Shell Programming and Scripting

hello all,

i have a text file and i want to remove the first 6 lines and last two lines of it

thanks for your help

What about searching these fora and adapting the solutions found?
Anyhow ... try this:

Code:

awk 'NR<=6 {next}
     {A[NR%3]=$0; j=(NR+1)%3; if (A[j]) print A[j]}
    ' file

thanks but i want easy way i have an idea in my mind to count the lines and use head and tail combination to get what i want but i need a very simple code one way using sed or awk ?

What could be simpler than the awk script proposed?

Code:

# cat file
line1
line2
line3
line4
line5
line6
line7
line8
line9
line10
line11
line12
line13
line14
line15
line16
line17
line18
line19
line20
line21
line22

Code:

# sed -e "1,6 d" -e 'N;$!P;$!D;$d' file
line7
line8
line9
line10
line11
line12
line13
line14
line15
line16
line17
line18
line19
line20

perfect solution thanks alot

could you explain it for me to be understand

Code:

sed -e "1,6 d" -e 'N;$!P;$!D;$d' file

what each symbol of these mean N;$!P;$!D;$d

thanks alot again

From the man of sed:

-e => add the script to the commands to be executed

1,6d => deletes a sequence of 1 to 6 lines. d stands for delete
N=> Read the next line in Pattern space
$!P=> If the current line is not the last line, print the current line
$!D=> If the current line is not the last line, delete the current line
$d=> If the current line is the last line, delete everything in the pattern-space

combination of all this "N;$!P;$!D;$d" will delete the last two lines