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.
Related
regex in VBscript has 3 methods, Test, Extract and Replace, but I can only seem to turn capture groups from Extract into variable.
However what I want to do is use capturing groups from 'Replace' as a variable. I can get a Regex.Replace working with no problems using $1 $2 etc for capturing groups, however I want multiply one of the capture groups.
In an xml file, I want to extract a value, times it by 15, and insert it back in. In this example the tag.
e.q.
strText = "
<rte>
<name>gpx.studio church 2 reduced.gpx</name>
<rtept lat='-33.482652' lon='150.159134'>
<ele>938.4</ele>
<desc>076</desc>
</rtept>
<rtept lat='-33.4825698175265' lon='150.159515440464'>
<ele>942.3</ele>
<desc>162</desc>
</rtept>
<rtept lat='-33.4828785376496' lon='150.159633457661'>
<ele>943.4</ele>
<desc>098</desc>
</rtept>
</rte>
</gpx>"
Dim oRegExp
Set oRegExp = New RegExp
oRegExp.Global=True
oRegExp.Multiline = True
oRegExp.Pattern = strPattern
strPattern = "(<rtept(?:(?:.|\n|\r)*?))<desc>(.*?)<\/desc>((?:(?:.|\n|\r)*?)<\/rtept>)"
strReplace = "$1<desc>$2<\/desc>$3"
' so on this line above, I want to turn the $2 into an integer and multiply it by 15 before putting back into replace.
' I have not done it here because I know it doesnt work as "$2"x1000
strNewText = oRegExp.Replace(strText, strReplace)
I want to turn the $2 into an integer and multiply it by 15 before putting back into replace.
I have tried to get the capture groups as SubMatches(1) which work with Regex.Extract method but it doesnt seem to work in Regex.Replace method, unless I am missing something....
help appreciated
i want to put a comma after 3 digits in a numeric variable in vbscript
w_orimpo = getvalue(rsmodifica , "w_orimpo")
w_orimpo = FormatNumber(w_orimpo,2)
The initial value of w_orimpo is 21960.
If I use FormatNumber I get the value 21,960.
But I would like to get the following one -> 219,60
We can handle this via a regex replacement:
Dim input, output, regex1, regex2
Set input = "21960"
Set regex1 = New RegExp
Set regex2 = New RegExp
regex1.Pattern = "(\d{3})"
regex1.Global = True
regex2.Pattern = ",$"
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 ran into an issue with the Classic ASP VbScript InStr() function. As shown below, the second call to InStr() returns 1 when searching for an empty string in a non empty string. I'm curious why this is happening.
' InStr Test
Dim someText : someText = "So say we all"
Dim emptyString : emptyString = ""
'' I expect this to be true
If inStr(1,someText,"so",1) > 0 Then
Response.write ( "I found ""so""<br />" )
End If
'' I expect this to be false
If inStr(1, someText, emptyString, 1) > 0 Then
Response.Write( "I found an empty string<br />" )
End If
EDIT:
Some additional clarification: The reason for the question came up when debugging legacy code and running into a situation like this:
Function Go(value)
If InStr(1, "Option1|Option2|Option3", value, 1) > 0 Then
' Do some stuff
End If
End Function
In some cases function Go() can get called with an empty string. The original developer's intent was not to check whether value was empty, but rather, whether or not value was equal to one of the piped delimited values (Option1,Option2, etc.).
Thinking about this further it makes sense that every string is created from an empty string, and I can understand why a programming language would assume a string with all characters removed still contains the empty string.
What doesn't make sense to me is why programming languages are implementing this. Consider these 2 statements:
InStr("so say we all", "s") '' evaluates to 1
InStr("so say we all", "") '' evaluates to 1
The InStr() function will return the position of the first occurrence of one string within another. In both of the above cases, the result is 1. However, position 1 always contains the character "s", not an empty string. Furthermore, using another string function like Len() or LenB() on an empty string alone will result in 0, indicating a character length of 0.
It seems that there is some inconsistency here. The empty string contained in all strings is not actually a character, but the InStr() function is treating it as one when other string functions are not. I find this to be un-intuitive and un-necessary.
The Empty String is the Identity Element for Strings:
The identity element I (also denoted E, e, or 1) of a group or related
mathematical structure S is the unique element such that Ia=aI=a for
every element a in S. The symbol "E" derives from the German word for
unity, "Einheit." An identity element is also called a unit element.
If you add 0 to a number n the result is n; if you add/concatenate "" to a string s the result is s:
>> WScript.Echo CStr(1 = 1 + 0)
>> WScript.Echo CStr("a" = "a" & "")
>>
True
True
So every String and SubString contains at least one "":
>> s = "abc"
>> For p = 1 To Len(s)
>> WScript.Echo InStr(p, s, "")
>> Next
>>
1
2
3
and Instr() reports that faithfully. The docs even state:
InStr([start, ]string1, string2[, compare])
...
The InStr function returns the following values:
...
string2 is zero-length start
WRT your
However, position 1 always contains the character "s", not an empty
string.
==>
Position 1 always contains the character "s", and therefore an empty
string too.
I'm puzzled why you think this behavior is incorrect. To the extent that asking Does 'abc' contain ''? even makes sense, the answer has to be "yes": All strings contain the empty string as a trivial case. So the answer to your "why is this happening" question is because it's the only sane thing to do.
It is s correct imho. At least it is what I expect that empty string is part of any other string. But maybe this is a philosophical question. ASP does it so, so live with it. Practically speaking, if you need a different behavior write your own Method, InStrNotEmpty or something, which returns false on empty search string.
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