I have an ASCII decimal that i want to convert to a character in vb 6. Is it possible? if yes, how do i do that? In other words, how do i convert vb 6 ASCII decimal to vb 6 character?
Dim myChar As Char
Dim myBtye As byte
myChar = ? myByte
The Chr() function is the slow Variant version of the deprecated Chr$() function.
Use ChrW$() instead, which was introduced over 15 years ago and is far faster.
Dim myByte As byte
Dim myChar As Char
myChar = Chr(myByte)
Chr is returning the string value of the given argument.
btw if you want to convert it back to ascii, you can use the Asc function.
Related
I need help to split string which contains a HEX character.
How to do it?
The code I have got doesnt work.
dim itr
itr="343434 XX7777777" ' SI
msgbox itr
dim scrns
scrns=Split(itr,"SI",-1,1)
msgbox scrns(1)
Define the character by its numeric value (0x0f for "shift in"):
scrns = Split(itr, Chr(&h0f))
In VBScript you define hexadecimal numbers by prefixing them with &h. The Chr function turns the number into the corresponding character.
As #JNevill pointed out in the comments you could also use the decimal instead of the hexadecimal value:
scrns = Split(itr, Chr(15))
I am trying to convert letters to numbers.
I have a sub which ensures only numbers are put into the textbox.
My questions is will the following code work. I have a textbox(for numbers) and combobbox(for letters)
Dim sha As String
Dim stringposition As Long
Dim EngNumber As Long
sha = "abcdefghifjklmnopqrstuvwxyz"
stringposition = InStr(1, sha, Mid(1, cmbEngletter.Text, 1))
MsgBox "stringposition"
EngNumber = (txtManuNo.Text * 10) + stringposition
My only question above would be will the multiplication work with a .text. I believe it won't because it is a string. Please advise then on how to deal with a situation.
You can use CLng() to convert a string to a Long variable
CLng() will throw an error though if it doesn't like the contents of the string (for example if it contains a non-numeric character), so only use it when you are certain your string will only contain numbers
More forgiving is it to use Val() to convert a string into a numeric variable (a Double by default)
I also suggest you look into the following functions:
Asc() : returns the ASCII value of a character
Chr$() : coverts an ASCII value into a character
Left$() : returns the first characters of a string
CStr() : convert a number into a string
I think in your code you mean to show the contents of your variable stringposition instead of the word "stringposition", so you should remove the ""
I do wonder though what you are trying to accomplish with your code, but applying the above to your code gives:
Dim sha As String
Dim stringposition As Long
Dim EngNumber As Long
sha = "abcdefghifjklmnopqrstuvwxyz"
stringposition = InStr(1, sha, Left$(cmbEngletter.Text, 1))
MsgBox CStr(stringposition)
EngNumber = (Val(txtManuNo.Text) * 10) + stringposition
I used Val() because I am not certain your txtManuNo will contain only numbers
To ensure an user can only enter numbers you can use the following code:
Private Sub txtManuNo_KeyPress(KeyAscii As Integer)
Select Case KeyAscii
Case vbKeyBack
'allowe backspace
Case vbKey0 To vbKey9
'allow numbers
Case Else
'refuse any other input
KeyAscii = 0
End Select
End Sub
An user can still input non-numeric charcters with other methods though, like copy-paste via mouse actions, but it is a quick and easy first filter
I am trying to convert an integer to hex in ruby but i am having trouble. I need it to be in the format \x00 but it seems this is not possible if you do not set it manually?
count = 5
hex = "\x0#{count}"
puts hex.inspect # "\x005" but i need it to be "\x05"
Thanks!
["\x00".unpack("C").first + count].pack("C")
somebody in #ruby # irc.freenode.net answered it
I need to write some code in VBScript and have a version number string in a text file that I need to compare against. If I write this code as a test:
option explicit
Dim VersionString
VersionString = "6.2.1"
Dim Version
Version = CDbl (VersionString)
Version = Version * 100
I get an error on the CDbl line:
Microsoft VBScript runtime error: Type mismatch: 'CDbl'
How should I read and compare this string value?
"6.2.1" is not a Double formatted as a String. So CDbl() can't convert it. Your options are:
treat versions as strings; ok if you only need to compare for equality, bad if you need "6.9.1" to be smaller that "6.10.2"
Split() the string on "." and deal with the parts (perhaps converted to Integer/Long) separately; you'll need to write a comparison function for such arrays
Remove the "."s and CLng the resulting string; will break for versions like "6.10.2"
Split() the string on "*" and multiply + add the 'digits' to get one (integer) version number (6 * 100 + 2 * 10 + 1 * 1 = 621 for your sample); may be more complex for versions like "15.00.30729.01"
The conversion to a double isn't working because there are two decimal points in your string. To convert the string, you will have to remove one or both of them.
For this, you can use the Replace function. The syntax for Replace is
Replace(string, find, replacewith [, start [, count [, compare]]])
where string is the string to search, find is the substring to find, replacewith is the substring to replace find with, start is an optional parameter specifying the index to start searching at, count is an optional parameter specifying how many replaces to make, and compare is an optional parameter that is either 0 (vbBinaryCompare) to perform a binary comparison, or 1 (vbTextCompare) to perform a textual comparison
' Remove all decimals
Version = CDbl(Replace(VersionString, ".", "")
' Remove only the first decimal
Version = CDbl(Replace(VersionString, ".", "", 1, 1)
' Remove only the second decimal
Version = CDbl(Replace(VersionString, ".", "", 3, 1)
I have some picture named like this . (like "2.jpeg", "1234.gif" etc.)
how can I get the integer from the string?
I mean, is there something like the C function:
sscanf(myString,"%d", myInteger);
thanks
Alessandro
You will need to remove the non-numeric portion and then call TryParse.
Dim numericPortion As String
Dim result As Integer
numericPortion = myString.Substring(0, myString.IndexOf('.'))
If (Not Int32.TryParse(numericPortion, ByRef result)) Then
''// Handle Error
Else
''// Use Result
End If