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
Related
Path = split(wscript.scriptFullName, wscript.scriptname)(0)
CreateObject("wscript.shell").run(Path & "Name.txt")
The above script works fine if both the file path and file name contain no spaces.
If either contains a space, the result will be;
Error: The system cannot find the file specified.
How can I fix the error?
The rules are fairly simple:
All strings have to start and end with double quotes to be a valid string.
Dim a
a = "Hello World" 'Valid string.
a = "Hello World 'Not valid and will produce an error.
Any use of variables must use the String Concatenation character & to combine them with strings.
Dim a: a = "Hello"
Dim b
b = a & " World" 'Valid concatenated string.
b = a " World" 'Not valid and will produce an error.
As double quotes are used to define a string, all instances of double quotes inside a string must be escaped by doubling the quotes "" but Rule 1. still applies.
Dim a: a = "Hello"
Dim b
b = """" & a & " World""" 'Valid escaped string.
b = """ & a & " World""" 'Not valid, start of string is not complete
'after escaping the double quote
'producing an error.
Follow these three rules and you won't go far wrong.
With those in mind the above line would need to be;
CreateObject("wscript.shell").run("""" & Path & "Name.txt""")
to generate a string surrounded by literal double quotes.
Useful Links
VBS with Space in File Path
Adding quotes to a string in VBScript
Breaking a String Across Multiple Lines (More on string concatenation).
CreateObject("wscript.shell").run(""""Path & "Name.txt""")
is how.
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 want to replace a particular value followed by TRN* to some other value.How to do the coding for the same?..Please provide an example.
For example :
TRN*12345*34444~
This is a segment in my file(like this I have many TRN* segments in my file).I want to replace the segment after TRN* and before next * (ie.12345)with some other value.
Is there any way to do this by using vbscript?
Thanks in advance
The regular expression way of safetyOtter is the way to go, you only have to fiddle with the pattern and replacement string:
originalString = "TRN*12345*34444~"
replaceValue = "78910"
Set re = new RegExp
re.Pattern = "(.*TRN\*)([^*]+)(.*)"
re.IgnoreCase = False
' This keeps the first and last part between parenthesis, but replaces the middle part
newString = re.Replace(originalString, "$1" & replaceValue & "$3")
msgbox newString
' result: TRN*78910*34444~
Something like this should work, may have to tweak, can't test it at the moment
Set regEx = New RegExp
regEx.Pattern = "TRN\*.*?\*"
regEx.IgnoreCase = True
replStr = "TRN*" & SomeOtherValue& "*"
ReplaceTest = regEx.Replace(YourStringHere, replStr)
Edit:
if you are looking to replace a specific number with another, a simple:
YourStringHere = Replace(YourStringHere,"TRN*12345*","TRN*67890*")
Is enough
I know I can split a string into multiple substrings by giving a delimiter. I know I can also choose a substring based on character position like this:
sAddressOverflow = Right(sAddressLine1,5)
What I would like to do though is split an input string like this:
"123 South Main Street Apt. 24B"
But I only want to end up with two substrings which are split based on the first space to the left of the 25th character. So my desired output using the above input would be:
Substring1 = "123 South Main Street"
Substring2 = "Apt. 24B"
Is this possible?
Regular expressions have the advantage that you can configure your pattern independently from the location where you use it and that they are highly adaptable, so I prefer to do string manipulation with regular expressions. Unfortunately the pattern of Ansgar Wiechers does not exactly match your requirement. Here is one that does:
myString = "1234 6789A 234567 9B12 4567 890"
Set re = new RegExp
re.Pattern = "^(.{1,25}) (.*)$"
Set matches = re.Execute(myString)
wscript.echo "leftpart: " & matches(0).submatches(0)
wscript.echo "rightpart: " & matches(0).submatches(1)
There is no such inbuilt function available,
but you might want to try this,
add = "123 South Main Street Apt. 24B"
valid = Left(add,25)
arr = Split(valid)
char= InStrRev(add,arr(UBound(arr)))-1
address1 = Left(add,char)
address2= Right(add,Len(add)-char)
Wscript.echo address1
Wscript.echo address2
this might not be the perfect way, but it works !!!
You can do this with a regular expression, but you need a well defined format:
addr = "123 South Main Street Apt. 24B"
Set re = New RegExp
re.Pattern = "^(\d+ .*) +(apt\. +\d+(.*?))$"
re.IgnoreCase = True
Set m = re.Execute(addr)
If m.Count > 0 Then
WScript.Echo m(0).SubMatches(0)
WScript.Echo m(0).SubMatches(1)
End If
By "well-defined format" I mean that you need some "anchors" (or fix points) in your expression to identify the parts in the string. In the example the anchor is the substring "apt." followed by one or more digits.
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 "".