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.
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$(temp_str, Len(temp_str) - 1)
End Function
That's the fairly easy part. If you are only going to handle people's names it still needs more work to handle names like MacFarland, O'Connor, etc.
Business names get more complicated with since they can have a name like "Village on the Lake Apartments" where some words are not capitalized. It's a legal business name so the capitalization is important.
Professional and business suffixes can also be problematic if everything is in lower case - like phd should be PhD, llc should be LLC, and iii, as in John Smith III, would come out Iii.
There is also a VB6 function that will capitalize the first letter of each word. It is StrConv(string,vbProperCase) but it also sets everything that is not the first letter to lower case. So PhD becomes Phd and III becomes Iii. Where as the above code does not change the trailing portion to lower case so if it is entered correctly it remains correct.
Try this
Option Explicit
Private Sub Form_Load()
MsgBox capitalize("abdirahman abdirisaq ali")
MsgBox capitalize("abdirahman abdirisaq ali")
End Sub
Private Function capitalize(txt As String) As String
Dim Names() As String
Dim NewNames() As String
Dim i As Integer
Dim j As Integer
Names = Split(txt, " ")
j = 0
For i = 0 To UBound(Names)
If Names(i) <> "" Then
Mid(Names(i), 1, 1) = UCase(Left(Names(i), 1))
ReDim Preserve NewNames(j)
NewNames(j) = Names(i)
j = j + 1
End If
Next
capitalize = Join(NewNames, " ")
End Function
Use the VB6 statement
Names = StrConv(Names, vbProperCase)
it's all you need (use your own variable instead of Names)
Suppose I have a variable set to the path of an image.
Let img = "C:\Users\Example\Desktop\Test\Stuff\Icons\test.jpg"
I want to slice everything before "\Icons" using Vb6. So after slicing the string it would be "\Icons\test.jpg" only.
I have tried fiddling with the Mid$ 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
I want to reduce the decimal length
text1.text = 2137.2198231578
From the above, i want to show only first 2 digit decimal number
Expected Output
text1.text = 2137.21
How to do this.
Format("2137.2198231578", "####.##")
I was about to post use Format() when I noticed p0rter comment.
Format(text1.text, "000.00")
I guess Int() will round down for you.
Been many years since I used VB6...
This function should do what you want (inline comments should explain what is happening):
Private Function FormatDecimals(ByVal Number As Double, ByVal DecimalPlaces As Integer) As String
Dim NumberString As String
Dim DecimalLocation As Integer
Dim i As Integer
Dim LeftHandSide As String
Dim RightHandSide As String
'convert the number to a string
NumberString = CStr(Number)
'find the decimal point
DecimalLocation = InStr(1, NumberString, ".")
'check to see if the decimal point was found
If DecimalLocation = 0 Then
'return the number if no decimal places required
If DecimalPlaces = 0 Then
FormatDecimals = NumberString
Exit Function
End If
'not a floating point number so add on the required number of zeros
NumberString = NumberString & "."
For i = 0 To DecimalPlaces
NumberString = NumberString & "0"
Next
FormatDecimals = NumberString
Exit Function
Else
'decimal point found
'split out the string based on the location of the decimal point
LeftHandSide = Mid(NumberString, 1, DecimalLocation - 1)
RightHandSide = Mid(NumberString, DecimalLocation + 1)
'if we don't want any decimal places just return the left hand side
If DecimalPlaces = 0 Then
FormatDecimals = LeftHandSide
Exit Function
End If
'make sure the right hand side if the required length
Do Until Len(RightHandSide) >= DecimalPlaces
RightHandSide = RightHandSide & "0"
Loop
'strip off any extra didgits that we dont want
RightHandSide = Left(RightHandSide, DecimalPlaces)
'return the new value
FormatDecimals = LeftHandSide & "." & RightHandSide
Exit Function
End If
End Function
Usage:
Debug.Print FormatDecimals(2137.2198231578, 2) 'outputs 2137.21
Looks fairly simple, but I must be missing something subtle here. What about:
Option Explicit
Private Function Fmt2Places(ByVal Value As Double) As String
Fmt2Places = Format$(Fix(Value * 100#) / 100#, "0.00")
End Function
Private Sub Form_Load()
Text1.Text = Fmt2Places(2137.2198231578)
End Sub
This also works in locales where the decimal point character is a comma.
Using VB6
Code.
Dim posn As Integer, i As Integer
Dim fName As String
posn = 0
For i = 1 To Len(flname)
If (Mid(flname, i, 1) = "\") Then posn = i
Next i
fName = Right(flname, Len(flname) - posn)
posn = InStr(fName, ".")
If posn <> 0 Then
fName = Left(fName, posn - 1)
End If
GetFileName = fName
FileName: Clockings8.mis06042009 120442.fin
But it is showing a filename is “Clockings8”. It should show “Clockings8.mis06042009 120442”
How to modify a code?
Need vb6 code Help
It is a bit cleaner to use the Scripting.FileSystemObject component. Try:
Dim fso as New Scripting.FileSystemObject
GetFileName = fso.GetBaseName(fname)
The reason why your code stops short is that InStr works from the beginning of the string to the end and stops wherever it finds a match. The filename "Clockings8.mis06042009 120442.fin" contains two periods. For this reason, you should use InStrRev instead to start the search from the end of the string.
Going with FileSystemObject's GetBaseName like David suggests is a good idea. If you can't or don't want to (and there are reasons why you might not want to) work with the FileSystemObject there's a simple solution: Remove all characters from a filename string starting with the last dot in the name.
Here's what I mean:
Dim fn As String
fn = "Clockings8.mis06042009 120442.fin"
Dim idx As Integer
idx = InStrRev(fn, ".")
GetFileName = Mid(fn, 1, idx - 1)
If your filename does not have an extenstion but has a dot somewhere in the filename string, then this method will return bad results.