display a specific line from a file

[mooi-kuan]

Hi,
I need to read a specific line from an input file. Is there any unix command syntax like,

unixcmd 4 file_name

where 4 is the line number to be read.

Thank you.

[mooi-kuan]

Sorry, I found a way for this using
sed -n 4p file_name

but if you have any other suggestion, please let me know too. Thank you.

[Madhu Sudhan_1]

Mooi-kuan lim:

One of the other way to do the same is:

# awk -v line=5 'NR == line { print $0 }' /etc/passwd

You can replace 5 with any shell variable if you need to. Here I have taken the file /etc/passwd.

Enjoy !
......Madhu

[federico_3]

type simply ( if you want to get the line 5 ):

cat file | awk '{ if (NR==5) {print $0}}'


regards,

federico

[Stefan Schulz]

Using sed is the fastest solution to this porblem. Of course you can also use awk or perl or .....

But if you have to select a known line by number sed is the fastest.

[Stefan Schulz · Accepted solution]

Using sed is the fastest solution to this porblem. Of course you can also use awk or perl or .....

But if you have to select a known line by number sed is the fastest.

[Bruce Regittko_1]

...And just to be different:

head -X | tail -1

where X is the line number to be displayed.

Of course though, sed is the way to go.

[Dan Hetzel]

Hi,

There are many ways to achieve what you want, including a for or while loop, but 'sed' is definitely the way to go.

sed -n ${LINENUM}p < filename

even allows you to have your line number in a variable

sed -n ${FIRST},${LAST}p < filename
will print lines from FIRST to LAST

There should be a space after the '-n' and no space before 'p'

Dan

[Igot ToMuch Time]

Try this:

ex junk.txt << EOF
5p
EOF

[Ivan Ferreira]

This will print the 4th line:

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

[Ivan Ferreira]

You can create an script called showline like this:

showline.sh:

awk -v LINE=$1 'NR==LINE {print $0}' $2

Usage:

showline.sh