What are the rules that vb6 uses to find the apostrophe that marks the beginning of a commented
out portion of a line(s)?
I don't feel confident about my ability to define such rules because of :
apostrophes inside string literals
apostrophes inside nested double quotes in string operations
the fact that double quotes can occur in comments
the fact that a line can be continued over multiple lines
colons breaking up a single line
possible other rules that I'm unaware of
If you really want to know, the VBA language specification is online on MSDN with a BNF grammar. VBA is 99% equivalent to VB6 - certainly the rules about identifying comments must be identical.
And if you are trying to eliminate dead code - just get MZ-Tools! It's free. Use the code review tool. "MZ-Tools can review your source code at project-group, project or file level (through context menus) detecting unused variables, constants, parameters, private procedures, and so on." To eliminate unused subs and functions, use the MZ-Tools feature that lists all callers.
EDIT There is discussion in the comments about how to eliminate items which have been unecessarily declared as Public: MZ Tools does not help with this.
Apostrophes in string literals are ignored. Apostrophes in comments also ignored (because they are already in a comment). You cannot put as apostrophe in a multi-line statement unless it is on the last line of the statement -- in which case it just follows the normal rules.
Why are you so concerned?
This seems to work on any examples I tried
The only rule it's using is to ignore apostrophes inside string literals
and it doesn't account for the use of the word rem to start a comment.
To my shame I never realised until yesterday that """" (4 double quotes) was a string literal containing a single double quote.
Public Function CommentStripOut(ByVal strLine As String) As String
Dim InLiteral As Boolean
Dim strReturn As String
Dim LenLine As Long
Dim counter As Long
Dim s1 As String
Dim s2 As String
strReturn = strLine
LenLine = Len(strLine)
InLiteral = False
For counter = 1 To LenLine
s1 = Mid$(strLine, counter, 1)
If counter < LenLine Then
s2 = Mid$(strLine, counter + 1, 1)
Else
s2 = ""
End If
If s1 = """" Then
If Not InLiteral Then
InLiteral = True
Else
If s2 = """" Then
counter = counter + 1
'skip on by 1 because
'want to treat escaped
'double quote as a single
'character
Else
InLiteral = False
End If
End If
Else
If Not InLiteral Then
If s1 = "'" Then
strReturn = Left$(strLine, counter - 1)
Exit For
End If
End If
End If
Next counter
CommentStripOut = strReturn
End Function
Related
hi all i have this question as bellow
how capitalize full in one vb6 Vb6 string variable
‘example
‘my fullname
Dim fullname as string
Fullname = “abdirahman abdirisaq ali”
Msgbox capitalize(fullname)
it prints abdirahmanAbdirisaq ali that means it skips the middle name space even if I add more spaces its same .
this is my own code and efforts it takes me at least 2 hours and still .
I tired it tired tired please save me thanks more.
Please check my code and help me what is type of mistakes I wrote .
This is my code
Private Function capitalize(txt As String) As String
txt = LTrim(txt)
temp_str = ""
Start_From = 1
spacing = 0
For i = 1 To Len(txt)
If i = 1 Then
temp_str = UCase(Left(txt, i))
Else
Start_From = Start_From + 1
If Mid(txt, i, 1) = " " Then
Start_From = i
spacing = spacing + 1
temp_str = temp_str & UCase(Mid(txt, Start_From + 1, 1))
Start_From = Start_From + 1
Else
temp_str = temp_str & LCase(Mid(txt, Start_From, 1))
End If
End If
Next i
checkName = temp_str
End Function
It's far simpler than that. In VB6 you should use Option Explicit to properly type your variables. That also requires you to declare them.
Option Explicit
Private Function capitalize(txt As String) As String
Dim temp_str as String
Dim Names As Variant
Dim Index As Long
'Remove leading and trailing spaces
temp_str = Trim$(txt)
'Remove any duplicate spaces just to be sure.
Do While Instr(temp_str, " ") > 0
temp_str = Replace(temp_str, " ", " ")
Loop
'Create an array of the individual names, separating them by the space delimiter
Names = Split(temp_str, " ")
'Now put them, back together with capitalisation
temp_str = vbnullstring
For Index = 0 to Ubound(Names)
temp_str = temp_str + Ucase$(Left$(Names(Index),1)) + Mid$(Names(Index),2) + " "
Next
'Remove trailing space
capitalize = Left$(temp_str, Len(temp_str) - 1)
End Function
That's the fairly easy part. If you are only going to handle people's names it still needs more work to handle names like MacFarland, O'Connor, etc.
Business names get more complicated with since they can have a name like "Village on the Lake Apartments" where some words are not capitalized. It's a legal business name so the capitalization is important.
Professional and business suffixes can also be problematic if everything is in lower case - like phd should be PhD, llc should be LLC, and iii, as in John Smith III, would come out Iii.
There is also a VB6 function that will capitalize the first letter of each word. It is StrConv(string,vbProperCase) but it also sets everything that is not the first letter to lower case. So PhD becomes Phd and III becomes Iii. Where as the above code does not change the trailing portion to lower case so if it is entered correctly it remains correct.
Try this
Option Explicit
Private Sub Form_Load()
MsgBox capitalize("abdirahman abdirisaq ali")
MsgBox capitalize("abdirahman abdirisaq ali")
End Sub
Private Function capitalize(txt As String) As String
Dim Names() As String
Dim NewNames() As String
Dim i As Integer
Dim j As Integer
Names = Split(txt, " ")
j = 0
For i = 0 To UBound(Names)
If Names(i) <> "" Then
Mid(Names(i), 1, 1) = UCase(Left(Names(i), 1))
ReDim Preserve NewNames(j)
NewNames(j) = Names(i)
j = j + 1
End If
Next
capitalize = Join(NewNames, " ")
End Function
Use the VB6 statement
Names = StrConv(Names, vbProperCase)
it's all you need (use your own variable instead of Names)
I need help on how to remove spaces/emtpy in data without compromising spaces on other data. Here's my sample data.
12345," ","abcde fgh",2017-06-06,09:00,AM," ", US
expected output:
12345,,"abcde fgh",2017-06-06,09:00,AM,, US
since " " should be considered as null.
I tried the Trim() function but it did not work. I also tried Regex pattern but still no use.
Here's my sample function.
Private Sub Transform(delimiter As String)
Dim sFullPath As String
Dim strBuff As String
Dim re As RegExp
Dim matches As Object
Dim m As Variant
If delimiter <> "," Then
strBuff = Replace(strBuff, delimiter, ",")
Else
With re
.Pattern = "(?!\B""[^""]*)" & delimiter & "(?![^""]*""\B)"
.IgnoreCase = False
.Global = True
End With
Set matches = re.Execute(strBuff)
For Each m In matches
strBuff = re.Replace(strBuff, ",")
Next
Set re = Nothing
Set matches = Nothing
End If
End Sub
I think you're on the right track. Try using this for your regular expression. The two double quotes in a row are how a single double quote is included in a string literal. Some people prefer to use Chr(34) to include double quotes inside a string.
\B(\s)(?!(?:[^""]*""[^""]*"")*[^""]*$)
Using that expression on your example string
12345," ","abcde fgh",2017-06-06,09:00,AM," ", US
yields
12345,"","abcde fgh",2017-06-06,09:00,AM,"", US
Example function
Private Function Transform(ByVal strLine As String) As String
Dim objRegEx As RegExp
On Error GoTo ErrTransForm
Set objRegEx = New RegExp
With objRegEx
.Pattern = "\B(\s)(?!(?:[^""]*""[^""]*"")*[^""]*$)"
.IgnoreCase = False
.Global = True
Transform = .Replace(strLine, "")
End With
ExitTransForm:
If Not objRegEx Is Nothing Then
Set objRegEx = Nothing
End If
Exit Function
ErrTransForm:
'error handling code here
GoTo ExitTransForm
End Function
And credit where credit is due. I used this answer, Regex Replace Whitespaces between single quotes as the basis for the expression here.
I would add an output string variable and have a conditional statement saying if the input is not empty, add it on to the output string. For example (VB console app format with the user being prompted to enter many inputs):
Dim input As String
Dim output As String
Do
input = console.ReadLine()
If Not input = " " Then
output += input
End If
Loop Until (end condition)
Console.WriteLine(output)
You can throw any inputs you don't want into the conditional to remove them from the output.
Your CSV file isn't correctly formatted.
Double quote shouldn't exists, then open your CSV with Notepad and replace them with a null string.
After this, you now have a real CSV file that you can import whitout problems.
I'm trying to make a program that can take the letters of a string and invert their case. I know that in vb.net there exists the IsUpper() command, but I don't know of such a thing in vb6.
What can I use in place of it?
Thanks!
Something like this should work:
Private Function Invert(strIn As String) As String
Dim strOut As String
Dim strChar As String
Dim intLoop As Integer
For intLoop = 1 To Len(strIn)
strChar = Mid(strIn, intLoop, 1)
If UCase(strChar) = strChar Then
strChar = LCase(strChar)
Else
strChar = UCase(strChar)
End If
strOut = strOut + strChar
Next
Invert = strOut
End Function
This loops through the supplied string, and extracts each character. It then tries to convert it to upper case and checks it against the extracted character. If it's the same then it was already upper case, so it converts it to lower case.
It handles non alpha characters just fine as UCase/LCase ignores those.
Suppose I have a variable set to the path of an image.
Let img = "C:\Users\Example\Desktop\Test\Stuff\Icons\test.jpg"
I want to slice everything before "\Icons" using Vb6. So after slicing the string it would be "\Icons\test.jpg" only.
I have tried fiddling with the Mid$ function in VB6, but I haven't really had much success. I am aware of the fact that Substring isn't a function available in vb6, but in vb.net only.
After the first \icons
path = "C:\Users\Example\Desktop\Test\Stuff\Icons\test.jpg"
?mid$(path, instr(1, path, "\icons\", vbTextCompare))
> \Icons\test.jpg
Or after the last should there be > 1
path = "C:\Users\Example\Desktop\Test\Icons\Stuff\Icons\test.jpg"
?right$(path, len(path) - InStrRev(path, "\icons\", -1, vbTextCompare) + 1)
> \Icons\test.jpg
This is pretty easy to do generically using the Split function. I wrote a method to demonstrate it's use and for grins it takes an optional parameter to specify how many directories you want returned. Passing no number returns a file name, passing a very high number returns a full path (either local or UNC). Please note there is no error handling in the method.
Private Function GetFileAndBasePath(ByVal vPath As String, Optional ByVal baseFolderLevel = 0) As String
Dim strPathParts() As String
Dim strReturn As String
Dim i As Integer
strPathParts = Split(vPath, "\")
Do While i <= baseFolderLevel And i <= UBound(strPathParts)
If i > 0 Then
strReturn = strPathParts(UBound(strPathParts) - i) & "\" & strReturn
Else
strReturn = strPathParts(UBound(strPathParts))
End If
i = i + 1
Loop
GetFileAndBasePath = strReturn
End Function
I have a string like X5BC8373XXX. Where X = a special character equals a Square.
I also have some special characters like \n but I remove them, but I can't remove the squares...
I'd like to know how to remove it.
I Found this method:
Dim Test As String
Test = Replace(Mscomm1.Input, Chr(160), Chr(64) 'Here I remove some of the special characters like \n
Test = Left$(Test, Len(Test) -2)
Test = Right$(Test, Len(Test) -2)
This method DOES remove those special characters, but it's also removing my first character 5.
I realize that this method just remove 2 characters from the left and the right,
but how could I work around this to remove these special characters ?
Also I saw something with vblF, CtrlF something like this, but I couldn't work with this ;\
You can use regular expressions. If you want to remove everything that's not a number or letter, you can use the code below. If there are other characters you want to keep, regular expressions are highly customizable, but can get a little confusing.
This also has the benefit of doing the whole string at once, instead of character by character.
You'll need to reference Microsoft VBScript Regular Expressions in your project.
Function AlphaNum(OldString As String)
Dim RE As New RegExp
RE.Pattern = "[^A-Za-z0-9]"
RE.Global = True
AlphaNum = RE.Replace(OldString, "")
End Function
Cleaning out non-printable characters is easy enough. One brute-force but easily customizable method might be:
Private Function Printable(ByVal Text As String) As String
Dim I As Long
Dim Char As String
Dim Count As Long
Printable = Text 'Allocate space, same width as original.
For I = 1 To Len(Text)
Char = Mid$(Text, I, 1)
If Char Like "[ -~]" Then
'Char was in the range " " through "~" so keep it.
Count = Count + 1
Mid$(Printable, Count, 1) = Char
End If
Next
Printable = Left$(Printable, Count)
End Function
Private Sub Test()
Dim S As String
S = vbVerticalTab & "ABC" & vbFormFeed & vbBack
Text1.Text = S 'Shows "boxes" or "?" depending on the font.
Text2.Text = Printable(S)
End Sub
This will remove control characters (below CHR(32))
Function CleanString(strBefore As String) As String
CleanString = ""
Dim strAfter As String
Dim intAscii As Integer
Dim strTest As String
Dim dblX As Double
Dim dblLen As Double
intLen = Len(strBefore)
For dblX = 1 To dblLen
strTest = Mid(strBefore, dblX, 1)
If Asc(strTest) < 32 Then
strTest = " "
End If
strAfter = strAfter & strTest
Next dblX
CleanString = strAfter
End Function