I currently have a cope snippet that allows me to extract the first 6 characters of a filename to insert in the right of a document footer and the file title (not including extension) in the left of the footer. It was all based on character position and had served its purpose. However, a complexity has been introduced.
Our previous file naming structure was solely "###### - [Title]" (where we had 6 digits, a space, dash, a space, a title). We now have additional file names of the format "######.## - [TITLE]". I don't believe I can rely solely on character position to achieve extraction and insertion. I'm looking for some assistance in how I might be able to setup the code to determine whether the 7th character is a decimal or a space and proceed accordingly with inserting ##### versus ######.## into the footer. The flip side would also be true where I would need to determine the actual title (not relying on the position since it would be dependent on the first case.
Any assistance is appreciated. Below is the current code. It might not be the cleanest, but it gets the job done:
Sub FooterFields(myFooterHL, myTotalPageCount, myFooterBold)
sTitle = ActiveDocument.Name
J = InStrRev(sTitle, ".")
If J > 0 Then
sTitle = Left(sTitle, J - 9)
If Len(sTitle) > 5 Then
sTitle = Left(sTitle, 9)
End If
Dim specTitle As String
Dim specTitle2 As String
Dim specTitleInt As String
specTitle = ActiveDocument.Name
If Right(specTitle, 5) = ".docx" Then
specTitleInt = Left(specTitle, Len(specTitle) - 5)
specTitle2 = Right(specTitleInt, Len(specTitleInt) - 9)
ElseIf Right(specTitle, 4) = ".doc" Then
specTitleInt = Left(specTitle, Len(specTitle) - 4)
specTitle2 = Right(specTitleInt, Len(specTitleInt) - 9)
Else
End If
sDiv = ActiveDocument.Name
K = InStrRev(sDiv, ".")
If K > 0 Then
sDivIntermediate = Right(sDiv, K - 6)
sDivFinal = specTitle2
If Len(sDiv) > 5 Then
sDivIntermediate = Right(sDiv, K - 6)
sDivFinal = specTitle2
End If
If ActiveWindow.View.SplitSpecial <> wdPaneNone Then
ActiveWindow.Panes(2).Close
End If
If ActiveWindow.ActivePane.View.Type = wdNormalView Or ActiveWindow. _
ActivePane.View.Type = wdOutlineView Then
ActiveWindow.ActivePane.View.Type = wdPrintView
End If
ActiveWindow.ActivePane.View.SeekView = wdSeekCurrentPageFooter
Selection.WholeStory
Selection.Delete
Selection.Font.AllCaps = True
With Selection
.Range.Text = sTitle
.Range.InsertAlignmentTab wdRight, wdMargin
.EndKey Unit:=wdLine
.Fields.Add Range:=Selection.Range, Type:=wdFieldEmpty, Text:="PAGE ",
PreserveFormatting:=True
If myTotalPageCount = vbYes Then
.TypeText Text:="/"
.Fields.Add Range:=Selection.Range, Type:=wdFieldEmpty,
Text:="NUMPAGES ", PreserveFormatting:=True
Else
End If
.HomeKey Unit:=wdLine
.Range.Text = sDivFinal
End With
PageNumberAlignment:=wdAlignPageNumberRight, FirstPage:=True
ActiveWindow.ActivePane.View.SeekView = wdSeekMainDocument
Else
MsgBox "Document has no filename extension."
End If
End If
End Sub
I would suggest you consider using regular expressions for this. (The link is for using regular expressions in VBA hosted under Excel, but most of the information applies to VBA hosted under Word as well.)
A short introduction to regular expressions:
Define a pattern
Find one or more matches to the pattern in a string
You could use the following pattern to match all the variations:
^\d{6}(?:\.\d{2}) - .*(?:\.doc|\.docx)$
Add a reference (Tools -> References...) to the Microsoft VBScript Regular Expressions 5.5 library. Then you can use something like the following:
Sub FooterFields2(myFooterHL, myTotalPageCount, myFooterBold)
Dim re As New RegExp
re.Pattern = "^(\d{6}(?:\.\d{2})) - (.*)(?:\.doc|\.docx)$"
re.IgnoreCase = True
Dim matches As MatchCollection
Set matches = re.Execute(ActiveDocument.Name)
If matches.Count = 0 Then
MsgBox "Document name doesn't match"
Exit Sub
End If
Dim match As match
Set match = matches(0)
Dim code As String, title As String
code = match.SubMatches(0)
title = match.SubMatches(1)
'insert code and title in the appropriate location
End Sub
Explanation
\d -- matches a single digit character
\d{6} -- matches exactly 6 digits
\. -- matches a .. Because the . is a part of regular expression syntax, it has to be escaped with a \
\.\d{2} -- matches a . followed by (as before) exactly two digits
(?:\.\d{2}) -- groups \.\d{2} so we can apply an operator. Grouping with (?: (as opposed to with () does not store the value within the group to be referenced separately
(?:\.\d{2})?-- matches zero or one instances of (?:\.\d{2}), because the title might have the decimal and two successive digits, or it might not
- .* -- matches a space, a dash, and a space, followed by any number (*) of any character (.)
\.doc|\.docx -- matches either .doc or .docx
(?:\.doc|\.docx) -- groups together the different variants of the extension, but doesn't save the extension for further use
In order to extract the digits and the title after we've matched against the regular epression, we need to use capturing groups, which are indicated by parentheses -- (). In this case we want the numeric code (first capturing group) and the title (second capturing group):
(\d{6}(?:\.\d{2})) - (.*)(?:\.doc|\.docx)
Related
I am looking for a solution to validate and highlight my cell in case false.
I tried the most promising solution: Regex. But still can not find the pattern I need.
My latest attempt was this pattern: "[A-Z-0-9_.]" This works only if the cell contains only a symbol and nothing else, if the symbol is part of a string it does not work.
Problem is that it does not catch cells that have an odd character in a string of text: Example C4UNIT| or B$GROUP.
Specification Cell can contain only capital characters and two allowed symbols Dash - and Underbar _
This is my complete code:
Function ValidateCellContent()
Sheets("MTO DATA").Select
Dim RangeToCheck As Range
Dim CellinRangeToCheck As Range
Dim CollNumberFirst As Integer
Dim CollNumberLast As Integer
Dim RowNumberFirst As Integer
Dim RowNumberLast As Integer
'--Start on Column "1" and Row "3"
CollNumberFirst = 1
RowNumberFirst = 3
'--Find last Column used on row "2" (Write OMI Headings)
CollNumberLast = Cells(2, Columns.count).End(xlToLeft).Column
RowNumberLast = Cells(Rows.count, 1).End(xlUp).Row
'--Set value of the used range of cell addresses like: "A3:K85"
Set RangeToCheck = Range(Chr(64 + CollNumberFirst) & RowNumberFirst & ":" & Chr(64 + CollNumberLast) & RowNumberLast)
Debug.Print "Cells used in active Range = " & (Chr(64 + CollNumberFirst) & RowNumberFirst & ":" & Chr(64 + CollNumberLast) & RowNumberLast)
For Each CellinRangeToCheck In RangeToCheck
Debug.Print "CellinRangeToCheck value = " & CellinRangeToCheck
If Len(CellinRangeToCheck.Text) > 0 Then
'--Non Printables (Space,Line Feed,Carriage Return)
If InStr(CellinRangeToCheck, " ") _
Or InStr(CellinRangeToCheck, Chr(10)) > 0 _
Or InStr(CellinRangeToCheck, Chr(13)) > 0 Then
CellinRangeToCheck.Font.Color = vbRed
CellinRangeToCheck.Font.Bold = True
'--Allowed Characters
ElseIf Not CellinRangeToCheck.Text Like "*[A-Z-0-9_.]*" Then
CellinRangeToCheck.Font.Color = vbRed
CellinRangeToCheck.Font.Bold = True
Else
CellinRangeToCheck.Font.Color = vbBlack
CellinRangeToCheck.Font.Bold = False
End If
End If
Next CellinRangeToCheck
End Function
Try this:
Option Explicit
Private Sub Worksheet_Change(ByVal Target As Range)
'we want only validate when cell content changed, if whole range is involved (i.e. more than 1 cell) then exit sub
If Target.Cells.Count > 1 Then Exit Sub
'if there is error in a cell, also color it red
If IsError(Target) Then
Target.Interior.ColorIndex = 3
Exit Sub
End If
'validate cell with our function, if cell content is valid, it'll return True
'if it i s not valid, then color cell red
If Not ValidateText(Target.Value) Then
Target.Interior.ColorIndex = 3
End If
End Sub
Function ValidateText(ByVal txt As String) As Boolean
Dim i As Long, char As String
'loop through all characters in string
For i = 1 To Len(txt)
char = Mid(txt, i, 1)
If Not ((Asc(char) >= 65 And Asc(char) <= 90) Or char = "-" Or char = "_") Then
'once we come upon invalid character, we can finish the function with False result
ValidateText = False
Exit Function
End If
Next
ValidateText = True
End Function
I've originally assumed you wanted to use RegEx to solve your problem. As per your comment you instead seem to be using the Like operator.
Like operator
While Like accepts character ranges that may resemble regular expressions, there are many differences and few similarities between the two:
Like uses ! to negate a character range instead of the ^ used in RegEx.
Like does not allow/know quantifiers after the closing bracket ] and thus always matches a single character per pair of brackets []. To match multiple characters you need to add multiple copies of your character range brackets.
Like does not understand advanced concepts like capturing groups or lookahead / lookbehind
probably more differences...
The unavailability of quantifiers leaves Like in a really bad spot for your problem. You always need to have one character range to compare to for each character in your cell's text. As such the only way I can see to make use of the Like operator would be as follows:
Private Function IsTextValid(ByVal stringToValidate As String) As Boolean
Dim CharValidationPattern As String
CharValidationPattern = "[A-Z0-9._-]"
Dim StringValidationPattern As String
StringValidationPattern = RepeatString(CharValidationPattern, Len(stringToValidate))
IsTextValid = stringToValidate Like StringValidationPattern
End Function
Private Function RepeatString(ByVal stringToRepeat As String, ByVal repetitions As Long) As String
Dim Result As String
Dim i As Long
For i = 1 To repetitions
Result = Result & stringToRepeat
Next i
RepeatString = Result
End Function
You can then pass the text you want to check to IsTextValid like that:
If IsTextValid("A.ASDZ-054_93") Then Debug.Print "Hurray, it's valid!"
As per your comment, a small Worksheet_Change event to place into the worksheet module of your respective worksheet. (You will also need to place the above two functions there. Alternatively you can make them public and place them in a standard module.):
Private Sub Worksheet_Change(ByVal Target As Range)
Dim ValidationRange As Range
Set ValidationRange = Me.Range("A2:D5")
Dim TargetCell As Range
For Each TargetCell In Target.Cells
' Only work on cells falling into the ValidationRange
If Not Intersect(TargetCell, ValidationRange) Is Nothing Then
If IsTextValid(TargetCell.Text) Then
TargetCell.Font.Color = vbBlack
TargetCell.Font.Bold = False
Else
TargetCell.Font.Color = vbRed
TargetCell.Font.Bold = True
End If
End If
Next TargetCell
End Sub
Regular Expressions
If you want to continue down the RegEx road, try this expression:
[^A-Z0-9_-]+
It will generate a match, whenever a passed-in string contains one or more characters you don't want. All cells with only valid characters should not return a match.
Explanation:
A-Z will match all capital letters,
0-9 will match all numbers,
_- will match underscore and dash symbols.
The preceding ^ will negate the whole character set, meaning the RegEx only matches characters not in the set.
The following + tells the RegEx engine to match one or more characters of the aforementioned set. You only want to match your input, if there is at least one illegal char in there. And if there are more than one, it should still match.
Once in place, adapting the system to changing requirements (different chars considered legal) is as easy as switching out a few characters between the [brackets].
See a live example online.
I created a macro for removing all whitespace in a string, specifically an email address. However it only removes about 95% of the whitespace, and leaves a few.
My code:
Sub NoSpaces()
Dim w As Range
For Each w In Selection.Cells
w = Replace(w, " ", "")
Next
End Sub
Things I have tried to solve the issue include:
~ Confirmed the spaces are indeed spaces with the Code function, it is character 32 (space)
~ Used a substitute macro in conjuction with the replace macro
~ Have additional macro utilizing Trim function to remove leading and trailing whitespace
~ Made a separate macro to test for non-breaking spaces (character 160)
~ Used the Find and Replace feature to search and replace spaces with nothing. Confirmed working.
I only have one cell selected when I run the macro. It selects and goes through all the cells because of the Selection.Cells part of the code.
A few examples:
1 STAR MOVING # ATT.NET
322 TRUCKING#GMAIL.COM
ALEZZZZ#AOL. COM.
These just contain regular whitespace, but are skipped over.
Just use a regular expression:
'Add a reference to Microsoft VBScript Regular Expressions 5.5
Public Function RemoveWhiteSpace(target As String) As String
With New RegExp
.Pattern = "\s"
.MultiLine = True
.Global = True
RemoveWhiteSpace = .Replace(target, vbNullString)
End With
End Function
Call it like this:
Sub NoSpaces()
Dim w As Range
For Each w In Selection.Cells
w.Value = RemoveWhiteSpace(w.Value)
Next
End Sub
Try this:
Sub NoSpaces()
Selection.Replace " ", ""
End Sub
Use "Substitute"
Example...
=SUBSTITUTE(C1:C18," ","")
Because you assume that Selection.Cells includes all cells on the sheet.
Cells.Replace " ", ""
And to add to the excellent advice from all the great contributors, try the
TRIM or LTRIM, or RTRIM and you can read more about these functions here:
https://msdn.microsoft.com/en-us/library/office/gg278916.aspx
Now this does not remove embedded spaces (spaces in between the letters) but it will remove any leading and trailing spaces.
Hope this helps.
Space Problem with Excel
ok, the only way i see this two types of space is by converting their Ascii code value of which I do it here
now to explain this function i made, it will just filter the string character by character checking if its equal to the two types of space i mentioned. if not it will concatenate that character into the string which will be the final value after the loop. hope this helps. Thanks.
Function spaceremove(strs) As String
Dim str As String
Dim nstr As String
Dim sstr As String
Dim x As Integer
str = strs
For x = 1 To VBA.Len(str)
sstr = Left(Mid(str, x), 1)
If sstr = " " Or sstr = " " Then
Else
nstr = nstr & "" & sstr
End If
Next x
spaceremove = nstr
End Function
I copied a HTML table with data and pasted in excel but the cells were filled with unwanted space and all methods posted here didn't work so I debugged and I discovered that it wasn't actually space chars (ASCII 32) it was Non-breaking space) (ASCII 160) or HTML
So to make it work with that Non-breaking space char I did this:
Sub NoSpaces()
Dim w As Range
For Each w In Selection.Cells
w.Value = Replace(w.Value, " ", vbNullString)
w.Value = Replace(w.Value, Chr(160), vbNullString)
Next
End Sub
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
In this case, I want to extract the beginning text in a cell and leave the remainder intact.
e.g. a series of cells contain:
2nd Unit. Miami
3rd Production Staff. Toronto
1st Ad. San Francisco
I want to break this up without using Text to columns as previous rows are formatted differently and these last few rows are outliers that I want to handle.
I thought Regular Expressions might do it, but that seems a bit complex.
My algorithm idea is:
1. grab the wanted text (what function or custom sub would do that?)
2. Past the text to it's new location
3. Cut the text from the cell, leaving the remaining text.
Seems simple but I'm still wending my way through VBA forest, and at the rate I'm going it's going to end up faster doing it by hand. But this seems like a good opportunity to learn some VBA tricks.
TIA
Update:
I want to take the text up to the ".\ " and move it to a different column, keeping the remainder where it is.
VBA is unnecessary. To get the text after .\ in cell A1: =MID(A1,FIND(".\",A1,1)+2,LEN(A1)) to get the text before .\ in A1: =LEFT(A1,FIND(".\",A1,1)-1).
As additional information, Find returns the placement in the string where .\ appears. It is the equivalent of InStr in VBA. If .\ is not in the cell, it will display #VALUE, because I didn't bother to add error checking.
Since you seem to want to modify the cell text in place, VBA will be required.
Inside a loop that sets cl to the cell to be processed:
str = cl.value
i = Instr(str, ".\")
cl = Trim(Mid$(str, i + 2)) ' assuming you want to exclude the ".\"
cl.Offset(0, 1) Trim(Left$(str, i - 1)) ' Places the original first part one cell to the right
For the sake of anyone who had this same question, here is the fully tested, working code.
Function RE6(strData As String) As String
Dim RE As Object, REMatches As Object
Set RE = CreateObject("vbscript.regexp")
With RE
.MultiLine = False
.Global = False
.IgnoreCase = True
.Pattern = "[0-9][0-9][0-9][0-9]B"
RE6 = .test(strData)
End With
Set REMatches = RE.Execute(strData)
If REMatches.Count > 0 Then
RE6 = True
Else
RE6 = False
End If
End Function
Sub territory()
Dim strTest As String, str As String, cl As Range
strTest = ActiveCell.Value
Set cl = ActiveCell
If RE6(strTest) = True Then
str = cl.Value
i = InStr(str, ". ")
cl = Trim(Mid$(str, i + 2))
cl.Offset(0, 1) = Trim(Left(str, i - 1))
cl.Offset(0, 2) = "Instance"
MsgBox RE6(strTest)
End If
End Sub
I have 10 lines of array that are first name space last name space zip code. All the zip codes start with different numbers. Is there a way to replace the #1 in the indexof below so that it searches for any number character instead?
'open file
inFile = IO.File.OpenText("Names.txt")
'process the loop instruct until end of file
intSubscript = 0
Do Until inFile.Peek = -1 OrElse intSubscript = strLine.Length
strLine(intSubscript) = inFile.ReadLine
intSubscript = intSubscript + 1
Loop
inFile.Close()
intSubscript = 0
strFound = "N"
Do Until strFound = "Y" OrElse intSubscript = strLine.Length
intIndex = strLine(intSubscript).IndexOf("1")
strName = strLine(intSubscript).Substring(0, intIndex - 1)
If strName = strFullname Then
strFound = "Y"
strZip = strLine(intSubscript).Substring(strLine(intSubscript).Length - 5, 5)
txtZip.Text = strZip
End If
Loop
End Sub
use a regular expression.
Regular expressions allow you to do pattern matching on text. It's like String.IndexOf() with wildcard support.
For example, suppose your source data looks like this:
James Harvey 10939
Madison Whittaker 33893
George Keitel 22982
...and so on.
Expressed in English, the pattern each line follows is this:
the beginning of the string, followed by
a sequence of 1 or more alphabetic characters, followed by
a sequence of one or more spaces, followed by
a sequence of 1 or more alphabetic characters, followed by
a sequence of one or more spaces, followed by
a sequence of 5 numeric digits, followed by
the end of the string
You can express that very precisely and succintly in regex this way:
^([A-Za-z]+) +([A-Za-z]+) +([0-9]{5})$
Apply it in VB this way:
Dim sourcedata As String = _
"James Harvey 10939" & _
vbcrlf & _
"Madison Whittaker 33893" & _
vbcrlf & _
"George Keitel 22982"
Dim regex = "^([A-Za-z]+) +([A-Za-z]+) +([0-9]{5})$"
Dim re = New Regex(regex)
Dim lineData As String() = sourceData.Split(vbcrlf.ToCharArray(), _
StringSplitOptions.RemoveEmptyEntries )
For i As Integer = 0 To lineData.Length -1
System.Console.WriteLine("'{0}'", lineData(i))
Dim matchResult As Match = re.Match(lineData(i))
System.Console.WriteLine(" zip: {0}", matchResult.Groups(3).ToString())
Next i
To get that code to compile, you must import the System.Text.RegularExpressions namespace at the top of your VB module, to get the Regex and Match types.
If your input data follows a different pattern, then you will need to adjust your regex.
For example if it could be "Chris McElvoy III 29828", then you need to adjust the regex accordingly, to handle the name suffix.