Search Field For Specific Text & Select Characters After Said Text

Legacy Poster

• February 04, 2016 07:22

Hi all

 

I'm a noob to BarTender VB Scripting and would like to know how I would go about setting up a script to search through the text of a field and select a specific combination of letters plus "x" nubmer of characters after that text.

 

eg. of text I might encounter:   'The samples were placed in container DNA 128:E08 for processing post assay setup...'  What I'm interested in pulling from this text is purely "DNA 128:E08".

 

I initially created an action under "Search and Replace: " in the Transforms field of the Text Properties using Search and Delete Everything Before (search criteria is "DNA") which gives me what I want but also all text following the 11 characters I desire.  I then tried combining a limitation of 11 characters under "Number of Characters:" transformation but that still recognized the data being deleted by the Search and Delete Everything Before criteria, thus nothing appears in my field.

 

I figured I could use VB Script to find text DNA and then only take the number of characters post that search criteria I desire, but I'm having a little trouble understanding the syntax.

 

Any help would be greatly appreciated.

 

Kind regards,

 

Nickolas.

 

 

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  Legacy Poster

  ★ BarTender Hero ★

  • February 04, 2016 11:39

  Hello,

   

  I would suggest looking into InString in combination with the Mid function.

  Say for instance I want to know what the position is of the first occerence of "DNA" in the string "I eat DNA for breakfast".

  My code would look like:

  Dim startPosition
  Dim myString
  Dim secondString
  
  myString = "I eat DNA for breakfast."
  
  'Check if the search string is found
  If Not InStr(myString, "DNA") = 0 Then
      'If found then add 3 to the start position
      startPositionNumber = InStr(myString, "DNA") + 3
      'return everything after the start position
      secondString = Mid(myString, startPositionNumber)
      MsgBox(secondString)
  End If
  

   

  Hope this helps.

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  Ian Cummings

  Moderator

  ★ BarTender Hero ★

  • February 04, 2016 12:03

  Without going into the specifics, you could also make use of the Regular Expression object of VB script.  Here is an example for validating a UK postal code.

   

  UK Postcode validation:

   

  For this I created a new named data source, not contained in any label object, in order to make use of the OnPostPrompt event for the postcode validation.  I created a function called RegExpTest() which creates the regular expression object to execute against an industry standard UK postcode regular expression pattern.  I then call the function on the OnPostPrompt event cancelling the print job if the post code is found to be invalid.

   

  Functions and Subs:

   

  Function RegExpTest(strng)
       
       'Create variables.
       Dim patrn, regEx, Match, Matches   
       
       'Define the regular expression pattern string.
       patrn = "(GIR 0AA)|((([ABCDEFGHIJKLMNOPRSTUWYZ][0-9][0-9]?)|(([ABCDEFGHIJKLMNOPRSTUWYZ][ABCDEFGHKLMNOPQRSTUVWXY][0-9][0-9]?)|" & _
                      "(([ABCDEFGHIJKLMNOPRSTUWYZ][0-9][ABCDEFGHJKSTUW])|([ABCDEFGHIJKLMNOPRSTUWYZ][ABCDEFGHKLMNOPQRSTUVWXY][0-9][ABEHMNPRVWXY]))))" & _
                      " [0-9][ABDEFGHJLNPQRSTUWXYZ]{2})"
  
       'Create a regular expression object and define its properties.
       Set regEx = New RegExp
       regEx.Pattern = patrn
       regEx.IgnoreCase = True
       regEx.Global = False
  
       'Execute the search.
       Set Matches = regEx.Execute(strng)
  
       'Handle any invalid matches returned on the search.
       If Matches.Count = 0 Then
            Call Format.CancelPrinting("Postcode: " & strng & " is invalid!","")
       Else
            If Matches.item(0).FirstIndex > 0 Or Matches.item(0).Length > Len(strng) Then
                 Call Format.CancelPrinting("Postcode: " & strng & " is invalid!","")
            End if
       End If
  
  End Function
  
  

   

  OnPostPrompt:

   

  Call RegExpTest (Format.NamedSubStrings("btPostcode").Value)
  

   

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  Legacy Poster

  ★ BarTender Hero ★

  • February 05, 2016 01:58

  Thank you @WvDriel & @Ian C - Technical Ninja at Seagull for you quick response.  Tried both your solutions and they were able to obtain the information I wanted but I'd found an "easier" path (by easier I mean for someone with not much scripting experince).

   

  Because I'd already "Transformed" the data in the Text Properties box using a "Search & Replace:" action by deleting everything before the text "DNA", I only had to concentrate on removing all data after the characters I desired.  I wrote the simple script value = left(value,11) and this seemd to do the job. 

   

  For the following text in the variable field :  "The_Hoff_is_always_right_about_DNA_123:A01_being_where_it_should_be...." I was able to delete "The_Hoff_is_always_right_about_" and then pick only the next 11 characters to display i.e. DNA_123:A01.

   

  I haven't tested it to see what happens if the text block contains 2 separate versions of the search criteria "DNA" though.  I assume it picks the 1st one in the line and runs with that, in which case I may need to implement your more advance scripts and include the pattern condition detailed in Ian C - Technical Ninja at Seagull response.

   

  Either way, thanks for both for your responses.  They are very much appreciated.

   

  Regards,

   

  Nickolas.

   

   

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