UNIX for Dummies Questions & Answers

Hai Friends

Can anyone provide me a grep command to print x to y lines in a file.

For example:
grep command to display 15th line to 21st

Thanks in advance
Collins

Using grep, no. Maybe one of the others on the forum knows how.

Using awk, you can find a string and print out the lines between it and another string ( awk /"first string"/ /"second string"/ )

Using wc command, you can find the total number of lines in a file, then cat the file, pipe to tail, pipe to head - that combination can give you line 15-21.

Something like ...
cat hosts |tail -200|head -6

I'd say sed is easiest.

Try
sed -n '5,10p' manylines

To print lines 5 thru 10 of the file "manylines"

Cheers
ZB

If I'm dealing with a huge file, and the lines I want spread in between the lines, how can I just pull those lines? I will print the lines to a file and use vi, but vi is not opening a huge file. For example, I have a file with 11163948 lines and the lines I want are between 10085200-10089210, 10090990-10091260 and 10604550-10604900.

Use sed -n 'x,yp' where x is the starting line & y is the last line..

Just as explained in post above ur post..

Also to open a huge file you can use ISPF editor.