find string and print first and last characters of line












1














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
















1














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














1












1








1







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


















2














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





















1














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































    0














    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









      2














      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


















      2














      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
















      2












      2








      2






      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

      299k54564913




      299k54564913








      • 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















      1














      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




























        1














        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


























          1












          1








          1






          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

          22.8k12346




          22.8k12346























              0














              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


























                0














                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
























                  0












                  0








                  0






                  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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