line addressing using grep

Could someone help me by using grep to address specific line number. Assuming line 37 in a shell script has a syntax error and I want to display the line on the terminal using grep, what grep option and/or regular expression should I use? Tried using

grep -n 37 foo.sh

but it didn't work.
Any help will be useful. Thanks

j2dizzo

Do you have to use grep?? sed would be my choice:

$ sed -n '2p' <file> => Show second line, ranges are also possibel. 2,5 will list lines 2 to 5.

To my knowledge, you cannot give a range (line numbers) to grep, it start searching the file and shows hits. You can put the line number in front of a hit (-n) and you can limit the amount of hits (-m).

I know sed and awk will do the task easier but I'm required to use grep. Could you ellaborate on what you mention about using a number before a hit (-n) and limiting it with (-m) 'cos I don't understand. Moreover, I looked up on the man document and it doens't mention of the -m option.

First things first: Which grep version are you using, and on which OS?

$grep -V
grep (GNU grep) 2.5

The above version, on linux, does have the -m option (also mentioned in the man page).

Ok, the -n option: Show the linenumber of the line that has a hit:

$ grep -n root /etc/passwd
1:root:x:0:0::/root:/bin/bash

The -m option: Only show set amount of hits.

$ grep -m 2 PATH /etc/profile (show only first 2 hits):
export MANPATH=/usr/local/man:/usr/man:/usr/X11R6/man
export PKG_CONFIG_PATH="/usr/lib/pkgconfig:/usr/local/lib/pkgconfig"

$ grep -n -m 1 PATH /etc/profile (show first hit an put linenumber in front):
7:export MANPATH=/usr/local/man:/usr/man:/usr/X11R6/man

If I understand you problem correctly, you want to show line X (or maybe a range X,Y) and do that using grep.

Like I stated before, this cannot be done using grep by itself. Grep searches complete files (or a variable string).