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)
Related
i want to put a comma after 3 digits in a numeric variable in vbscript
w_orimpo = getvalue(rsmodifica , "w_orimpo")
w_orimpo = FormatNumber(w_orimpo,2)
The initial value of w_orimpo is 21960.
If I use FormatNumber I get the value 21,960.
But I would like to get the following one -> 219,60
We can handle this via a regex replacement:
Dim input, output, regex1, regex2
Set input = "21960"
Set regex1 = New RegExp
Set regex2 = New RegExp
regex1.Pattern = "(\d{3})"
regex1.Global = True
regex2.Pattern = ",$"
output = regex1.Replace(StrReverse(input), "$1,")
output = StrReverse(regex2.Replace(output, ""))
Rhino.Print output
Note that two regex replacements are needed here because VBScript's regex engine does not support lookarounds. There is a single regex pattern which would have gotten the job done here:
(\d{3})(?!$)
This would match (and capture) only groups of three digits at a time, and only if those three digits are not followed by the end of the input. This is needed to cover the following edge case:
123456 -> 123,456
We don't want a comma after the final group of three digits. My answer gets around this problem by doing another regex replacement to trim off any trailing comma.
Or without regex:
Mid(CStr(w_orimpo), 1, 3) & "," & Mid(CStr(w_orimpo), 4)
Or
Dim divider
divider = 10 ^ (Len(CStr(w_orimpo)) - 3)
w_orimpo = FormatNumber(w_orimpo / divider, 2)
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
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 have 6400+ records which I am looping through. For each of these: I check that the address is valid by testing it against something similar to what the Post Office uses (find address). I need to double check that the postcode I have pulled back matches.
The only problem is that the postcode may have been inputted in a number of different formats for example:
OP6 6YH
OP66YH
OP6 6YH.
If Replace(strPostcode," ","") = Replace(xmlAddress.selectSingleNode("//postcode").text," ","") Then
I want to remove all spaces from the string. If I do the Replace above, it removes the space for the first example but leave one for the third.
I know that I can remove these using a loop statement, but believe this will make the script run really slow as it will have to loop through 6400+ records to remove the spaces.
Is there another way?
I didn't realise you had to add -1 to remove all spaces
Replace(strPostcode," ","",1,-1)
Personally I've just done a loop like this:
Dim sLast
Do
sLast = strPostcode
strPostcode = Replace(strPostcode, " ", "")
If sLast = strPostcode Then Exit Do
Loop
However you may want to use a regular expression replace instead:
Dim re : Set re = New RegExp
re.Global = True
re.Pattern = " +" ' Match one or more spaces
WScript.Echo re.Replace("OP6 6YH.", "")
WScript.Echo re.Replace("OP6 6YH.", "")
WScript.Echo re.Replace("O P 6 6 Y H.", "")
Set re = Nothing
The output of the latter is:
D:\Development>cscript replace.vbs
OP66YH.
OP66YH.
OP66YH.
D:\Development>
This is the syntax Replace(expression, find, replacewith[, start[, count[, compare]]])
it will default to -1 for count and 1 for start. May be some dll is corrupt changing the defaults of Replace function.
String.Join("", YourString.Split({" "}, StringSplitOptions.RemoveEmptyEntries))
Because you get all strings without spaces and you join them with separator "".
Hey all, i am trying to replace large spaces between text with just one. My output looks like this right now:
5964215">
This is just the first example of the spaces
5964478">
This would be the 2nd example of showing how many spaces this thing has in each sentence.
5964494">
That comes from a textbox that has multi-line to true. Here is what it looks like when it doesn't have multi-line to true.
http://www.june3rdsoftware.com/forums/vb6.jpg
I can not seem to get the spaces to go away! BTW, this text is from a webpage if that makes any difference.
David
According to the suggestion of MvanGeest, here is some VB code to replace blocks of white spaces:
Sub test()
Dim x As String, y As String
x = "abcd defg 1233"
Dim re As New RegExp
re.Pattern = "\s+"
re.Global = True
y = re.Replace(x, " ")
Debug.Print y
End Sub
To make this work, you will have to add a reference to "Microsoft VBScript Regular Expresssions" to your project.
Assuming no regex support, why not set up a simple state machine that will set the state=1 when a space is found and set state=0 once a non-space is encountered. You can move char by char when state=0 (thus copying over only 1 space per series of spaces).
Also assuming no regex, you could try something like
str = "long text with spaces "
i = LenB(str)
str = Replace(str, " ", " ")
Do While LenB(str) <> i
i = LenB(str)
str = Replace(str, " ", " ")
Loop
Of course this code could be optimized for long space sequences but it might be all you need as well