2 Then
oPartList.PartsListRows.Item(i3).Item(1).Value = oNumCheck2
''''''''This is where I want it to exit the loop and grab the next original value'''''''
End If
'For virtual components get the following comparison values
ElseIf oPartList2.PartsListRows.Item(i6).ReferencedFiles.Count = 0 Then
oNumCheck2 = oPartList2.PartsListRows.Item(i6).Item(oItem2).Value
oDescripCheck2 = oPartList2.PartsListRows.Item(i6).Item(oDescription2).Value
'Compare the descriptions and if they match change the part list item number for the original to that of the match
If oDescripCheck = oDescripCheck2 Then
oPartList.PartsListRows.Item(i3).Item(1).Value = oNumCheck2
''''''''This is where I want it to exit the loop and grab the next original value'''''''
End If
Else
''''''''This is where if no matches were found I want it to continue going through the comparison loop'''''''
End If
Next
Next
Next
Next
Next
Next
'MsgBox ("Matching Numbers has been finished")
End Sub
For escape from nested for loop you can use GoTo and specify where.
Sub GoToTest()
Dim a, b, c As Integer
For a = 0 To 1000 Step 100
For b = 0 To 100 Step 10
For c = 0 To 10
Debug.Print vbTab & b + c
If b + c = 12 Then
GoTo nextValueForA
End If
Next
Next
nextValueForA:
Debug.Print a + b + c
Next
End Sub
Here are a few examples that demonstrate (1) breaking out of (exiting) a loop and (2) finding the values in arrays.
The intersection of 2 arrays example can be modified to meet your need to "Create a comparison loop to go through the drawing that checks the oRefPart against other BOM items and see if there is a match." Note, you may find multiple matches between 2 arrays.
Option Explicit
Option Base 0
' Example - break out of loop when condition met.
Public Sub ExitLoopExample()
Dim i As Integer, j As Integer
' let's loop 101 times
For i = 0 To 100:
j = i * 2
'Print the current loop number to the Immediate window
Debug.Print i, j
' Let's decide to break out of the loop is some
' condition is met. In this example, we exit
' the loop if j>=10. However, any condition can
' be used.
If j >= 10 Then Exit For
Next i
End Sub
' Example - break out of inner loop when condition met.
Public Sub ExitLoopExample2()
Dim i As Integer, j As Integer
For i = 1 To 5:
For j = 1 To 5
Debug.Print i, j
' if j >= 2 then, exit the inner loop.
If j >= 2 Then Exit For
Next j
Next i
End Sub
Public Sub FindItemInArrayExample():
' Find variable n in array arr.
Dim intToFind As Integer
Dim arrToSearch As Variant
Dim x, y
intToFind = 4
arrToSearch = Array(1, 2, 3, 4, 5, 6, 7, 8, 9)
x = FindItemInArray(FindMe:=intToFind, _
ArrayToSearch:=arrToSearch)
If IsEmpty(x) Then
Debug.Print intToFind; "not found in arrToSearch"
Else
Debug.Print "found "; x
End If
intToFind = 12
y = FindItemInArray(FindMe:=intToFind, _
ArrayToSearch:=arrToSearch)
If IsEmpty(y) Then
Debug.Print intToFind; "not found in arrToSearch"
Else
Debug.Print "found "; y
End If
End Sub
Public Function FindItemInArray(FindMe, ArrayToSearch As Variant):
Dim i As Integer
For i = LBound(ArrayToSearch) To UBound(ArrayToSearch)
If FindMe = ArrayToSearch(i) Then
FindItemInArray = ArrayToSearch(i)
Exit For
End If
Next i
End Function
' Create a comparison loop to go through the drawing that checks
' the oRefPart against other BOM items and see if there is a match.
Public Sub ArrayIntersectionExample():
Dim exampleArray1 As Variant, exampleArray2 As Variant
Dim arrIntersect As Variant
Dim i As Integer
' Create two sample arrays to compare
exampleArray1 = Array(1, 2, 3, 4, 5, 6, 7)
exampleArray2 = Array(2, 4, 6, 8, 10, 12, 14, 16)
' Call our ArrayIntersect function (defined below)
arrIntersect = ArrayIntersect(exampleArray1, exampleArray2)
' Print the results to the Immediate window
For i = LBound(arrIntersect) To UBound(arrIntersect)
Debug.Print "match " & i + 1, arrIntersect(i)
Next i
End Sub
Public Function ArrayIntersect(arr1 As Variant, arr2 As Variant) As Variant:
' Find items that exist in both arr1 and arr2 (intersection).
' Return the intersection as an array (Variant).
Dim arrOut() As Variant
Dim matchIndex As Long
Dim i As Long, j As Long
' no matches yet
matchIndex = -1
' begin looping through arr1
For i = LBound(arr1) To UBound(arr1)
' sub-loop for arr2 for each item in arr1
For j = LBound(arr2) To UBound(arr2)
' check for match
If arr1(i) = arr2(j) Then
' we found an item in both arrays
' increment match counter, which we'll
' use to size our output array
matchIndex = matchIndex + 1
' resize our output array to fit the
' new match
ReDim Preserve arrOut(matchIndex)
' now store the new match our output array
arrOut(matchIndex) = arr1(i)
End If
Next j
Next i
' Have the function return the output array.
ArrayIntersect = arrOut
End Function
I need to separate following strings into Name and Number: e.g.
evil333 into evil and 333
bili454 into bili and 454
elvis04 into elvis and 04
Split(String, "#") ' don't work here because numbers are unknown
similarly
Mid(String, 1, String - #) ' don't work because Numbers length is unknown
so what should be the best way to start? Just want to keep it simple as possible
Update:
For further info follow - https://youtu.be/zjF7oLLgtms
Two more ways for solving this:
Sub test()
Dim sInputString As String
Dim i As Integer
Dim lFirstNumberPos As Long
sInputString = "evil333"
'loop through text in input string
'if value IsNumeric (digit), stop looping
For i = 1 To Len(sInputString)
If IsNumeric(Mid(sInputString, i, 1)) Then
lFirstNumberPos = i
Exit For
End If
Next i
Dim Name As String
Dim Number As String
'return result
Name = Left$(sInputString, lFirstNumberPos - 1)
Number = Mid$(sInputString, lFirstNumberPos)
End Sub
Or another method:
Sub test2()
'if you are going to have too long string it would maybe better to use "instr" method
Dim sInputString As String
Dim lFirstNumberPos As Long
Dim i As Integer
sInputString = "evil333"
Dim lLoopedNumber as Long
LoopedNumber = 0
lFirstNumberPos = Len(sInputString) + 1
'loop through digits 0-9 and stop when any of the digits will be found
For i = 0 To 9
LoopedNumber = InStr(1, sInputString, cstr(i), vbTextCompare)
If LoopedNumber > 0 Then
lFirstNumberPos = Application.Min(LoopedNumber,lFirstNumberPos)
End If
Next i
Dim Name As String
Dim Number As String
'return result
Name = Left$(sInputString, lFirstNumberPos - 1)
Number = Mid$(sInputString, lFirstNumberPos)
End Sub
You should regular expressions (regex) to match the two parts of your strings. The following regex describes how to match the two parts:
/([a-z]+)([0-9]+)/
Their use in VBA is thorougly explained in Portland Runner's answer to How to use Regular Expressions (Regex) in Microsoft Excel both in-cell and loops
How do I get a sentence with multiple commas in MS Word with VBA that the cursor is in?
All the posts I've found said to get the sentence the cursor is in then use the code:
Selection.Sentences(1)
The above works well with a sentence with only 1 comma. But if I have a sentence with multiple commas like this:
For example, tomorrow is Tuesday(e.g., not Wednesday) or Thursday.
where the cursor is set somewhere in "For example" then "Selection.Sentences(1)" returns between the bars "...(e.g.|, |n...".
I'm using the latest version of Word. I plan on launching the code on an older version (I think 2013) that I first noticed the problem on.
This code is better suited to explain why MS didn't solve your problem than it is to actually solve it. However - depending upon your circumstances - you may like to play with it.
Option Explicit
Sub SelectSentence()
' 30 Jan 2018
' list abbreviations containing periods only
' in sequence of their expected frequency of occurrance
Const Abbs As String = "e.g.,f.i.,etc.,i.e."
Dim Fun As String ' sentence to select
Dim Para As Range
Dim SelStart As Long ' location of selection
Dim Sp() As String ' array of Abbs
Dim Cp() As String ' array of encoded Abbs
With Selection
Set Para = .Paragraphs(1).Range
SelStart = .Start
End With
Sp = Split(Abbs, ",")
With Para
Application.ScreenUpdating = False
.Text = CleanString(.Text, Sp, Cp)
Fun = ActiveDocument.Range(SelStart, SelStart + 1).Sentences(1).Text
SelStart = InStr(.Text, Fun) + .Start - 1
.Text = OriginalString(.Text, Cp)
.SetRange SelStart, SelStart + Len(Fun) - 1
Application.ScreenUpdating = True
.Select
End With
Fun = Selection.Text
Debug.Print Fun
End Sub
Private Function CleanString(ByVal Txt As String, _
Abbs() As String, _
Cp() As String) As String
' 30 Jan 2018
Dim i As Integer
ReDim Cp(UBound(Abbs))
For i = 0 To UBound(Abbs)
If InStr(Txt, ".") = 0 Then Exit For
Cp(i) = AbbToTxt(Abbs(i))
Txt = Replace(Txt, Abbs(i), Cp(i))
Next i
ReDim Preserve Cp(i)
CleanString = Txt
End Function
Private Function AbbToTxt(ByVal Abb As String) As String
' 30 Jan 2018
' use a character for Chr(92) not occurring in your document.
' Apparently it must be a character with a code below 128.
' use same character as function 'AbbToTxt'
AbbToTxt = Replace(Abb, ".", Chr(92))
End Function
Private Function OriginalString(ByVal Txt As String, _
Cp() As String) As String
' 30 Jan 2018
Dim i As Integer
For i = 0 To UBound(Cp) - 1
Txt = Replace(Txt, Cp(i), TxtToAbb(Cp(i)))
Next i
OriginalString = Txt
End Function
Private Function TxtToAbb(ByVal Txt As String) As String
' 30 Jan 2018
' use same character as function 'AbbToTxt'
TxtToAbb = Replace(Txt, Chr(92), ".")
End Function
For one, the code will only handle abbreviations which you program into it (see Const Abbs at the top of the code). For another, it will fail to recognise a period with dual meaning, such as "etc." found at the end of a sentence.
If you are allowed to edit the documents you work with, the better way of tackling your problem may well be to remove the offending periods with Find > Replace. After all, whoever understands "e.g." is also likely to understand "eg". Good Luck!
This is the output of a program.
I have specified what shall be width of each cell in the program and my program shows correct output.
What I want to do is cell content shall be written from right to left. E.g highlighted figure 9983.54 has width of 21. Text file has used first 7 columns. But I want it to use last 7 columns of text file.
Please see expected output image.
I am not getting any clue how to do this. I am not a very professional programmer but I love coding. This text file is used as input to some other program and i am trying to automate writing text file from excel VBA.
Can anyone suggest a way to get this output format?
Here is the code which gave me first output
Option Explicit
Sub CreateFixedWidthFile(strFile As String, ws As Worksheet, s() As Integer)
Dim i As Long, j As Long
Dim strLine As String, strCell As String
'get a freefile
Dim fNum As Long
fNum = FreeFile
'open the textfile
Open strFile For Output As fNum
'loop from first to last row
'use 2 rather than 1 to ignore header row
For i = 1 To ws.Range("a65536").End(xlUp).Row
'new line
strLine = ""
'loop through each field
For j = 0 To UBound(s)
'make sure we only take chars up to length of field (may want to output some sort of error if it is longer than field)
strCell = Left$(ws.Cells(i, j + 1).Value, s(j))
'add on string of spaces with length equal to the difference in length between field length and value length
strLine = strLine & strCell & String$(s(j) - Len(strCell), Chr$(32))
Next j
'write the line to the file
Print #fNum, strLine
Next i
'close the file
Close #fNum
End Sub
'for example the code could be called using:
Sub CreateFile()
Dim sPath As String
sPath = Application.GetSaveAsFilename("", "Text Files,*.txt")
If LCase$(sPath) = "false" Then Exit Sub
'specify the widths of our fields
'the number of columns is the number specified in the line below +1
Dim s(6) As Integer
'starting at 0 specify the width of each column
s(0) = 21
s(1) = 9
s(2) = 15
s(3) = 11
s(4) = 12
s(5) = 10
s(6) = 186
'for example to use 3 columns with field of length 5, 10 and 15 you would use:
'dim s(2) as Integer
's(0)=5
's(1)=10
's(2)=15
'write to file the data from the activesheet
CreateFixedWidthFile sPath, ActiveSheet, s
End Sub
Something like this should work:
x = 9983.54
a = Space(21-Len(CStr(x))) & CStr(x)
Then a will be 14 spaces followed by x:
a = " 9983.54"
Here 21 is the desired column width --- change as necessary. CStr may be unnecessary for non-numeric x.
If you're going to right-justify a lot of different data to different width fields you could write a general purpose function:
Function LeftJust(val As String, width As Integer) As String
LeftJust = Space(width - Len(val)) & val
End Function
The you call it with LeftJust(CStr(9983.54), 21).
Also note that VBA's Print # statement has a Spc(n) parameter that you can use to produce fixed-width output, e.g., Print #fNum, Spc(n); a; before this statement you calculate n: n = 21-Len(CStr(a)).
Hope that helps
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