I have a file with the following inputs
"SMEGOLD 1312",20131127,"C","11606233E","SMX","C",20131009,170028,"SMX","70207",0,1,4699,0,469.9,"USD",0,"",0,"",0,"",0,"",0,0,0,8062696,"",0,20131009,170028,"SYSTEM","25228","","166","121328200000223",785,0,"","","","","","","","","","","","",0,0,0,"",20131009,170028,"ADVMEE"
"SMEGOLD 1312",20131127,"C","11606233E","SMX","C",20131009,170030,"SMX","70207",0,1,4699,0,469.9,"USD",0,"",0,"",0,"",0,"",0,0,0,8062697,"",0,20131009,170031,"SYSTEM","25228","","167","121328200000223",786,0,"","","","","","","","","","","","",0,0,0,"",20131009,170028,"ADVMEE"
What i would like to achieve is to only obtain the first quote text of the line. Example "SMEGOLD 1312". Then i would like to append the first 3 characters and the last 4 characters of that extracted text to the back of line.
And move to the next line to carry on with the procedure till the end of the file.
Any advise will be a great help. I tried using the objRegEx.Pattern but to no avail
Thanks
Use a pattern that cuts/groups the first field in toto (1), its first three (2) and its last four (3) characters, and all the rest (4). Then .Replace creatively:
>> s = Replace("'SMEGOLD 1312','whatever','ADVMEE'", "'", """")
>> WScript.Echo s
>> Set r = New RegExp
>> r.Pattern = "^((""[^""]{3})[^""]*?([^""]{4}""))(.*)"
>> WScript.Echo r.Replace(s,"$1$4,$2$3")
>>
"SMEGOLD 1312","whatever","ADVMEE"
"SMEGOLD 1312","whatever","ADVMEE","SME1312"
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 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 "".
I'm new to vbscript and Stack Overflow, and could really use some help.
Currently I'm trying to format a phone number that is read from an image and stored in a variable. Because the images are "dirty" extra characters find their way in, such as periods or parenthesis. I've already restricted the field as much as possible to help prevent picking up extra characters, but alas!
For example I want to turn ",123.4567890" into "123 456 7890" (not including the double quotes). Trouble is I won't know what extra characters could potentially be picked up, ruling out a simple replace.
My logic is remove any non numeric characters, start from the left, insert a space after the third number, insert a space after the sixth number.
Any help would be great, and please feel free to ask for more information if needed.
Welcome to Stack Overflow. You can remove non-digits using Regex, and concatenate the parts using Mid function.
For example:
Dim sTest
sTest = ",123.4567890"
With (New RegExp)
.Global = True
.Pattern = "\D" 'matches all non-digits
sTest = .Replace(sTest, "") 'all non-digits removed
End With
WScript.Echo Mid(sTest, 1, 3) & " "& Mid(sTest, 4, 3) & " "& Mid(sTest, 7, 4)
Or fully using Regex (via a second grouping pattern):
Dim sTest
sTest = ",123.4567890"
With (New RegExp)
.Global = True
.Pattern = "\D" 'matches all non-digits
sTest = .Replace(sTest, "") 'all non-digits removed
.Pattern = "(.{3})(.{3})(.{4})" 'grouping
sTest = .Replace(sTest, "$1 $2 $3") 'formatted
End With
WScript.Echo sTest
Use a first RegExp to clean non-digits from the input and second one using groups for the layout:
Function cleanPhoneNumber( sSrc )
Dim reDigit : Set reDigit = New RegExp
reDigit.Global = True
reDigit.Pattern = "\D"
Dim reStruct : Set reStruct = New RegExp
reStruct.Pattern = "(.{3})(.{3})(.+)"
cleanPhoneNumber = reStruct.Replace( reDigit.Replace( sSrc, "" ), "$1 $2 $3" )
End Function ' cleanPhoneNumber
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