find string and print first and last characters of line











up vote
1
down vote

favorite












I have files with hundreds of lines of varying length. I want to find each line with the string "New" and print the first 7 characters and the 10th from the last character.



For example, cat file1.txt



1234567 New line with irrelevant info x end line

2345678 irrelevant line

3456789 New line with different irrelevant info y end line

4567890 irrelevant line

5678901 New line with yet more irrelevant info z end line


And my output would be:



1234567 x 

3456789 y

5678901 z





sed grep cut







share|improve this question




















  • 1




    Expected output seems wrong .... second line starting with 2345678 is irrelevant.
    – George Vasiliou
    Dec 1 at 21:36






  • 1




    ... as is the third.
    – RudiC
    Dec 1 at 21:45










  • thanks, typo on my part. fixed now.
    – user2535719
    Dec 1 at 21:56















up vote
1
down vote

favorite












I have files with hundreds of lines of varying length. I want to find each line with the string "New" and print the first 7 characters and the 10th from the last character.



For example, cat file1.txt



1234567 New line with irrelevant info x end line

2345678 irrelevant line

3456789 New line with different irrelevant info y end line

4567890 irrelevant line

5678901 New line with yet more irrelevant info z end line


And my output would be:



1234567 x 

3456789 y

5678901 z





sed grep cut







share|improve this question




















  • 1




    Expected output seems wrong .... second line starting with 2345678 is irrelevant.
    – George Vasiliou
    Dec 1 at 21:36






  • 1




    ... as is the third.
    – RudiC
    Dec 1 at 21:45










  • thanks, typo on my part. fixed now.
    – user2535719
    Dec 1 at 21:56













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have files with hundreds of lines of varying length. I want to find each line with the string "New" and print the first 7 characters and the 10th from the last character.



For example, cat file1.txt



1234567 New line with irrelevant info x end line

2345678 irrelevant line

3456789 New line with different irrelevant info y end line

4567890 irrelevant line

5678901 New line with yet more irrelevant info z end line


And my output would be:



1234567 x 

3456789 y

5678901 z





sed grep cut







share|improve this question















I have files with hundreds of lines of varying length. I want to find each line with the string "New" and print the first 7 characters and the 10th from the last character.



For example, cat file1.txt



1234567 New line with irrelevant info x end line

2345678 irrelevant line

3456789 New line with different irrelevant info y end line

4567890 irrelevant line

5678901 New line with yet more irrelevant info z end line


And my output would be:



1234567 x 

3456789 y

5678901 z





sed grep cut




sed grep cut






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 1 at 21:55

























asked Dec 1 at 21:09









user2535719

62




62








  • 1




    Expected output seems wrong .... second line starting with 2345678 is irrelevant.
    – George Vasiliou
    Dec 1 at 21:36






  • 1




    ... as is the third.
    – RudiC
    Dec 1 at 21:45










  • thanks, typo on my part. fixed now.
    – user2535719
    Dec 1 at 21:56














  • 1




    Expected output seems wrong .... second line starting with 2345678 is irrelevant.
    – George Vasiliou
    Dec 1 at 21:36






  • 1




    ... as is the third.
    – RudiC
    Dec 1 at 21:45










  • thanks, typo on my part. fixed now.
    – user2535719
    Dec 1 at 21:56








1




1




Expected output seems wrong .... second line starting with 2345678 is irrelevant.
– George Vasiliou
Dec 1 at 21:36




Expected output seems wrong .... second line starting with 2345678 is irrelevant.
– George Vasiliou
Dec 1 at 21:36




1




1




... as is the third.
– RudiC
Dec 1 at 21:45




... as is the third.
– RudiC
Dec 1 at 21:45












thanks, typo on my part. fixed now.
– user2535719
Dec 1 at 21:56




thanks, typo on my part. fixed now.
– user2535719
Dec 1 at 21:56










3 Answers
3






active

oldest

votes

















up vote
1
down vote













Choose one you like:



awk solution:



awk '/New/{ print substr($0, 1, 7), substr($0, length-9, 1) }' file1.txt



sed solution:



sed -rn '/New/ s/^(.{7}).*(.).{9}$/1 2/p' file1.txt



Sample output (for both approaches):



1234567 x

3456789 y

5678901 z





share|improve this answer






























    up vote
    1
    down vote













    POSIXly:



    Assuming the lines contain at least 10 characters (if not, the behaviour is unspecified for the second substr(), you can add a && length >= 10 or && length >= 17 after /New/ to skip the lines that have fewer than 10 or 17 characters):



    awk '/New/ {print substr($0, 1, 7), substr($0, length - 9, 1)}'


    or assuming the lines contain at least 17 characters (the lines that don't will be skipped):



    sed -n '/New/ s/^(.{7}).*(.).{9}$/1 2/p'





    share|improve this answer

















    • 1




      What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
      – George Vasiliou
      Dec 1 at 21:53










    • @GeorgeVasiliou, he edited the awk solution in after I had started writing mine and I hadn't seen his edit before I posted mine. His sed one is GNU specific, my intention was to offer a standard solution.
      – Stéphane Chazelas
      Dec 1 at 21:56












    • ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
      – George Vasiliou
      Dec 1 at 22:00




















    up vote
    0
    down vote













    This is one "brutal" gawk solution that does the job, using null as field separator FS and output field separator OFS , meaning each char of inputfile is considered to be a field for awk.



    awk '/New/{print $1,$2,$3,$4,$5,$6,$7," ",$(NF-9)}' FS="" OFS="" file1
    
1234567 x
    
3456789 y
    
5678901 z


    More solutions with grep / sed will follow.






    share|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote













      Choose one you like:



      awk solution:



      awk '/New/{ print substr($0, 1, 7), substr($0, length-9, 1) }' file1.txt



      sed solution:



      sed -rn '/New/ s/^(.{7}).*(.).{9}$/1 2/p' file1.txt



      Sample output (for both approaches):



      1234567 x
      
3456789 y
      
5678901 z





      share|improve this answer



























        up vote
        1
        down vote













        Choose one you like:



        awk solution:



        awk '/New/{ print substr($0, 1, 7), substr($0, length-9, 1) }' file1.txt



        sed solution:



        sed -rn '/New/ s/^(.{7}).*(.).{9}$/1 2/p' file1.txt



        Sample output (for both approaches):



        1234567 x
        
3456789 y
        
5678901 z





        share|improve this answer

























          up vote
          1
          down vote










          up vote
          1
          down vote









          Choose one you like:



          awk solution:



          awk '/New/{ print substr($0, 1, 7), substr($0, length-9, 1) }' file1.txt



          sed solution:



          sed -rn '/New/ s/^(.{7}).*(.).{9}$/1 2/p' file1.txt



          Sample output (for both approaches):



          1234567 x
          
3456789 y
          
5678901 z





          share|improve this answer














          Choose one you like:



          awk solution:



          awk '/New/{ print substr($0, 1, 7), substr($0, length-9, 1) }' file1.txt



          sed solution:



          sed -rn '/New/ s/^(.{7}).*(.).{9}$/1 2/p' file1.txt



          Sample output (for both approaches):



          1234567 x
          
3456789 y
          
5678901 z






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Dec 1 at 21:46

























          answered Dec 1 at 21:37









          RomanPerekhrest

          22.8k12246




          22.8k12246
























              up vote
              1
              down vote













              POSIXly:



              Assuming the lines contain at least 10 characters (if not, the behaviour is unspecified for the second substr(), you can add a && length >= 10 or && length >= 17 after /New/ to skip the lines that have fewer than 10 or 17 characters):



              awk '/New/ {print substr($0, 1, 7), substr($0, length - 9, 1)}'


              or assuming the lines contain at least 17 characters (the lines that don't will be skipped):



              sed -n '/New/ s/^(.{7}).*(.).{9}$/1 2/p'





              share|improve this answer

















              • 1




                What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
                – George Vasiliou
                Dec 1 at 21:53










              • @GeorgeVasiliou, he edited the awk solution in after I had started writing mine and I hadn't seen his edit before I posted mine. His sed one is GNU specific, my intention was to offer a standard solution.
                – Stéphane Chazelas
                Dec 1 at 21:56












              • ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
                – George Vasiliou
                Dec 1 at 22:00

















              up vote
              1
              down vote













              POSIXly:



              Assuming the lines contain at least 10 characters (if not, the behaviour is unspecified for the second substr(), you can add a && length >= 10 or && length >= 17 after /New/ to skip the lines that have fewer than 10 or 17 characters):



              awk '/New/ {print substr($0, 1, 7), substr($0, length - 9, 1)}'


              or assuming the lines contain at least 17 characters (the lines that don't will be skipped):



              sed -n '/New/ s/^(.{7}).*(.).{9}$/1 2/p'





              share|improve this answer

















              • 1




                What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
                – George Vasiliou
                Dec 1 at 21:53










              • @GeorgeVasiliou, he edited the awk solution in after I had started writing mine and I hadn't seen his edit before I posted mine. His sed one is GNU specific, my intention was to offer a standard solution.
                – Stéphane Chazelas
                Dec 1 at 21:56












              • ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
                – George Vasiliou
                Dec 1 at 22:00















              up vote
              1
              down vote










              up vote
              1
              down vote









              POSIXly:



              Assuming the lines contain at least 10 characters (if not, the behaviour is unspecified for the second substr(), you can add a && length >= 10 or && length >= 17 after /New/ to skip the lines that have fewer than 10 or 17 characters):



              awk '/New/ {print substr($0, 1, 7), substr($0, length - 9, 1)}'


              or assuming the lines contain at least 17 characters (the lines that don't will be skipped):



              sed -n '/New/ s/^(.{7}).*(.).{9}$/1 2/p'





              share|improve this answer












              POSIXly:



              Assuming the lines contain at least 10 characters (if not, the behaviour is unspecified for the second substr(), you can add a && length >= 10 or && length >= 17 after /New/ to skip the lines that have fewer than 10 or 17 characters):



              awk '/New/ {print substr($0, 1, 7), substr($0, length - 9, 1)}'


              or assuming the lines contain at least 17 characters (the lines that don't will be skipped):



              sed -n '/New/ s/^(.{7}).*(.).{9}$/1 2/p'






              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Dec 1 at 21:48









              Stéphane Chazelas

              297k54562908




              297k54562908








              • 1




                What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
                – George Vasiliou
                Dec 1 at 21:53










              • @GeorgeVasiliou, he edited the awk solution in after I had started writing mine and I hadn't seen his edit before I posted mine. His sed one is GNU specific, my intention was to offer a standard solution.
                – Stéphane Chazelas
                Dec 1 at 21:56












              • ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
                – George Vasiliou
                Dec 1 at 22:00
















              • 1




                What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
                – George Vasiliou
                Dec 1 at 21:53










              • @GeorgeVasiliou, he edited the awk solution in after I had started writing mine and I hadn't seen his edit before I posted mine. His sed one is GNU specific, my intention was to offer a standard solution.
                – Stéphane Chazelas
                Dec 1 at 21:56












              • ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
                – George Vasiliou
                Dec 1 at 22:00










              1




              1




              What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
              – George Vasiliou
              Dec 1 at 21:53




              What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
              – George Vasiliou
              Dec 1 at 21:53












              @GeorgeVasiliou, he edited the awk solution in after I had started writing mine and I hadn't seen his edit before I posted mine. His sed one is GNU specific, my intention was to offer a standard solution.
              – Stéphane Chazelas
              Dec 1 at 21:56






              @GeorgeVasiliou, he edited the awk solution in after I had started writing mine and I hadn't seen his edit before I posted mine. His sed one is GNU specific, my intention was to offer a standard solution.
              – Stéphane Chazelas
              Dec 1 at 21:56














              ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
              – George Vasiliou
              Dec 1 at 22:00






              ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
              – George Vasiliou
              Dec 1 at 22:00












              up vote
              0
              down vote













              This is one "brutal" gawk solution that does the job, using null as field separator FS and output field separator OFS , meaning each char of inputfile is considered to be a field for awk.



              awk '/New/{print $1,$2,$3,$4,$5,$6,$7," ",$(NF-9)}' FS="" OFS="" file1
              
1234567 x
              
3456789 y
              
5678901 z


              More solutions with grep / sed will follow.






              share|improve this answer

























                up vote
                0
                down vote













                This is one "brutal" gawk solution that does the job, using null as field separator FS and output field separator OFS , meaning each char of inputfile is considered to be a field for awk.



                awk '/New/{print $1,$2,$3,$4,$5,$6,$7," ",$(NF-9)}' FS="" OFS="" file1
                
1234567 x
                
3456789 y
                
5678901 z


                More solutions with grep / sed will follow.






                share|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  This is one "brutal" gawk solution that does the job, using null as field separator FS and output field separator OFS , meaning each char of inputfile is considered to be a field for awk.



                  awk '/New/{print $1,$2,$3,$4,$5,$6,$7," ",$(NF-9)}' FS="" OFS="" file1
                  
1234567 x
                  
3456789 y
                  
5678901 z


                  More solutions with grep / sed will follow.






                  share|improve this answer












                  This is one "brutal" gawk solution that does the job, using null as field separator FS and output field separator OFS , meaning each char of inputfile is considered to be a field for awk.



                  awk '/New/{print $1,$2,$3,$4,$5,$6,$7," ",$(NF-9)}' FS="" OFS="" file1
                  
1234567 x
                  
3456789 y
                  
5678901 z


                  More solutions with grep / sed will follow.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Dec 1 at 21:33









                  George Vasiliou

                  5,59531028




                  5,59531028






























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