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Regular Expressions

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  • Regular Expressions

    Hi all,

    been trying to extract dates from text fields, the date is not neccessarily in the same place or same format in each record. Below is my trial-and-error attempt.
    I want to replace each dot with a forward-slash and ensure that day and month are always two digits and the year is 4 digits.

    Code:
    testData1 = "Engine cut on approach and undershot practice forced landing Lords Bridge Cambs. 13.11.42 DBF"
    SearchExpr1 ="([0-9]{2})[.]([0-9]{2})[.]([0-9]{2})"
    testsplit1 = regex_split(testData1 ,SearchExpr1)
    testresult1 = crlf_to_comma(testsplit1)
    trace.writeln("1) "+Testresult1+" = "+padl(word(testresult1,1,","),2,"0")+"/"+padl(word(testResult1,2,","),2,"0")+"/19"+word(testresult1,3,","))
    
    testData2 = "Engine cut on approach and undershot practice forced landing Lords Bridge Cambs. 1.11.42 DBF"
    SearchExpr2 = "([0-9])[.]([0-9]{2})[.]([0-9]{2})"
    testsplit2 = regex_split(testData2 ,SearchExpr2)
    testresult2 = crlf_to_comma(testsplit2)
    trace.writeln("2) "+Testresult2 + " = "+padl(word(testresult2,1,","),2,"0")+"/"+padl(word(testResult1,2,","),2,"0")+"/19"+word(testresult2,3,","))
    
    testData3 = "Engine cut on approach and undershot practice forced landing Lords Bridge Cambs. 1.1.42 DBF"
    SearchExpr3 = "([0-9])[.]([0-9])[.]([0-9]{2})"
    testsplit3 = regex_split(testData3 ,SearchExpr3)
    testresult3 = crlf_to_comma(testsplit3)
    trace.writeln("3) "+Testresult3 +" = "+padl(word(testresult3,1,","),2,"0")+"/"+padl(word(testResult3,2,","),2,"0")+"/19"+word(testresult3,3,","))
    
    testData4 = "Engine cut on approach and undershot practice forced landing Lords Bridge Cambs. 13.1.42 DBF"
    SearchExpr4 = "([0-9]{2})[.]([0-9])[.]([0-9]{2})"
    testsplit4 = regex_split(testData4 ,SearchExpr4)
    testresult4 = crlf_to_comma(testsplit4)
    trace.writeln("4) "+Testresult4 + " = "+padl(word(testresult4,1,","),2,"0")+"/"+padl(word(testResult4,2,","),2,"0")+"/19"+word(testresult4,3,","))
    
    testData5 = "Engine cut on approach and undershot practice forced landing Lords Bridge Cambs. 1.11.42 DBF"
    SearchExpr5 = "([0-9])[.]([0-9]{2})[.]([0-9]{2})"
    testsplit5 = regex_split(testData5 ,SearchExpr5)
    testresult5 = crlf_to_comma(testsplit5)
    trace.writeln("1) "+Testresult5+" = "+padl(word(testresult5,1,","),2,"0")+"/"+padl(word(testResult5,2,","),2,"0")+"/19"+word(testresult5,3,","))
    
    testData6 = "Engine cut on approach and undershot practice forced landing Lords Bridge Cambs. .42 DBF"
    SearchExpr6 = "[.]([0-9]{2})"
    testsplit6 = regex_split(testData6,SearchExpr6)
    testresult6 = crlf_to_comma(testsplit6)
    trace.writeln("6) "+Testresult6+" = "+"19"+word(testresult6,1,","))
    
    testData7 = "Engine cut on approach and undershot practice forced landing Lords Bridge Cambs. 1.42 DBF"
    SearchExpr7 = "([0-9])[.]([0-9]{2})"
    testsplit7 = regex_split(testData7 ,SearchExpr7)
    testresult7 = crlf_to_comma(testsplit7)
    trace.writeln("7) "+Testresult7 +" = "+padl(word(testresult7,1,","),2,"0")+"/19"+word(testresult7,2,","))
    
    testData8 = "Engine cut on approach and undershot practice forced landing Lords Bridge Cambs. 11.42 DBF"
    SearchExpr8 = "([0-9]{2})[.]([0-9]{2})"
    testsplit8 = regex_split(testData8 ,SearchExpr8)
    testresult8 = crlf_to_comma(testsplit8)
    trace.writeln("8) "+Testresult8+" = "+padl(word(testresult8,1,","),2,"0")+"/19"+word(testresult8,2,","))
    I can do it for each of the 8 formats present in the text field , but how do I combine them into one solution. Going around in cricles so need a helping hand.

    is there a more simpler way of doing this using regular expressions, where I won't have to know which format the date is in before transformation . the result will be put into another field in the record. ?


    thx





    --
    Support your local Search and Rescue Unit, Get Lost!

    www.westrowops.co.uk

  • #2
    Hi Graham,
    I'm going to swing at the low-hanging fruit first, using testData8 as the example.
    I can do more, but I would need to know what basic format the dates are in (m/d/y or ?)

    Gregg

    Code:
    testData8 = strtran(testData8,".","/")
    testData8 = strtran(testData8,"/ ",". ")
    Gregg Schmidt

    Comment


    • #3
      came up with one solution, although still a bit long..

      Code:
      FUNCTION NewDateFormat AS C (text_data AS C )
      
      Srch_Expr1 = "([0-9]{1})[.]([0-9]{1})[.]([0-9]{2})"  'Search for date format 1.1.42
      Repl_Expr1 = "0\1\/0\2\/19\3"                        'convert to date format 01/01/1942
      
      Srch_Expr2 = "([0-9]{2})[.]([0-9]{1})[.]([0-9]{2})"  'Search for date format 11.1.42
      Repl_Expr2 = "\1\/0\2\/19\3"                         'convert to date format 11/01/1942
      
      Srch_Expr3 = "([0-9]{1})[.]([0-9]{2})[.]([0-9]{2})"  'Search for date format 1.11.42
      Repl_Expr3 = "0\1\/\2\/19\3"                         'convert to date format 01/11/1942
      
      Srch_Expr4 = "([0-9]{2})[.]([0-9]{2})[.]([0-9]{2})"  'Search for date format 11.11.42
      Repl_Expr4 = "\1\/\2\/19\3"                          'convert to date format 11.11.1942
      
      Srch_Expr5 = "([0-9]{1})[.]([0-9]{2})"   'Search for date format 1.42
      Repl_Expr5 = "0\1\/19\2"                 'convert to date format 01/1942
      
      Srch_Expr6 = "([0-9]{2})[.]([0-9]{2})"   'Search for date format 11.42
      Repl_Expr6 = "\1\/19\2"                  'convert to date format 11/1942
      
      Srch_Expr7 = "[.]([0-9]{2})"  'Search for date format .42
      Repl_Expr7 = "19\1"           'convert to date format 1942
      
      for x = 1 to w_count(Text_Data," ")
      Sample_Text = word(Text_Data,x," ")
      
      select
      
      case regex_match(Sample_Text,Srch_Expr1)
              NewDateFormat = regex_merge(Sample_Text,Srch_Expr1,Repl_Expr1)
              goto Finish
      
      case regex_match(Sample_Text,Srch_Expr2)
              NewDateFormat = regex_merge(Sample_Text,Srch_Expr2,Repl_Expr2)
              goto Finish
      
      case regex_match(Sample_Text,Srch_Expr3)
              NewDateFormat = regex_merge(Sample_Text,Srch_Expr3,Repl_Expr3)
              goto Finish
      
      case regex_match(Sample_Text,Srch_Expr4)
              NewDateFormat = regex_merge(Sample_Text,Srch_Expr4,Repl_Expr4)
              goto Finish
      
      case regex_match(Sample_Text,Srch_Expr5)
              NewDateFormat = regex_merge(Sample_Text,Srch_Expr5,Repl_Expr5)
              goto Finish
      
      case regex_match(Sample_Text,Srch_Expr6)
              NewDateFormat = regex_merge(Sample_Text,Srch_Expr6,Repl_Expr6)
              goto Finish
      
      case regex_match(Sample_Text,Srch_Expr7)
              NewDateFormat = regex_merge(Sample_Text,Srch_Expr7,Repl_Expr7)
              goto Finish
      
      end select
      next x
      Finish:
      END FUNCTION
      Last edited by Graham Wickens; 04-11-2020, 02:42 AM.
      --
      Support your local Search and Rescue Unit, Get Lost!

      www.westrowops.co.uk

      Comment


      • #4
        I could only do basic testing, but I think this shortens the code.
        Code:
        FUNCTION NewDateFormat AS C (text_data AS C )
        dim fixit[0] as p
        
        fixit[].Srch_Expr = "([0-9]{1})[.]([0-9]{1})[.]([0-9]{2})" ' find date format 1.1.42
        fixit[..].Repl_Expr = "0\1\/0\2\/19\3" 'convert to date format 01/01/1942
        
        fixit[].Srch_Expr = "([0-9]{2})[.]([0-9]{1})[.]([0-9]{2})" ' find date format 11.1.42
        fixit[..].Repl_Expr = "\1\/0\2\/19\3" 'convert to date format 11/01/1942
        
        fixit[].Srch_Expr = "([0-9]{1})[.]([0-9]{2})[.]([0-9]{2})" ' find date format 1.11.42
        fixit[..].Repl_Expr = "0\1\/\2\/19\3" 'convert to date format 01/11/1942
        
        fixit[].Srch_Expr = "([0-9]{2})[.]([0-9]{2})[.]([0-9]{2})" 'find date format 11.11.42
        fixit[..].Repl_Expr = "\1\/\2\/19\3" 'convert to date format 11.11.1942
        
        fixit[].Srch_Expr = "([0-9]{1})[.]([0-9]{2})" ' find date format 1.42
        fixit[..].Repl_Expr = "0\1\/19\2" 'convert to date format 01/1942
        
        fixit[].Srch_Expr = "([0-9]{2})[.]([0-9]{2})" 'find date format 11.42
        fixit[..].Repl_Expr = "\1\/19\2" 'convert to date format 11/1942
        
        fixit[].Srch_Expr = "[.]([0-9]{2})" 'find date format .42
        fixit[..].Repl_Expr = "19\1" 'convert to date format 1942
        
        for x = 1 to w_count(Text_Data," ")
        Sample_Text = word(Text_Data,x," ")
        
        for each option in fixit
        select
        case regex_match(Sample_Text,option.Srch_Expr)
        NewDateFormat = regex_merge(Sample_Text,option.Srch_Expr,option.Repl_Expr)
        end select
        
        next 'each option
        next x
        
        END FUNCTION
        Gregg Schmidt

        Comment


        • #5
          Thanks Gregg, got there in the end.
          --
          Support your local Search and Rescue Unit, Get Lost!

          www.westrowops.co.uk

          Comment


          • #6
            Glad to hear. Now for the kicker, if you really want to shorten you can bury the Srch_Expr and Repl_Expr statements in a sql table,
            then populate the array with a relatively simple query, although you might have to make accommodations for and \s.
            Gregg Schmidt

            Comment


            • #7
              I'm a bit late to the party on this one, but have coded a single line solution that matches all the test options...
              Code:
              ?if(val(word(testdata1,-4))>0,padl(word(testdata1,-4),2,"0")+"/","")+if(val(word(testdata1,-3))>0,padl(word(testdata1,-3),2,"0")+"/","")+"19"+word(testdata1,-2)
              This works if the date value(s) is always followed by a single word and not preceded by a number when day of month is not specified.

              Comment


              • #8
                Originally posted by madtowng View Post
                Glad to hear. Now for the kicker, if you really want to shorten you can bury the Srch_Expr and Repl_Expr statements in a sql table,
                then populate the array with a relatively simple query, although you might have to make accommodations for and \s.
                gulp! how do I do that !!!!!!!!
                --
                Support your local Search and Rescue Unit, Get Lost!

                www.westrowops.co.uk

                Comment


                • #9
                  Originally posted by armasoft View Post
                  I'm a bit late to the party on this one, but have coded a single line solution that matches all the test options...
                  Code:
                  ?if(val(word(testdata1,-4))>0,padl(word(testdata1,-4),2,"0")+"/","")+if(val(word(testdata1,-3))>0,padl(word(testdata1,-3),2,"0")+"/","")+"19"+word(testdata1,-2)
                  This works if the date value(s) is always followed by a single word and not preceded by a number when day of month is not specified.
                  Interesting, but unfortunately wont work in my scenario as the date can be anywhere in the field, beginning middle or end!
                  --
                  Support your local Search and Rescue Unit, Get Lost!

                  www.westrowops.co.uk

                  Comment


                  • #10
                    Is it possible there could be other numeric data in the record?
                    If so, will the date always be the last numeric value in the record?

                    Comment


                    • #11
                      Suggest you get rid of the text and work with the numbers only.
                      Graham said it could be anywhere.
                      Ted Giles
                      Example Consulting - UK
                      .

                      sigpichttp://ec12.example-software.com//
                      See our site for Alpha Support, Conversion and Upgrade.

                      Comment


                      • #12
                        Hi Graham,

                        The code for this assumes you have a MySQL table named regexpressions setup as shown at http://www.sqlfiddle.com/#!9/717182c/1 , and a named::connection of AAlab.
                        You can change those but obviously will need to adjust the code if you do.
                        Let me know if (or how) this works for you.

                        Gregg

                        Code:
                        FUNCTION NewDateFormat AS C (text_data AS C )
                        dim myCN as sql::connection
                        dim sqlCode as c
                        dim itsOpen as l
                        dim itWorked as l
                        dim fixit[0] as p
                        
                        sqlCode = <<%txt%
                        select * from regexpressions
                        %txt%
                        
                        if .not. myCN.isopen then
                        itsOpen = myCN.open("::name::AAlab")
                        end if
                        if .not. itsOpen then
                        NewDateFormat = "Could not Open Table"
                        exit function
                        end if
                        
                        itWorked = myCN.ToPropertyArray(sqlCode, fixit)
                        
                        for x = 1 to w_count(Text_Data," ")
                        Sample_Text = word(Text_Data,x," ")
                        
                        for each option in fixit
                        select
                        case regex_match(Sample_Text,option.Srch_Expr)
                        NewDateFormat = regex_merge(Sample_Text,option.Srch_Expr,option.Repl_Expr)
                        exit function
                        end select
                        
                        next 'each option
                        next x
                        
                        END FUNCTION
                        Gregg Schmidt

                        Comment


                        • #13
                          Originally posted by Ted Giles View Post
                          Suggest you get rid of the text and work with the numbers only.
                          Graham said it could be anywhere.
                          I think the solution wioll be quite easy if there's no other numerics, but not going to spend time on it until I know

                          Comment


                          • #14
                            Agree. My approach would be to strip the chars and count the numbers left.
                            If there were more than 8, you would have to find a pattern.
                            Nn.nn.nnnn would be fine (10.01.1942) so would the other mixes.
                            Nnn would not (500) - as in 500 feet.

                            We'll see.
                            Ted Giles
                            Example Consulting - UK
                            .

                            sigpichttp://ec12.example-software.com//
                            See our site for Alpha Support, Conversion and Upgrade.

                            Comment

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