How to print third column to last column

Question

I m trying to remove the first two columns  of which I m not interested in  from a DbgView log file  I can t seem to find an example that prints from column 3 onwards until the end of the line  Note that each line has variable number of columns

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In AWK columns are called fields  hence NF is the key  all rows   awk -F   lt column separator gt     print   NF-2     lt filename gt    first row only   awk -F   lt column separator gt    NR lt  1 print   NF-2     lt filename gt

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A bit late here  but none of the above seemed to work   Try this  using printf  inserts spaces between each   I chose to not have newline at the end   awk   for i 3 i lt  NF   i  printf   s      i

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Well  you can easily accomplish the same effect using a regular expression  Assuming the separator is a space  it would look like   awk    sub                        print

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Perl solution   perl -lane  splice  F 0 2  print join      F  file   These command-line options are used    -n loop around every line of the input file  do not automatically print every line -l removes newlines before processing  and adds them back in afterwards -a autosplit mode     split input lines into the  F array   Defaults to splitting on whitespace -e execute the perl code   splice  F 0 2 cleanly removes columns 0 and 1 from the  F array  join      F joins the elements of the  F array  using a space in-between each element  If your input file is comma-delimited  rather than space-delimited  use -F  -lane    Python solution   python -c  import sys  sys stdout write     join line split   2        n   for line in sys stdin    lt  file

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or a simpler solution  cut -f 3- INPUTFILE  just add the correct delimiter  -d  and you got the same effect

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awk    print substr  0  index  0  3        solution found here  http   www linuxquestions org questions linux-newbie-8 awk-print-field-to-end-and-character-count-179078

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awk   for i 3 i lt  NF   i print  i

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In Bash you can use the following syntax with positional parameters   while read -a cols  do echo   cols    2   done  lt  file txt   Learn more  Handling positional parameters at Bash Hackers Wiki

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If you want to print the columns after the 3rd for example in the same line  you can use   awk   for i 3  i lt  NF    i  printf   s     i  print        For example   Mar 09 39 20180301 123131 jpg Mar 13 28 20180301 124304 jpg Mar 13 35 20180301 124358 jpg Feb 09 45 Cisco WebEx Add-On dmg Feb 12 49 Docker dmg Feb 09 04 Grammarly dmg Feb 09 20 Payslip 10459  2828-02-2018 29 pdf   It will print   20180301 123131 jpg 20180301 124304 jpg 20180301 124358 jpg Cisco WebEx Add-On dmg Docker dmg Grammarly dmg Payslip 10459  2828-02-2018 29 pdf   As we can see  the payslip even with space  shows in the correct line

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What about following line   awk    1  2  3     print   file  Based on  ghostdog74 suggestion  Mine should behave better when you filter lines  i e    awk    exim4-config    1     print    file

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awk -v m   x0a  -v N  3     N m N  print substr  0  index  0 m  1      This chops what is before the given field nr   N  and prints all the rest of the line  including field nr N and maintaining the original spacing  it does not reformat   It doesn t mater if the string of the field appears also somewhere else in the line  which is the problem with daisaa s answer   Define a function   fromField       awk -v m   x0a  -v N   1     N m N  print substr  0 index  0 m  1        And use it like this     echo    bat   bi       iru   lau bost       fromField 3 iru   lau bost      echo    bat   bi       iru   lau bost       fromField 2 bi       iru   lau bost    Output maintains everything  including trailing spaces  Works well for files where   n  is the record separator so you don t have that new-line char inside the lines  If you want to use it with other record separators then use   awk -v m   x01  -v N  3     N m N  print substr  0  index  0 m  1      for example  Works well with almost all files as long as they don t use hexadecimal char nr  1 inside the lines

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Jonathan Feinberg s answer prints each field on a separate line  You could use printf to rebuild the record for output on the same line  but you can also just move the fields a jump to the left   awk   for  i 1  i lt  NF-2  i     i     i 2   NF- 2  print   logfile

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awk    1  2  3    1  file   NB  this method will leave  blanks  in 1 2 3 fields but not a problem if you just want to look at output

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awk   print     for i 3 i lt  NF   i printf  i

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The following awk command prints the last N fields of each line and at the end of the line prints a new line character   awk   for  i 6  i lt  NF  i     printf    s     i     printf    n         Find below an example that lists the content of the  usr bin directory and then holds the last 3 lines and then prints the last 4 columns of each line using awk     ls -ltr  usr bin    tail -3 -rwxr-xr-x 1 root root       14736 Jan 14  2014 bcomps -rwxr-xr-x 1 root root       10480 Jan 14  2014 acyclic -rwxr-xr-x 1 root root    35868448 May 22  2014 skype    ls -ltr  usr bin    tail -3   awk   for  i 6  i lt  NF  i     printf    s     i     printf    n       Jan 14 2014 bcomps  Jan 14 2014 acyclic  May 22 2014 skype

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awk    1  2    1  FILENAME   sed  s  s    g    First two columns are cleared  sed removes leading spaces

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awk   a match  0   3   print substr  0 a      First you find the position of the start of the third column   With substr you will print the whole line   0  starting at the position in this case a  to the end of the line

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awk   for  i 4  i lt  NF  i   printf   c    i   printf   n        prints records starting from the 4th field to the last field in the same order they were in the original file

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If its only about ignoring the first two fields and if you don t want a space when masking those fields  like some of the answers above do     awk   gsub  1    2         print    file

[linux] How to print third column to last column?

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