With the Linux "cat" command, how do I show only certain lines by number

Question

If I use cat -n text.txt to automatically number the lines, how do I then use the command to show only certain numbered lines.

Answer 1

Use sed

Usage

$ cat file
Line 1
Line 2
Line 3
Line 4
Line 5
Line 6
Line 7
Line 8
Line 9
Line 10

To print one line (5)

$ sed -n 5p file
Line 5

To print multiple lines (5 & 8)

$ sed -n -e 5p -e 8p file
Line 5
Line 8

To print specific range (5 - 8)

$ sed -n 5,8p file
Line 5
Line 6
Line 7
Line 8

To print range with other specific line (5 - 8 & 10)

$ sed -n -e 5,8p -e 10p file
Line 5
Line 6
Line 7
Line 8
Line 10

Answer 2

One way of doing it is by using sed:

cat -n text.txt | sed '11d'

where 11 is the number of the line you want removed.

Or to remove all but 11:

cat -n text.txt | sed '11!d'

Ranges are also possible:

cat -n text.txt | sed '9,12!d'

And cat -n isn't even needed:

sed '9,12!d' text.txt

Answer 3

You can use awk straight up.

awk 'NR==1' file.txt

replacing '1' with the desired line number.

Answer 4

Depending on goals I like head or grep

cat /var/log/syslog -n | head -n 50 | tail -n 10

will return lines 41 thru 50.

or

cat /var/log/syslog -n | grep " 50" -b10 -a10

will show lines 40 thru 60. The problem with the grep method is that you have to use account for padding of the line numbers (notice the space)

Both are quite handy for parsing log files.

Answer 5

As others have shown you, there is no need to use cat -n. Other programs will do it for you. If, however, you really need to parse the output of cat -n and show only specific lines (for example, 4-8, 12 and 42), you could do:

$ cat -n file | awk '$1>=4 && $1<=8 || $1==12 || $1==42'
 4  Line 4
 5  Line 5
 6  Line 6
 7  Line 7
 8  Line 8
12  Line 12
42  Line 42

In awk, $1 is the first field, so this command prints all lines whose first fields are i) between 4 and 8 (inclusive) or ii) 12 or iii) 42.

If you also want to remove the field added by cat -n to get the original lines from the file, you can do:

$ cat -n file | awk '$1>=4 && $1<=8 || $1==12 || $1==42{sub(/^\s*[0-9]+\s*/,""); print}'
Line 4
Line 5
Line 6
Line 7
Line 8
Line 12
Line 42

Answer 6

You can use sed to show only one number and loop this with a for loop:

for line in 1 3 7 11; do sed -n ${line}p text.txt; done