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Comment: Another method of printing a line using -n and p flag with sed.
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| The dirty (but not quick) way would be {{{sed -n ${n}p "$file"}}} but this reads the whole input file, even if you only wanted the third line. | One dirty (but not quick) way is: |
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| The following {{{sed}}} command line reads a file printing nothing (-n). At line $n the command "p" is run, printing it, with a "q" afterwards: quit the program. {{{ sed -n "$n{p;q;}" "$file" |
{{{#!highlight bash sed -n "${n}p" "$file" |
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| Another way, more obvious to some, is to grab the last line from a listing of the first n lines: {{{ head -n $n $file | tail -n 1 |
But this reads the entire file even if only the third line is desired, which can be avoided by using the `q` command to quit on line `$n`, and deleting all other lines with the `d` command: {{{#!highlight bash sed "${n}q;d" "$file" |
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| Using awk: {{{ awk 'NR==n{print;exit}' file |
There is yet another simple way of printing only the line you want using `sed`. The "-n" option will not print anything unless `p` flag is explicitly specified. For example to print the 4th line in a file use. {{{#!highlight bash sed -n "4p" "$file" |
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Another method is to grab lines starting at `n`, then get the first line of that. {{{#!highlight bash tail -n "+$n" "$file" | head -n 1 }}} Another approach, using [[AWK]]: {{{#!highlight bash awk "NR==$n{print;exit}" "$file" }}} If more than one line is needed, it's easy to adapt any of the previous methods: {{{#!highlight bash x=3 y=4 sed -n "$x,${y}p;${y}q;" "$file" # Print lines $x to $y; quit after $y. head -n "$y" "$file" | tail -n "$((y - x + 1))" # Same head -n "$y" "$file" | tail -n "+$x" # If your tail supports it awk "NR>=$x{print} NR==$y{exit}" "$file" # Same }}} Or a counter with a simple `read` loop: {{{#!highlight bash # Bash/ksh m=0 while ((m++ < n)) && read -r _; do : done head -n 1 }}} To read into a variable, it is preferable to use `read` or `mapfile` rather than an external utility. More than one line can be read into the given array variable or the default array `MAPFILE` by adjusting the argument to mapfile's -n option: {{{#!highlight bash # Bash4 mapfile -ts "$((n - 1))" -n 1 x <"$file" printf '%s\n' "$x" }}} === See Also === * [[BashFAQ/001]] * http://wiki.bash-hackers.org/commands/builtin/mapfile ---- CategoryShell |
How can I print the n'th line of a file?
One dirty (but not quick) way is:
1 sed -n "${n}p" "$file"
But this reads the entire file even if only the third line is desired, which can be avoided by using the q command to quit on line $n, and deleting all other lines with the d command:
1 sed "${n}q;d" "$file"
There is yet another simple way of printing only the line you want using sed. The "-n" option will not print anything unless p flag is explicitly specified. For example to print the 4th line in a file use.
1 sed -n "4p" "$file"
Another method is to grab lines starting at n, then get the first line of that.
1 tail -n "+$n" "$file" | head -n 1
Another approach, using AWK:
1 awk "NR==$n{print;exit}" "$file"
If more than one line is needed, it's easy to adapt any of the previous methods:
Or a counter with a simple read loop:
To read into a variable, it is preferable to use read or mapfile rather than an external utility. More than one line can be read into the given array variable or the default array MAPFILE by adjusting the argument to mapfile's -n option: