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    • gelle_marrisaG
      gelle_marrisa @PeterJones
      last edited by

      @peterjones
      suppose if there is only 1 format for abcd area(no slash), then is there possibility to get the solution? I can replace all the slashes in first place and then use the regex.
      1234567891
      abcd :
      111|22
      xyz :
      333
      product :
      blablabla 456
      code :
      01010
      serial :
      8888899999

      1 Reply Last reply Reply Quote 0
      • Neil SchipperN
        Neil Schipper @gelle_marrisa
        last edited by

        @gelle_marrisa The regex I provided you is tested on the text quoted just above it. All it does is match either of the 6 character strings in the quote. I provided it on the assumption that you were trying to learn techniques to help solve your overall problem. It was not intended as a complete solution.

        If you want help with a complete solution, you will need to read, with care, with attention, with seriousness, my remarks about the importance of being able to determine the start and the end of records in your data.

        gelle_marrisaG 1 Reply Last reply Reply Quote 0
        • gelle_marrisaG
          gelle_marrisa @Neil Schipper
          last edited by

          @neil-schipper
          lets forget the slash value, if there is only 111|22 in 2nd or 3rd line, can we get the desired result?

          \d{10}\R\d{3}/\d{2}\R\d{3} this shows pattern error,

          I am noob, \d{3}[/|]\d{2} not sure to use it as complete pattern or i have to merge it with any previous pattern that was mentioned above.

          Neil SchipperN 1 Reply Last reply Reply Quote 0
          • Neil SchipperN
            Neil Schipper @gelle_marrisa
            last edited by

            @gelle_marrisa

            To convert this:

            12
            abcd :
            115/22
            xyz :
            333
            product :
            blablabla 4567
            code :
            01010
            serial :
            34
            
            56
            abcd :
            116|22
            xyz :
            333
            product :
            blablabla 45678
            code :
            01010
            serial :
            78
            
            90
            abcd :
            117|22
            xyz :
            333
            product :
            blablabla 456789
            code :
            01010
            serial :
            12
            
            

            into this:

            12|115|22|333|4567|01010|34
            56|116|22|333|45678|01010|78
            90|117|22|333|456789|01010|12
            

            You can use:

            F: (\d+)\D+(\d+)\D+(\d+)\D+(\d+)\D+(\d+)\D+(\d+)\D+(\d+)\D+
            R: $1|$2|$3|$4|$5|$6|$7\r\n
            Set cursor to the left of first number of first record
            Execute Replace All

            It will only work on your whole file if every record has exactly 7 numbers.

            1 Reply Last reply Reply Quote 0
            • guy038G
              guy038
              last edited by guy038

              Hello, @neil-schipper, @gelle_marrisa, @peterjones and All,

              An other solution which does not depend on the number of lines of a section would be :

              • SEARCH \D+((^\r\n)+|\z)|\D+

              • REPLACE ?1\r\n:|

              • Tick the Wrap around option

              • Click on the Replace All button

              Of course, I assume that each section is separated by, at least, one pure empty line

              So, from this INPUT text :

              12
              abcd :
              115/22
              xyz :
              333
              product :
              blablabla 4567
              code :
              01010
              serial :
              34
              
              
              
              
              56
              abcd :
              116|22
              xyz :
              
              
              
              
              90
              abcd :
              product :
              blablabla 456789
              code :
              serial :
              12
              

              You would obtain this expected text :

              12|115|22|333|4567|01010|34
              56|116|22
              90|456789|12
              

              Note that if we try to factorize the search regex expression as below :

              • SEARCH \D+(((^\r\n)+|\z)|)

              • REPLACE ?2\r\n:|

              This regex S/R does not work properly and gives this output :

              12|115|22|333|4567|01010|34|56|116|22|90|456789|12
              

              So, why, in this new regex, the case \D+(^\r\n)+ never occurs ? For instance, after the number 34, ending the first section of my exemple ? Well, we have this range of chars : 34\r\n\r\n\r\n\r\n\r\n56. So :

              • First, the regex \D+ matches \r\n\r\n\r\n\r\n\r\n but would need some backtraking process in order that the first alternative \D+(^\r\n)+ matches this same range

              • As the whole regex contains other alternatives, the regex engine, before backtracking, tries a match attempt with the second alternative. However, the regex \D+\z cannot be applied to, at this position !

              • Finally, the regex engine tries the last empty alternative \D+() which, of course, matches the range \r\n\r\n\r\n\r\n\r\n

              This explains why the gap between two sections is never detected with this second version of the regex S/R

              Best Regards,

              guy038

              Neil SchipperN 1 Reply Last reply Reply Quote 1
              • guy038G
                guy038
                last edited by guy038

                Hello, @neil-schipper, @gelle_marrisa, @peterjones and All,

                My reasoning, at the end of my previous post, about the second form of regex \D+(((^\r\n)+|\z)|) is not exact ! Indeed, I said :

                As the whole regex contains other alternatives, the regex engine, before backtracking, tries a match attempt with the second alternative

                But, in this case, the correct search regex of my previous post \D+((^\r\n)+|\z)|\D+, which also contains an alternation, should show the same behavior and always choose the second alternative \D+ ?!

                I’ve tried to find out an explanation, without any success :-( May be, one of yours will be able to find out a plausible one !


                In brief, even simplifying the first version by omitting the \z case , and given this INPUT text, with a blank line after the last 12 number

                12
                abcd :
                115/22
                xyz :
                333
                product :
                blablabla 4567
                code :
                01010
                serial :
                34
                
                
                
                
                56
                abcd :
                116|22
                xyz :
                
                
                
                
                90
                abcd :
                product :
                blablabla 456789
                code :
                serial :
                12
                
                

                Why the regex S/R :

                • SEARCH \D+(^\r\n)+|\D+

                • REPLACE ?1\r\n:|

                gives :

                12|115|22|333|4567|01010|34
                56|116|22
                90|456789|12|
                

                And this second equivalent S/R :

                • SEARCH \D+((^\r\n)+|)

                • REPLACE ?2\r\n:|

                gives this result :

                12|115|22|333|4567|01010|34|56|116|22|90|456789|12|
                

                ???

                BR

                guy038

                P.S. :

                The problem does not comes from the empty alternative. For instance, the regex abc(def|) does find, either, the strings abcdef and just abc !

                1 Reply Last reply Reply Quote 0
                • Neil SchipperN
                  Neil Schipper @guy038
                  last edited by

                  @guy038 said in How to find numbers in multiline in Notepad++:

                  solution which does not depend on the number of lines of a section

                  Very nice solution. I can see its applicability and am glad to know it so thanks for sharing.

                  I won’t be much help on the follow-up discussion. I’m not even clear on what motivated you to go in this direction:

                  if we try to factorize the search regex expression

                  However, in trying to understand one building block of your newer regex, which includes a null in an OR subexpression, I encountered something confusing. I wanted to know “does a captured null return true or false?”

                  So I ran ‘Replace All’ with F=(), R=?1dog:cat on a few cases.

                  In the case of a new empty file, there are 0 matches. This seems wrong, although I wouldn’t be surprised if a more experienced regex person would say it’s correct and expected (because maybe in the docs it says “no text ==> no matches” or maybe, “a zero-length null only occurs before or after a character”).

                  In the case of a file with the single character ‘p’ there are 2 matches and we get dogpdog which seems reasonable.

                  1 Reply Last reply Reply Quote 1
                  • guy038G
                    guy038
                    last edited by guy038

                    Hello, @Neil-Schipper and All,

                    I had never done this test :

                    SEARCH ()

                    REPLACE (?1dog:cat)

                    Interesting ! You said :

                    In the case of a new empty file, there are 0 matches. This seems wrong,…

                    Well, your assertion is a bit philosophical : does an empty file contains a single empty string ( or an infinity ! ) ?

                    Note that , in regex mode, the search of () ( an empty group 1 ) does show the ^ zero length match calltip, when applied to a new empty tab or a zero byte file !

                    However, as you said, even a simple replacement with a dummy string, as for instance Test, does not occur and no text is inserted !

                    Now, if I type in the phrase This is a test in a new tab and I use the regex S/R :

                    SEARCH ()

                    REPLACE ?1:|:x

                    I get, after clicking on the Replace All button, with the Wrap around option ticked, the text :

                    |T|h|i|s| |i|s| |a| |t|e|s|t|
                    

                    And, to my mind, all this is quite logic :

                    • The group 1 is defined and contains an empty string

                    • Technically, an empty string does exist between two characters, as well as before the first char and after the last. So each occurrence is changed into the | char

                    Note that we can obtain the same result with this other regex S/R :

                    SEARCH (.{0})

                    REPLACE ?1:|:x

                    and also with the more simple forms :

                    SEARCH ()

                    REPLACE |

                    or

                    SEARCH .{0}

                    REPLACE |


                    As we’re speaking about empty groups, I would like to mention a particular but important point when using conditional structures, in regex mode :

                    Let’s consider this list :

                    Ted=First Name
                    25=Age
                    Mary=First Name
                    75=Age
                    Elisabeth=First Name
                    47=Age
                    Bob=First Name
                    62=Age
                    

                    Let’s introduce the conditional regex structure (?(1)Age|First Name) which means : if a group 1 has been previously defined, in the search regex, searches for the string Age else searches for the string First Name

                    If we build the regex (?-si)^(\d*).*=(?(1)Age|First Name)$, you could say :

                    • If a line begins with a number, the part \d* matches this number, the part .* matches an empty string = matches the equal sign and the conditional bloc (?(1)Age|First Name) matches the string Age as the group 1 contains the number and is defined

                    • If a line does not begin with a number, the part \d* matches an empty string, the part .* matches the first name, = matches the equal sign and the conditional bloc (?(1)Age|First Name) matches the string First Name as the group 1 is not defined and empty

                    However, running this regex, against our text, it matches only the lines relative to the age and not all the lines. Why ?

                    Well, what really represents the (\d*) group, after the ^ assertion :

                    • If a line begins with some digits, no problem : group 1 is defined and contains the number

                    • Now, if a line does not begin with digits, the group 1 is ALSO defined but contains an empty string

                    Thus, in all cases the group 1 is defined;, breaking the normal behaviour of the conditional part (?(1)Age|First Name)

                    To get a functional overall regex, you need to change this non-optional group 1 (\d*) into an optional group, with a non-optional contents…, thanks to the syntax (\d+)?. Then, the search regex becomes :

                    (?-si)^(\d+)?.*=(?(1)Age|First Name)$

                    This time :

                    • If a line begins with a number, the optional part (\d+)? matches this number and the group 1 is clearly defined and contains this number

                    • But, if a line does not begin with a number the optional part (\d+)? matches nothing and the group 1 is not defined at all !

                    You can verify that this final regex find, as expected, all the lines of our text !

                    Remark : Of course, we could had simply used the regex (?-si)^(\d+=Age|.+=First Name)$, without any conditional block !


                    This reasoning can be applied, as well, to conditional replacements ! For instance, given this text :

                    Ted
                    25
                    Mary
                    75
                    Elisabeth
                    47
                    Bob
                    62
                    

                    The following regex S/R :

                    SEARCH (?-s)^(\d+)?.*$

                    REPLACE (?1Age:First Name) : $&

                    would gives :

                    First Name : Ted
                    Age : 25
                    First Name : Mary
                    Age : 75
                    First Name : Elisabeth
                    Age : 47
                    First Name : Bob
                    Age : 62
                    
                    • If a number begins a line, group 1 is defined and the string Age, followed with \x20:\x20, is inserted right before the number

                    • If a number does not begin a line, the group 1 is not defined at all. So the string First Name, followed with \x20:\x20, is inserted, this time, right before the first name

                    And you’ll verify, that the similar version, with the non-optional group 1 (\d*) :

                    SEARCH (?-s)^(\d*).*$

                    REPLACE (?1Age:First Name) : $&

                    gives wrong results, with the string "Age : " ALWAYS inserted :-((

                    Best Regards,

                    guy038

                    Neil SchipperN 1 Reply Last reply Reply Quote 0
                    • Neil SchipperN
                      Neil Schipper @guy038
                      last edited by

                      Good write up, @guy038. It’s good to know there’s a way to have a group conditionally defined as you showed.

                      To get a functional overall regex, you need to change this non-optional group 1 (\d*) into an optional group, with a non-optional contents…, thanks to the syntax (\d+)?

                      At first it seemed like this property of (spec+)? was an anomaly being exploited, or maybe an afterthought by the regex authors, but upon reflection there is some sense to it:

                      In cases where (spec) has no match…

                      • with (spec*) the (little man in the) machine says, "you asked for a capture group containing zero or more matches, so I’m giving you a capture group that contains null text; and a thing which contains surely must be defined.

                      • with (spec+)? the (little man in the) machine says, "you asked for zero or one capture groups containing matched text, so I give you zero such groups, and a thing of which there are zero (in compsci) has no memory allocated and no address, ie, is undefined

                      After realizing this, I wondered if some sticky situations might arise using this technique when there’s a sequence of these conditionally defined groups (ConDefGrps for short). Consider an expression in which all capture groups (CaptGrps) are also ConDefGrps, and, say the first ConDefGrp doesn’t match, so a CaptGrp isn’t defined, but, the second one does; since this latter one is the first CaptGrp that “comes to life”, wouldn’t its reference be 1 so that any conditional test on it (no matter if later in the same expression or in the substitution statement) would actually be testing for the existence of that second appearing, first defined CaptGrp?

                      So I set up a test to check this.

                      Consider a scheme in which a valid code consists of zero or more number 1’s, then 2’s, then 3’s, in that order, with at least one element present.

                      An expression that only matches lines completely filled by a valid code is: ^(?=\S)([1]+)?([2]+)?([3]+)?$ but that’s not so interesting.

                      Here’s an F/R pair that always captures the whole line whether it contains valid codes or not, and then writes it back with information about each group’s existence appended:

                      F: ^([1]+)?([2]+)?([3]+)?.*$
                      R: $0 - groups (?{1}1:.)(?{2}2:.)(?{3}3:.)

                      When applied to this test data:

                      1
                      2
                      3
                      
                      112
                      1222222223
                      2233111111
                      4
                      4123
                      12z3
                      1111111222223333
                      31
                      32
                      1133
                      

                      we obtain:

                      1 - groups 1..
                      2 - groups .2.
                      3 - groups ..3
                       - groups ...
                      112 - groups 12.
                      1222222223 - groups 123
                      2233111111 - groups .23
                      4 - groups ...
                      4123 - groups ...
                      12z3 - groups 12.
                      1111111222223333 - groups 123
                      31 - groups ..3
                      32 - groups ..3
                      1133 - groups 1.3
                      

                      What the above demonstrates is that when a ConDefGrp is encountered in an expression, even though it may remain “undefined” (and return False in an existence test) it still consumes a group number allocated in the normal fashion.

                      Thus, one need not worry that including multiple ConDefGrps might lead to ambiguity in the mapping of group to group number.

                      1 Reply Last reply Reply Quote 2
                      • guy038G
                        guy038
                        last edited by guy038

                        Hi, @neil-schipper and All,

                        To summarize :

                        • With the syntax ^(1+)?•••••, group 1 must contain some 1'. So, if no 1' can be found in text, the group 1 cannot be defined and is not used as optional
                          (? quantifier )

                        • With the syntax ^(1*)•••••, group 1 may or not contain some 1'. So, if no 1' can be found in text, the group 1 is still defined with empty contents
                          (* quantifier )

                        • With the syntax ^(1)*•••••, group 1 must contain one 1'. So, if no 1' can be found in text, the group 1 cannot be defined and is not used as optional
                          (* quantifier )


                        So, given the text :

                        000000 |
                        111111 |
                        222222 |
                        333333 |
                        111222 |
                        111133 |
                        223333 |
                        112233 |
                        

                        The regex S/R :

                        SEARCH (?-s)^(1+)?(2+)?(3+)?.+

                        REPLACE $0 groups (?{1}1:.)(?{2}2:.)(?{3}3:.)

                        gives :

                        000000 | groups ...
                        111111 | groups 1..
                        222222 | groups .2.
                        333333 | groups ..3
                        111222 | groups 12.
                        111133 | groups 1.3
                        223333 | groups .23
                        112233 | groups 123
                        

                        The regex S/R :

                        SEARCH (?-s)^(1*)(2*)(3*).+

                        REPLACE $0 groups (?{1}1:.)(?{2}2:.)(?{3}3:.)

                        gives :

                        000000 | groups 123
                        111111 | groups 123
                        222222 | groups 123
                        333333 | groups 123
                        111222 | groups 123
                        111133 | groups 123
                        223333 | groups 123
                        112233 | groups 123
                        

                        And the regex S/R :

                        SEARCH (?-s)^(1)*(2)*(3)*.+

                        REPLACE $0 groups (?{1}1:.)(?{2}2:.)(?{3}3:.)

                        gives :

                        000000 | groups ...
                        111111 | groups 1..
                        222222 | groups .2.
                        333333 | groups ..3
                        111222 | groups 12.
                        111133 | groups 1.3
                        223333 | groups .23
                        112233 | groups 123
                        

                        BR

                        guy038

                        1 Reply Last reply Reply Quote 0
                        • Alan KilbornA
                          Alan Kilborn
                          last edited by

                          So this is a good discussion thread, but the choice to use literal 1, 2, 3 in the examples IMO wasn’t the best for the utmost clarity. :-)

                          1 Reply Last reply Reply Quote 0
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