Print a file skipping the first X lines in Bash

Question

I have a very long file which I want to print  skipping the first 1 000 000 lines  for example  I looked into the cat man page  but I did not see any option to do this  I am looking for a command to do this or a simple Bash program

User · Answer

A less verbose version with AWK   awk  NR  gt  1e6  myfile txt   But I would recommend using integer numbers

User · Answer

Just to propose a sed alternative     To skip first one million lines  try  sed  1 1000000d    Example     perl -wle  print for  1  1 000 005   sed  1 1000000d  1000001 1000002 1000003 1000004 1000005

User · Answer

If you want to skip first two line   tail -n  3  lt filename gt    If you want to skip first x line   tail -n     x 1    lt filename gt

User · Answer

Use   sed -n  1d p    This command will delete the first line and print the rest

User · Answer

Use the sed delete command with a range address  For example   sed 1 100d file txt   Print file txt omitting lines 1-100    Alternatively  if you want to only print a known range  use the print command with the -n flag   sed -n 201 300p file txt   Print lines 201-300 from file txt   This solution should work reliably on all Unix systems  regardless of the presence of GNU utilities

User · Answer

You ll need tail   Some examples      tail great-big-file log  lt  Last 10 lines of great-big-file log  gt    If you really need to SKIP a particular number of  first  lines  use    tail -n   lt N 1 gt   lt filename gt   lt  filename  excluding first N lines   gt    That is  if you want to skip N lines  you start printing line N 1  Example     tail -n  11  tmp myfile  lt   tmp myfile  starting at line 11  or skipping the first 10 lines   gt    If you want to just see the last so many lines  omit the         tail -n  lt N gt   lt filename gt   lt  last N lines of file   gt

User · Answer

If you want to see the first 10 lines you can use sed as below   sed -n  1 10 p  myFile txt   Or if you want to see lines from 20 to 30 you can use   sed -n  20 30 p  myFile txt

User · Answer

cat  lt  File  gt    awk   if NR  gt  6  print  0

User · Answer

You can do this using the head and tail commands   head -n  lt num gt    tail -n  lt lines to print gt    where num is 1e6   the number of lines you want to print

User · Answer

This shell script works fine for me      bin bash awk -v initial line  1 -v end line  2        if  NR  gt   initial line  amp  amp  NR  lt   end line       print  0     3   Used with this sample file  file txt    one two three four five six   The command  it will extract from second to fourth line in the file    edu debian5     script sh 2 4 file txt   Output of this command   two three four   Of course  you can improve it  for example by testing that all argument values are the expected  -

User · Answer

Easiest way I found to remove the first ten lines of a file    sed 1 10d file txt  In the general case  where X is the number of initial lines to delete  credit to commenters and editors for this     sed 1 Xd file txt

User · Answer

I needed to do the same and found this thread   I tried  tail -n    but it just printed everything   The more  lines worked nicely on the prompt  but it turned out it behaved totally different when run in headless mode  cronjob    I finally wrote this myself   skip 5 FILE   tmp filetoprint  tail -n    cat    FILE     wc -l  - skip      FILE

User · Answer

If you have GNU tail available on your system  you can do the following   tail -n  1000001 huge-file log   It s the   character that does what you want   To quote from the man page      If  the  first  character of K  the number of bytes or lines  is a        print beginning with the Kth item from the start of each  file    Thus  as noted in the comment  putting  1000001 starts printing with the first item after the first 1 000 000 lines

[linux] Print a file, skipping the first X lines, in Bash

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