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))
Related
I tried almost all the methods (CLEAN,TRIM,SUBSTITUTE) trying to remove the character hiding in the beginning and the end of a text. In my case, I downloaded the bill of material report from oracle ERP and found that the item codes are a victim of hidden characters.
After so many findings, I was able to trace which character is hidden and found out that it's a question mark'?' (via VBA code in another thread) both at the front and the end. You can take this item code: 11301-21
If you paste the above into your excel and see its length =LEN(), you can understand my problem much better.
I need a good solution for this problem. Therefore please help!
Thank you very much in advance.
Thanks to Gary's Student, because his answer inspired me.
Also, I used this answer for this code.
This function will clean every single char of your data, so it should work for you. You need 2 functions: 1 to clean the Unicode chars, and other one to clean your item codes_
Public Function CLEAN_ITEM_CODE(ByRef ThisCell As Range) As String
If ThisCell.Count > 1 Or ThisCell.Count < 1 Then
CLEAN_ITEM_CODE = "Only single cells allowed"
Exit Function
End If
Dim ZZ As Byte
For ZZ = 1 To Len(ThisCell.Value) Step 1
CLEAN_ITEM_CODE = CLEAN_ITEM_CODE & GetStrippedText(Mid(ThisCell.Value, ZZ, 1))
Next ZZ
End Function
Private Function GetStrippedText(txt As String) As String
If txt = "–" Then
GetStrippedText = "–"
Else
Dim regEx As Object
Set regEx = CreateObject("vbscript.regexp")
regEx.Pattern = "[^\u0000-\u007F]"
GetStrippedText = regEx.Replace(txt, "")
End If
End Function
And this is what i get using it as formula in Excel. Note the difference in the Len of strings:
Hope this helps
You have characters that look like a space character, but are not. They are UniCode 8236 & 8237.
Just replace them with a space character (ASCII 32).
EDIT#1:
Based on the string in your post, the following VBA macro will replace UniCode characters 8236 amd 8237 with simple space characters:
Sub Kleanup()
Dim N1 As Long, N2 As Long
Dim Bad1 As String, Bad2 As String
N1 = 8237
Bad1 = ChrW(N1)
N2 = 8236
Bad2 = ChrW(N2)
Cells.Replace what:=Bad1, replacement:=" ", lookat:=xlPart
Cells.Replace what:=Bad2, replacement:=" ", lookat:=xlPart
End Sub
Could somebody help me to extract the string between dash (-) and dot (.) at the last of URL in ASP classic?
For example:
mypizza.com/this-is-my-special-6-pizza-this-week-3256.html
How can I extract the 3256 value?
PS: There are many dashes and some numbers appear in URL.
This works if you definitely have a dash before the number. If you could have a / before the number then add another replace for / into -.
dim s, aSplit
s = "mypizza.com/this-is-my-special-6-pizza-this-week-3256.html"
s = replace(s, ".", "-") ' replace any dots with dashes
aSplit = split(s, "-") ' break s into an array, splitting at dashes. Note it is a zero-based array.
dim sOut
sOut = aSplit(ubound(aSplit) - 1) ' get the penultimate entry in the array
Solved!
I found the answer:
Dim n, strPost
dashCount = len(urlPost)-len(replace(urlPost,"-",""))
n=dashCount
thisURL=split(urlPost,"-")
strPost=replace(thisURL(n),".html","")
response.write(strPost)
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.
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 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)