Some of us unfortunately are still supporting legacy app like VB6. I have forgotten how to parse a string.
Given a string:
Dim mystring As String = "1234567890"
How do you loop in VB6 through each character and do something like
for each character in mystring
debug.print character
next
In C# i would do something like
char[] myChars = mystring.ToCharArray();
foreach (char c in theChars)
{
//do something with c
}
Any ideas?
Thanks a lot
You can use the 'Mid' function to get at the individual characters:
Dim i As Integer
For i = 1 To Len(mystring)
Print Mid$(mystring, i, 1)
Next
Note this is untested.
There is no possibility to use foreach on strings.
Use
Dim i As Integer
For i = 1 To Len(YourString)
Result = Mid$(YourString, i, 1)
Next
note that the type of Result is a length-1 string, no char or byte type.
If performance is important, you'll have to convert the string to a bytearray fist (using StrConv) and then loop through it like this.
Dim i As Long
For i = 0 To UBound(Data)
Result = Data(i) ' Type is Byte '
Next
This is much more efficient.
The easiest way is to convert the string into an array of bytes and iterate over the byte array (converting each byte to a character).
Dim str As String
Dim bytArray() As Byte
Dim count As Integer
str = "This is a string."
bytArray = str
For count = 0 To UBound(bytArray)
Debug.Print Chr(bytArray(count))
Next
Don't loop; rather, set a reference to Microsoft VBScript Regular Expressions library and use regular expressions to achieve your 'do something' goal.
Related
I have a string that looks something like 'NS-BATHROOMS 04288'
I only want the numbers.
I hve searched for answers, but none so far even get pst the compiler.
How can I do this?
without regex you can do it with: (altough VB6/VBA Code really isn`t nice to look at)
Public Function ReturnNonAlpha(ByVal sString As String) As String
Dim i As Integer
For i = 1 To Len(sString)
If Mid(sString, i, 1) Like "[0-9]" Then
ReturnNonAlpha = ReturnNonAlpha + Mid(sString, i, 1)
End If
Next i
End Function
I'd personally use regex. You can match given regex patterns to achieve what you need. This function matches only digits.
For VB6 you'd do something like:
Dim myRegExp, ResultString
Set myRegExp = New RegExp
myRegExp.Global = True
myRegExp.Pattern = "[\d]"
Then you'd go against your String.
https://support.microsoft.com/en-us/kb/818802
You can use this function for extract numerical chr as string value:
Public Function Str_To_Int(MyString As Variant) As Long
Dim i As Integer
Dim X As Variant
If IsNull(MyString) Or MyString = "" Then
Str_To_Int = 0
Exit Function
End If
For i = 1 To Len(MyString)
If IsNumeric(Mid(MyString, i, 1)) = True Then
X = Nz(X, "") & Mid(MyString, i, 1)
End If
Next i
Str_To_Int = Nz(X, 0)
End Function
I want to check if an address starts with http://www.youtube.com.
If I have something like this
if rs("mainVideoName")="http://www.youtube.com*" then
This doesn't work, so how can I do it?
Try this:
Function UrlStartsWith(string1, string2)
UrlStartsWith = InStr(1, string1, string2, 1) = 1
End Function
If UrlStartsWith(rs("mainVideoName"), "http://www.youtube.com") Then
End If
Starts with is tested by using IntStr and ensuring that it returns 1 as the starting position that the search string is found. Since you are testing a URL the code above uses a TextCompare to make insensitive to case.
You can use the InStr() function for this:
Dim positionOfMatchedString
positionOfMatchedString= InStr(rs("mainVideoName"),"http://www.youtube.com")
If positionOfMatchedString > 0 Then
// Do your stuff here
End If
As Anthony points out, this tells you that string2 is contained in string1. You could write it as:
If positionOfMatchedString = 1 Then
to check for it beginning with.
How about...
Dim s: s = "http://www.youtube.com"
Dim l: l = Len(s)
If Left(rs("mainVideoName"), l) = s Then
' String begins with URL part '
End If
Working on converting a Visual Basic SHA-256 encryption routine to work in LotusScript.
Is going well except for the VB's AscB command.
Found: "Use the AscB function to return the first byte of a string containing byte data."
Not finding way to do same in LotusScript.
See the LS CByte command comes close: "CByte returns an expression that has been converted to Byte."
Don't see way to have it return just the first Byte of the expression.
Any suggestions?
Derek
AscB is only appropriate for strings in single-byte character encoding. All LotusScript string data is Unicode represented in UTF16 double-byte encoding.
The LotusScript Uni() function returns a Long containing the integer value of the Unicode character. Since the input is a double byte character, the value returned by Uni() ranges from 0 to 65535. If you want to get the values of each of the two bytes, code like this will do the trick:
Dim ws As New NotesUIWorkspace
Dim s1 As String
Dim u1 As Long
Dim u2 As Long
Dim lowbyte As Integer
Dim highbyte As Integer
Dim b1 As Byte
Dim b2 as Byte
s1 = "Ʃ"
u1 = Uni(s1)
lowbyte = u1 Mod 256
highbyte = (u1 - lowbyte) / 256
b1 = Cbyte(lowbyte)
b2 = Cbyte(highbyte)
Call ws.Prompt(prompt_ok,"test",s1 + " " + Cstr(Cint(b1)) + " " + Cstr(Cint(b2)))
Would Asc, LeftB and RightB do what you need?
In my testing...
Lenb("A") = 2
Leftb("A", 1) = "A"
Asc(Leftb("A", 1)) = 65
Leftb("A", 2) = "A"
Asc(Leftb("A", 2)) = 65
Asc(Rightb(Leftb("A", 2), 1)) = 0
I have unicode string passed to vbscript procedure (not visual basic 6, but vbscript). I want to iterate unicode string char by char, get code for every symbol, truncate code to byte range [0..255] and create array of bytes.
This way new array of bytes should be twice smaller in memory compared to original unicode string. I am going save this array to file via ADODB.Stream object further
How can I convert unicode string to bytes array with symbol code truncated to byte range?
Thank you in advance!
Firstly, translating unicode to ascii will only work if your string only contains ascii characters. Since unicode contains ascii, it is just a matter of removing every second character.
Look up unicode on the internet for details.
EDIT: In unicode, every ascii character is proceeded with a NULL (0) byte. Remove this byte to convert the string to ASCII.
It seems there is no way to create array of bytes in vbs (though it's very straightforward in visual basic) -- all arrays are arrays of variants.
The task was to send binary stream from server to vbs script via string type. I have found the solution by creating Xml Document on the server with CDATA section that contains base64 coded array of bytes as string data.
Client (vbs) do the following:
set xmlDoc = CreateObject("Microsoft.XmlDom")
xmlDoc.loadXML(dataFromServer)
base64str = xmlDoc.DocumentElement.Text ' it's base64 coded binary stream
arrayOfBytes = decodeBase64(base64str)
Function decodeBase64(base64)
set dm = CreateObject("Microsoft.XMLDOM")
set el = dm.createElement("tmp")
el.DataType = "bin.base64"
el.Text = base64
decodeBase64 = el.NodeTypedValue
set dm = Nothing
End Function
This function creates an array of bytes:
' http://www.motobit.com/tips/detpg_binarytostring/
Function MultiByteToBinary(MultiByte)
'� 2000 Antonin Foller, http://www.motobit.com
' MultiByteToBinary converts multibyte string To real binary data (VT_UI1 | VT_ARRAY)
' Using recordset
Dim RS, LMultiByte, Binary
Const adLongVarBinary = 205
Set RS = CreateObject("ADODB.Recordset")
LMultiByte = LenB(MultiByte)
If LMultiByte>0 Then
RS.Fields.Append "mBinary", adLongVarBinary, LMultiByte
RS.Open
RS.AddNew
RS("mBinary").AppendChunk MultiByte & ChrB(0)
RS.Update
Binary = RS("mBinary").GetChunk(LMultiByte)
End If
MultiByteToBinary = Binary
End Function
This function creates a multi-byte string.
' http://www.motobit.com/help/regedit/pa26.htm
'Converts unicode string to a multibyte string
Function StringToMB(S)
Dim I, B
For I = 1 To Len(S)
B = B & ChrB(Asc(Mid(S, I, 1)))
Next
StringToMB = B
End Function
Is it possible to test a string with IsNumeric() and for it to return true, but when you cast that same string to an integer using CInt() and assign it to a variable of type integer that it will give a type mismatch error?
I ask because I was getting a type mismatch error, so I used IsNumeric() to check the string was numeric before trying to cast it, but I still get the error.
I am tearing my hair out with this.
Here is the code in question.
iGlobalMaxAlternatives = CInt(strMaxAlternatives) is where the error is occuring.
Dim strMaxAlternatives As String
Dim iGlobalMaxAlternatives As Integer
iGlobalMaxAlternatives = 0
bSurchargeIncInFare = True
strMaxAlternatives = ReadStringKeyFromRegistry("Software\TL\Connection Strings\" & sConn & "\HH", "MaxAlt")
If IsNumeric(strMaxAlternatives) And strMaxAlternatives <> "" Then
iGlobalMaxAlternatives = CInt(strMaxAlternatives)
End If
You may have an overflow due the maximum integer size; the currency type actually does very well for large numbers (but beware of any regional issues). See edits below for Int64 discussion.
According to MSDN documentation on IsNumeric:
IsNumeric returns True if the data
type of Expression is Boolean, Byte,
Decimal, Double, Integer, Long,
SByte, Short, Single, UInteger,
ULong, or UShort, or an Object that
contains one of those numeric types.
It also returns True if Expression is
a Char or String that can be
successfully converted to a number.
IsNumeric returns False if Expression
is of data type Date or of data type
Object and it does not contain a
numeric type. IsNumeric returns False
if Expression is a Char or String
that cannot be converted to a number.
Since you are getting a Type Mismatch, perhaps a Double is interfering with the conversion. The IsNumeric does not guarantee it is an Integer, just that it matches one of the numeric possibilities. If the number is a double, perhaps regional settings (comma versus period and so on) are causing the exception.
You might try converting it to a double and then to an integer.
' Using a couple of steps
Dim iValue As Integer
Dim dValue As Double
dValue = CDbl(SourceValue)
iValue = CInt(iValue)
' Or in one step (might make debugging harder)
iValue = CInt(CDbl(SourceValue))
EDIT: After your clarification, it appears you are getting an overflow conversion. First try using a Long and CLng() instead of CInt(). There is still a chance the entry is Int64 though, which is more difficult using VB6.
I have used the following code for the LARGE_INTEGER and Integer8 types (both Int64), but it may not work for your situation:
testValue = CCur((inputValue.HighPart * 2 ^ 32) + _
inputValue.LowPart) / CCur(-864000000000)
This example was from an LDAP password expiration example, but like I said it may or may not work in your scenario. If you don't have the LARGE_INTEGER type, it looks like:
Private Type LARGE_INTEGER
LowPart As Long
HighPart As Long
End Type
Search for LARGE_INTEGER and VB6 for more information.
EDIT: For debugging, it may be useful to temporarily avoid error handling and then turn it back on after passing the troubling lines:
If IsNumeric(strMaxAlternatives) And strMaxAlternatives <> "" Then
On Error Resume Next
iGlobalMaxAlternatives = CInt(strMaxAlternatives)
If Err.Number <> 0 Then
Debug.Print "Conversion Error: " & strMaxAlternatives & _
" - " & Err.Description
EndIf
On Error Goto YourPreviousErrorHandler
End If
Yes, "3.41" would be numeric but not an integer.
VB6 doesn't provide good methods to guarantee CInt won't fail. I've found the simplest way is to just use error-handling.
function TryParse(byval text as string, byref value as integer) as boolean
on error resume next
value = CInt(text)
TryParse = (err.number = 0)
endfunction
Of course your error-handling preferences may vary.
Yes. Try this:
If IsNumeric("65537") Then
Dim i As Integer
i = CInt("65537") 'throws an error on this line!
End If
This one's an overflow, but I think it illustrates the unreliability of IsNumeric() in general (especially for ints - for doubles it's much more reliable).
According to the VB6 documentation, "IsNumeric returns True if the data type of Expression is Boolean, Byte, Decimal, Double, Integer, Long, SByte, Short, Single, UInteger, ULong, or UShort, or an Object that contains one of those numeric types. It also returns True if Expression is a Char or String that can be successfully converted to a number."
Many of those cannot be converted to an Integer. For example "1.5" is numeric but it's not an integer. So, you can convert it to a number, but not necessarily an integer.
The following code works without a Type Mismatch error in Visual BASIC 6
Dim I As Integer
I = CInt("3.41")
The same for this variant
Dim I As Integer
Dim TempS As String
TempS = "3.41"
I = CInt(TempS)
Posting the code in question would help answer your question. Basically there are several function in VB6 that are used to convert strings into number.
CInt and Int convert into number but handle rounding different. Direct assignment works and equivalent to using CInt. Howver I recommend continuing to use CInt to make the operation clear to you and your fellow developers in the future.
CInt works on number with commas like "3,041.41" However VB6 has problem handling region settings so if you are using notation other than standard American English you will get strange results and errors.
Your best bet is to start logging the errors with the actual values it's working with so you can figure out whats going on.
IsNumeric will return true if it can convert the string to a number. Even if there are non-numeric characters in the string. I always loop though the string one character at a time and test each character. If one character fails then I can return an error.
Just found this nugget. If you run the following, script #1 returns TRUE but script #2 & #3 will fail:
SELECT ISNUMERIC('98,0') AS isNum -- Fails
SELECT CONVERT(INT, '98,0') -- Fails
SELECT CONVERT(NUMERIC(11,4), '98,0') -- Fails
Two options...
Change
If IsNumeric(strMaxAlternatives) And strMaxAlternatives <> "" Then
iGlobalMaxAlternatives = CInt(strMaxAlternatives)
End If
To
If IsNumeric(strMaxAlternatives) And strMaxAlternatives <> "" Then
iGlobalMaxAlternatives = CDbl(strMaxAlternatives) ' Cast to double instead'
End If
Or
If IsNumeric(strMaxAlternatives) And strMaxAlternatives <> "" Then
If CDbl(strMaxAlternatives) Mod 1 = 0 Then ' Make sure there\'s no decimal points'
iGlobalMaxAlternatives = CInt(strMaxAlternatives)
End If
End If