I am a long time C# developer and have had to look at some VB6 code lately. After some digging, we found out that, somewhere, somehow, we are storing a number as a string. Now we are reading it back and our code is not handling culture differences very well. Here is an example:
Dim text as String
Dim myVal as Double
Set text = "1,123456"
'This sets the value of myVal to 1123456 on our system - incorrect
myVal = text
Set text = "1.123456"
'This sets the value of myVal to 1.123456 on our system - correct
myVal = text
Keeping in mind that this is VB6 and NOT VB.NET, are there any built-in methods, functions, whatever, that can convert a string to a double using a particular culture?
Or at the very least, hinting to the conversion that we may be dealing with a different format?
We are still digging how the value gets written and see if we can reverse engineer the process to give us a consistent result. However, our customer(s) already has(have) data in one format or the other (or both, we are checking...), so a proper conversion algorithm or implementation would be a better answer at this point.
I've used this quick and dirty function in the past to get a double from a text. Here in Argentina, people sometimes use the point and sometimes the comma to separate decimal values, so this function is ready to accept both formats. If only one punctuation is present, it's assumed that it's the decimal separator.
function ConvertDBL( texto )
dim retval, SepDecimal, SepMiles
If texto = "" then texto = "0"
retval = replace(trim(texto), " ", "")
If cdbl("3,24") = 324 then
SepDecimal = "."
SepMiles = ","
else
SepDecimal = ","
SepMiles = "."
end if
If InStr( Retval, SepDecimal ) > 0 then
If InStr( Retval, SepMiles ) > 0 then
If InStr( Retval, SepDecimal ) > InStr( Retval, SepMiles ) then
Retval = replace( Retval, SepMiles, "" )
else
Retval = replace( Retval, SepDecimal, "" )
Retval = replace( Retval, SepMiles, SepDecimal )
end if
end if
else
Retval = replace( Retval, SepMiles, SepDecimal )
end if
ConvertDbl = cdbl( retval )
end function
You cannot choose an arbitrary culture by passing a culture object like you can in .Net. You can choose:
Functions which always use the current regional settings. CStr, Format, CDbl etc. Implicit conversions (like the ones in your question) also use the current regional settings.
Functions which always use USA regional settings (similar to the .Net invariant culture). Str, Val
Usually the current regional settings are taken from Control Panel. There are rumours that you can call SetThreadLocale to change the regional settings from code at runtime - never tried it myself.
In Visual Basic tested works very fast and efficient.
Step 1: Conversion from a string to a value representation.
Step 2: Conversion from a value representation to a double integer value.
Dim Str as String
Dim mValue as double
str = "100,10"
mValue = CDbl(Val(str))
Related
I'm getting a error message in a VB6 .exe file running on Windows XP.
I compile and "make it" on Windows 7/8, but always get an Overflow error message when it executes this two lines on XP:
sUrl = "C:\Arquivos de Programas\Internet Explorer\IEXPLORE.EXE http://example.com/WebForms/send.aspx?id=" & intCodID & "&type=500&usr=" & intCodUser
openWeb = Shell(sUrl, vbMaximizedFocus)
sUrl is a String and OpenWeb is actually a Integer, but I already declared it as Double and as nothing (just Dim OpenWeb) and still get the overflow error.
UPDATE
Didn't managed to find out what was happening there, but another solution for calling IE:
Dim IE
sUrl = "http://www.google.com/"
Set IE = CreateObject("InternetExplorer.Application")
IE.Visible = True
IE.Navigate sUrl
While the VB6 documentation says Shell() returns a Variant Double... that appears to be obsolete information left over from manuals for earler versions of VB. Instead if you check the typelib info (i.e. look in the IDE's Object Browser) it actually returns a Double type result value.
As far as I can tell Shell() is a wrapper around a call to the WinExec() function.
The returned values are:
0 The system is out of memory or resources.
ERROR_BAD_FORMAT = 11 The .exe file is invalid.
ERROR_FILE_NOT_FOUND = 2 The specified file was not found.
ERROR_PATH_NOT_FOUND = 3 The specified path was not found.
or a Process ID
Also contrary to the documentation, Shell() turns those error values into exceptions ("File not found", "Invalid procedure call or argument," etc.). So if the call succeeds you always get back a PID value.
In all cases this is a DWORD. So it always fits in a Double without the possibility of an overflow. If you are seeing an overflow there is something else going wrong in your code.
Sadly a Double isn't particularly useful here, though it can at least hold the entire range of values. But you'd normally want to carefully convert it to a Long value:
Option Explicit
Function DDoubleToDLong(ByVal DDouble As Double) As Long
'Some functions like the intrinsic Shell() return a Double
'to get around the lack of a UI4 type (DWORD, i.e. unsigned
'Long) in VB. Of course this isn't clean to pass to API
'calls, making it sort of worthless so we need to do a type
'conversion such as this:
If DDouble > 2147483647# Then
DDoubleToDLong = CLng(DDouble - 2147483648#) Or &H80000000
Else
DDoubleToDLong = CLng(DDouble)
End If
End Function
Private Sub Form_Load()
Dim DD As Double
Dim DL As Long
AutoRedraw = True
Font.Name = "Courier New" 'Or other handy monospaced font.
Font.Size = 12#
DD = 0#: DL = DDoubleToDLong(DD): Print DD, DL, Hex$(DL)
DD = 1#: DL = DDoubleToDLong(DD): Print DD, DL, Hex$(DL)
DD = 2147483647#: DL = DDoubleToDLong(DD): Print DD, DL, Hex$(DL)
DD = 2147483648#: DL = DDoubleToDLong(DD): Print DD, DL, Hex$(DL)
DD = 4294967295#: DL = DDoubleToDLong(DD): Print DD, DL, Hex$(DL)
End Sub
Integer is worthless since overflows will be common. Long without the conversion could cause overflows now and then. String is just silly.
You also need to quote the values for the EXE and its arguments property, as in:
Option Explicit
Function DDoubleToDLong(ByVal DDouble As Double) As Long
If DDouble > 2147483647# Then
DDoubleToDLong = CLng(DDouble - 2147483648#) Or &H80000000
Else
DDoubleToDLong = CLng(DDouble)
End If
End Function
Private Sub Form_Load()
Dim sUrl As String
Dim PID As Long
sUrl = """C:\Arquivos de Programas\Internet Explorer\IEXPLORE.EXE"" " _
& """http://example.com/WebForms/send.aspx?id=" _
& intCodID _
& "&type=500&usr=" _
& intCodUser _
& """"
PID = DDoubleToDLong(Shell(sUrl, vbMaximizedFocus))
End Sub
Even this isn't quite "right" since exception handling should be added. And both intCodID and intCodUser may require "cleansing" (UrlEncoding) depending on what types they are and what values they really have. These might be Integers based on the names, with you relying on implicit String coercion? If so they might be Ok.
BTW, as we see above special folder names get localized. For that matter the system drive might not even be C:\ at all! So such paths should never be hard-coded but instead be built up based on values returned from calls to Shell32 to look up the special folder.
An integer can only be a whole number. No decimals.
You say it's declared as an integer therefore you cannot assign 1. anything, and you certainly can't assign anything like that to a number variable as it's not a valid number anyway as it has two decimal points.
You need to declare it as string.
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
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I want to make a search tool to find a specific word in the Arabic language that i can find just the word for example:
ذهب الولد إلى المدرسة من البيت و منهم الى البيت
If I try to find the word "من", the code not only finds the word "من", but also finds part of the word " منهم". I don't want the program to do that. I want to find the word " من" and every word like it and make whole case to the word.
To make things more clear (using an English example), if I were to search for the word 'to' in the following sentence, I would only want whole words to be found, and not words that contain the word 'to' such as 'toward' to become a part of the result.
Sentence: I want to go towards the bus.
Searches like this can be frustrating. What I usually do is to add a space at the front of the search string and also at the end and then search for SearchString.
So... "I want to go towards the bus." becomes " I want to go towards the bus. ". Then I search for " to ". The problem with this method is that punctuation can cause a problem. For example, if you wanted to search for "bus", you would use:
" I want to go towards the bus. " and search for " bus ". This would not be found because there is punctuation after the word bus.
I would encourage you to use regular expressions for this functionality. VB6 does not have regular expressions built in, but you can use the Microsoft VBScript Regular Expressions functionality to accomplish this. Please take a look at this page to help you get started: http://support.microsoft.com/kb/818802
Edit based on your comment
You have this line of code:
pos = InStr(start_at, txtBody.Text, target)
try changing that line to this:
pos = InStr(start_at, " " & txtBody.Text & " ", " " & target & " ", vbBinaryCompare)
By adding spaces in the code, you are actually checking for (space)target(space). So that you don't miss potential matches at the beginning or end of txtBody.Text, spaces are added there (for comparison purposes only). By adding vbBinaryCompare, InStr will now perform a case sensitive search.
The only completely thorough way of doing this is to use the Instr() function, and then check that the next character is a punctuation character, a newline, or the word is at the end of the string, e.g.
Option Explicit
Private Declare Function GetStringTypeW Lib "Kernel32.dll" ( _
ByVal dwInfoType As Long, _
ByVal lpSrcStr As Long, _
ByVal cchSrc As Long, _
ByRef lpCharType As Integer _
) As Long
Private Const CT_CTYPE1 As Long = &H1
Private Const C1_UPPER As Long = &H1 ' Uppercase
Private Const C1_LOWER As Long = &H2 ' Lowercase
Private Const C1_DIGIT As Long = &H4 ' Decimal digits
Private Const C1_SPACE As Long = &H8 ' Space characters
Private Const C1_PUNCT As Long = &H10 ' Punctuation
Private Const C1_CNTRL As Long = &H20 ' Control characters
Private Const C1_BLANK As Long = &H40 ' Blank characters
Private Const C1_XDIGIT As Long = &H80 ' Hexadecimal digits
Private Const C1_ALPHA As Long = &H100 ' Any linguistic character: alphabetical, syllabary, or ideographic
Private Const C1_DEFINED As Long = &H200 ' A defined character, but not one of the other C1_* types
Function FindFullWord(ByVal in_lStartPos As Long, ByRef in_sText As String, ByRef in_sSearch As String, Optional ByVal in_eCompareMethod As VbCompareMethod = vbBinaryCompare) As Long
Dim nLenText As Long
Dim nLenSearch As Long
Dim sNextChar As String
Dim iCharType As Integer
FindFullWord = InStr(in_lStartPos, in_sText, in_sSearch, in_eCompareMethod)
' Did we find the search string in the text?
If (FindFullWord > 0) Then
' Save the length of the text.
nLenText = Len(in_sText)
nLenSearch = Len(in_sSearch)
Do
' Does this position mean that the search is the end of the string?
If (FindFullWord + nLenSearch - 1) = nLenText Then
' If so, we can exit now - there are no following characters.
Exit Function
End If
' Look at the next character.
sNextChar = Mid$(in_sText, FindFullWord + nLenSearch, 1)
' Is this next char a space, punctuation character, or a blank?
If (GetStringTypeW(CT_CTYPE1, StrPtr(sNextChar), 1, iCharType)) Then
If (iCharType And C1_SPACE) = C1_SPACE Then
Exit Function
ElseIf (iCharType And C1_PUNCT) = C1_PUNCT Then
Exit Function
ElseIf (iCharType And C1_BLANK) = C1_BLANK Then
Exit Function
End If
End If
' Find the position of the search string in the text.
FindFullWord = InStr(FindFullWord + nLenSearch, in_sText, in_sSearch, in_eCompareMethod)
Loop Until FindFullWord = 0
End If
End Function
I originally started doing a test for every character which could follow a word and would not be part of that word, but the code started getting very long. And, of course, I know absolutely nothing about Arabic. So I wondered whether there was a standard way of finding out the general "type" of a character, regardless of language. And as it happens, there was.
The GetStringTypeW() method is documented in the Win32 documentation, and essentially can retrieve information about all the characters in a string. In my case, I am only looking at the character which follows the search word in a piece of text. The variable iCharType which returns the value from the string is a bitfield, and contains a number of values OR'ed together. I am using the AND operator to isolate just the values I am interested in.
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.
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