Grep a single line out of file?

I need a shell script that will return a single value from a file given a particular line.  For example a file contains:

10
20
30
33
44

I need a command that can return the value 33 when passed in the param 4, for line 4.

Thanks,
Ethan
epostAsked:
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geotigerConnect With a Mentor Commented:
One liner will do:

cat yourfile | awk '{if (NR==4) print $0}'

This will print the fourth line. If you want just print particular field in the line. Please let me know.
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epostAuthor Commented:
Cool, how can I turn the line number into a variable?   When I use $1 it thinks I mean a column.
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geotigerCommented:
Yes, say, if you want to print out all the lines with the first column eq 33, you can do it as

$ echo "33 34\n34 35\n33 36\n37 45" | awk '{if ($1==v) print $0}' v=33
33 34
33 36

If these lines are in a file, you can just use cat or just put it in the end.

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