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 "".
Related
I have a file "D:\test.log" that has either one of two styles. This will appear if the user is offline when the user received the message:
[02:19:47] Brother Aimbot (adama900): (Saved Thu Mar 31 05:15:09 2016)This is a test line
It will be like this if the user is online when the user received the message:
[02:19:47] Brother Aimbot (adama900): This is a test line
What I would like this to do is cut out the excess parts so it would look like this if it's either the first or second style:
Brother Aimbot (adama900) This is a test line
then place it into a message box.
Here is my code:
Sub main()
filename = "D:\Test.txt"
Set fso = CreateObject("Scripting.FileSystemObject")
Set f = fso.OpenTextFile(filename)
LNEVAL = f.ReadLine
LNENUM = 0
Do Until f.AtEndOfStream
For i = 1 To LNENUMs
f.ReadLine
Next
If InStr(LNEVAL, "(S") Then
LNEVAL = Left(LNEVAL, (Len("(S")+4))
MsgBox = LNEVAL
End If
Loop
f.Close
End Sub
This is what I have so far.
It's fairly simple to do what you want with a regular expression replacement. Basically what you want to do is remove three things from each line:
a substring between square brackets from the beginning of the string,
the colon separating the name from the message, and
an optional substring between parentheses after that colon.
A regular expression ^\[.*?\] matches an opening square bracket at the beginning of a string and the shortest number of characters up to a closing square bracket.
A regular expression \(Saved.*?\) matches an opening parenthesis followed by the word Saved and the shortest number of characters up to a closing parenthesis. However, since this part is optional you need to indicate that the expression can occur zero or one time by putting it in a non-capturing group and appending the ? modifier to it ((?:...)?).
Put the submatches that you do want to preserve in parentheses to create capturing groups
^\[.*?\] (.*?): (?:\(Saved.*?\))?(.*)
and replace each matching line with just the captured groups:
Set re = New RegExp
re.Pattern = ...
Set f = fso.OpenTextFile(filename)
Do Until f.AtEndOfStream
MsgBox re.Replace(f.ReadLine, "$1 $2")
Loop
f.Close
Some comments on your existing code:
For i = 1 To LNENUMs: this loop is always skipped over, because you set LNEUMs to 0. Since you only do f.ReadLine inside that For loop your outer Do loop becomes an infinite loop, since you never read the file to the end.
Len("(S")+4 always evaluates to 6, because the length of the string (S is not going to change, so you could just replace the expression with the numeric value.
MsgBox = LNEVAL: The MsgBox function doesn't work that way. Remove the = between function name and message.
A VBScript is in use to shorten the system path by replacing entries with the 8.3 versions since it gets cluttered with how much software is installed on our builds. I'm currently adding the ability to remove duplicates, but it's not working correctly.
Here is the relevant portion of code:
original = "apple;orange;apple;lemon\banana;lemon\banana"
shortArray=Split(original, ";")
shortened = shortArray(1) & ";"
For n=2 to Ubound(shortArray)
'If the shortArray element is not in in the shortened string, add it
If NOT (InStr(1, shortened, shortArray(n), 1)) THEN
shortened = shortened & ";" & shortArray(n)
ELSE
'If it already exists in the string, ignore the element
shortened=shortened
End If
Next
(Normally "original" is the system path, I'm just using fruit names to test...)
The output should be something like
apple;orange;lemon\banana
The issue is entries with punctuation, such as lemon\banana, seem to be skipped(?). I've tested it with other punctuation marks, still skips over it. Which is an issue, seeing how the system path has punctuation in every entry.
I know the basic structure works, since there are only one of each entry without punctuation. However, the real output is something like
apple;orange;lemon\banana;lemon\banana
I thought maybe it was just a character escape issue. But no. It still will not do anything with entries containing punctuation.
Is there something I am doing wrong, or is this just a "feature" of VBScript?
Thanks in advance.
This code:
original = "apple;orange;apple;lemon\banana;lemon\banana"
shortArray = Split(original, ";")
shortened = shortArray(0) ' array indices start with 0; & ";" not here
For n=1 to Ubound(shortArray)
'If the shortArray element is not in in the shortened string, add it
'i.e. If InStr() returns *number* 0; Not applied to a number will negate bitwise
' If 0 = InStr(1, shortened, shortArray(n), 1) THEN
If Not CBool(InStr(1, shortened, shortArray(n), 1)) THEN ' if you insist on Not
WScript.Echo "A", shortArray(n), shortened, InStr(1, shortened, shortArray(n), vbTextCompare)
shortened = shortened & ";" & shortArray(n)
End If
Next
WScript.Echo 0, original
WScript.Echo 1, shortened
WScript.Echo 2, Join(unique(shortArray), ";")
Function unique(a)
Dim d : Set d = CreateObject("Scripting.Dictionary")
Dim e
For Each e In a
d(e) = Empty
Next
unique = d.Keys()
End Function
output:
0 apple;orange;apple;lemon\banana;lemon\banana
1 apple;orange;lemon\banana
2 apple;orange;lemon\banana
demonstrates/explains your errors (indices, Not) and shows how to use the proper tool for uniqueness (dictionary).
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.
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