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Regex Too Complicated - replaceAll - large strings preprocessed before deserialize


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1















Given




  • A JSON string with Apex reserved words (like "currency":"USD" and otherwise undeserializable-into-Date type (like "someDate":"0001-01-01T00:00:00Z" )


  • An apex method that uses



    String s = s.replaceAll('"currency":','"currencyX":')
    .replaceAll('"0001\-01\-01T00:00:00Z"','null');



before deserializing into a custom Apex type with property String currencyX and Date someDate



When



The incoming Apex string is very large (> 1MB)



Then



Uncatchable error: Regex Too Complicated




What to do?











share|improve this question



























    1















    Given




    • A JSON string with Apex reserved words (like "currency":"USD" and otherwise undeserializable-into-Date type (like "someDate":"0001-01-01T00:00:00Z" )


    • An apex method that uses



      String s = s.replaceAll('"currency":','"currencyX":')
      .replaceAll('"0001\-01\-01T00:00:00Z"','null');



    before deserializing into a custom Apex type with property String currencyX and Date someDate



    When



    The incoming Apex string is very large (> 1MB)



    Then



    Uncatchable error: Regex Too Complicated




    What to do?











    share|improve this question

























      1












      1








      1








      Given




      • A JSON string with Apex reserved words (like "currency":"USD" and otherwise undeserializable-into-Date type (like "someDate":"0001-01-01T00:00:00Z" )


      • An apex method that uses



        String s = s.replaceAll('"currency":','"currencyX":')
        .replaceAll('"0001\-01\-01T00:00:00Z"','null');



      before deserializing into a custom Apex type with property String currencyX and Date someDate



      When



      The incoming Apex string is very large (> 1MB)



      Then



      Uncatchable error: Regex Too Complicated




      What to do?











      share|improve this question














      Given




      • A JSON string with Apex reserved words (like "currency":"USD" and otherwise undeserializable-into-Date type (like "someDate":"0001-01-01T00:00:00Z" )


      • An apex method that uses



        String s = s.replaceAll('"currency":','"currencyX":')
        .replaceAll('"0001\-01\-01T00:00:00Z"','null');



      before deserializing into a custom Apex type with property String currencyX and Date someDate



      When



      The incoming Apex string is very large (> 1MB)



      Then



      Uncatchable error: Regex Too Complicated




      What to do?








      apex






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 1 hour ago









      cropredycropredy

      36k441123




      36k441123






















          1 Answer
          1






          active

          oldest

          votes


















          3














          This is basically teaching an old dog (me) new tricks



          A cursory reading of the Apex String class documentation shows two methods




          • replace(target,replacement)

          • replaceAll(regExp,replacement)


          The naive developer (me) assumes that the first method, replace, only replaces a SINGLE instance of the target string and since the use case assumes you need to replace ALL occurrences of some pattern, you should use the more powerful replaceAll(..)



          But epistemic arrogance crept in and I did not realize replace(target,replacement) replaces ALL occurrences of target with replacement




          Replaces each substring of a string that matches the literal target sequence target with the specified literal replacement sequence replacement.




          So, if you have large strings and you are just doing simple text substitution, avoid replaceAll



          String s = s.replace('"currency":','"currencyX":')
          .replace('"0001-01-01T00:00:00Z"','null');





          share|improve this answer























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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            This is basically teaching an old dog (me) new tricks



            A cursory reading of the Apex String class documentation shows two methods




            • replace(target,replacement)

            • replaceAll(regExp,replacement)


            The naive developer (me) assumes that the first method, replace, only replaces a SINGLE instance of the target string and since the use case assumes you need to replace ALL occurrences of some pattern, you should use the more powerful replaceAll(..)



            But epistemic arrogance crept in and I did not realize replace(target,replacement) replaces ALL occurrences of target with replacement




            Replaces each substring of a string that matches the literal target sequence target with the specified literal replacement sequence replacement.




            So, if you have large strings and you are just doing simple text substitution, avoid replaceAll



            String s = s.replace('"currency":','"currencyX":')
            .replace('"0001-01-01T00:00:00Z"','null');





            share|improve this answer




























              3














              This is basically teaching an old dog (me) new tricks



              A cursory reading of the Apex String class documentation shows two methods




              • replace(target,replacement)

              • replaceAll(regExp,replacement)


              The naive developer (me) assumes that the first method, replace, only replaces a SINGLE instance of the target string and since the use case assumes you need to replace ALL occurrences of some pattern, you should use the more powerful replaceAll(..)



              But epistemic arrogance crept in and I did not realize replace(target,replacement) replaces ALL occurrences of target with replacement




              Replaces each substring of a string that matches the literal target sequence target with the specified literal replacement sequence replacement.




              So, if you have large strings and you are just doing simple text substitution, avoid replaceAll



              String s = s.replace('"currency":','"currencyX":')
              .replace('"0001-01-01T00:00:00Z"','null');





              share|improve this answer


























                3












                3








                3







                This is basically teaching an old dog (me) new tricks



                A cursory reading of the Apex String class documentation shows two methods




                • replace(target,replacement)

                • replaceAll(regExp,replacement)


                The naive developer (me) assumes that the first method, replace, only replaces a SINGLE instance of the target string and since the use case assumes you need to replace ALL occurrences of some pattern, you should use the more powerful replaceAll(..)



                But epistemic arrogance crept in and I did not realize replace(target,replacement) replaces ALL occurrences of target with replacement




                Replaces each substring of a string that matches the literal target sequence target with the specified literal replacement sequence replacement.




                So, if you have large strings and you are just doing simple text substitution, avoid replaceAll



                String s = s.replace('"currency":','"currencyX":')
                .replace('"0001-01-01T00:00:00Z"','null');





                share|improve this answer













                This is basically teaching an old dog (me) new tricks



                A cursory reading of the Apex String class documentation shows two methods




                • replace(target,replacement)

                • replaceAll(regExp,replacement)


                The naive developer (me) assumes that the first method, replace, only replaces a SINGLE instance of the target string and since the use case assumes you need to replace ALL occurrences of some pattern, you should use the more powerful replaceAll(..)



                But epistemic arrogance crept in and I did not realize replace(target,replacement) replaces ALL occurrences of target with replacement




                Replaces each substring of a string that matches the literal target sequence target with the specified literal replacement sequence replacement.




                So, if you have large strings and you are just doing simple text substitution, avoid replaceAll



                String s = s.replace('"currency":','"currencyX":')
                .replace('"0001-01-01T00:00:00Z"','null');






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 1 hour ago









                cropredycropredy

                36k441123




                36k441123






























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