extract a line from a file using the line number

grandtheftander · #1 (**permalink**) 08-02-2006

Hello,
I am having trouble extracting a specific line from a file when the line number is known. My first attempt involved grep -n 'hi' (the word 'hi will always be there) to get the line number before the line that I actually want (line 4).
Extra Notes:
-I am working in a bash script.
-The file size is always variable, so I couldn't use head or tail.
#file that I am evaluating

Code:

hello
hey
hi
there <-line that i want
how
are
you 
doing

blowtorch · #2 (**permalink**) 08-02-2006

Use sed. Check the 'p' command in particular.

System Shock · #3 (**permalink**) 08-02-2006

If you want a particular line number, say line 4, using head and tail commands:

Code:

 head -4 filename | tail -1

Using sed:

Code:

 sed 4!d filename

vish_indian · #4 (**permalink**) 08-03-2006

You can use awk

awk 'NR==4{print $0}' filename

would print the 4th line for you.

grandtheftander · #5 (**permalink**) 08-03-2006

Thank you everyone for responding. I went with the head to tail command. I previously didn't think the head/tail command would work but thank you for showing me the correct usage.

tmarikle · #6 (**permalink**) 08-03-2006

No need to explicitly print in the awk solution, it can simply be: awk 'NR==4' filename

ashishdugar · #7 (**permalink**) 02-06-2008

Quote:

Originally Posted by grandtheftander

Hello,
I am having trouble extracting a specific line from a file when the line number is known. My first attempt involved grep -n 'hi' (the word 'hi will always be there) to get the line number before the line that I actually want (line 4).
Extra Notes:
-I am working in a bash script.
-The file size is always variable, so I couldn't use head or tail.
#file that I am evaluating

Code:

hello
hey
hi
there <-line that i want
how
are
you 
doing

head -4 | awk '{ print $1 }' < filename | tail -1

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