bash read a given line number from text

bendeco13 · 12-09-2004, 06:33 PM

I'm writing a bash script that reads entries from a fext file (kinda like a database)
The script is scanning in directories off of the text file.

I tried to use this but I this don't work with awk.

Code:

DCOUNT=1
      DCNT=cont
      
      while [ $DCNT != no ]
      do
         D[$DCOUNT]=`awk '{ printf $'$DCOUNT' }' data/dirlist.dat`  
	 # DOEASN'T WORK WITH SPACES. NEEDS SOME OTHER WAY OF RESTORING THE DIRECTORIES.
	 if [ ! "'${D[$DCOUNT]}'" ];
         then
            DCNT=no
         else
            echo TEST
         fi
      done

I tried using cat and grep, but they read in all the lines.
Is there a way to get either of these to only read in a specified line or another program similar that would do this?

THANKS IN ADVANCE

wapcaplet · 12-09-2004, 07:58 PM

The sed command for printing line 52 is:

Code:

sed -n '52p' (file)

(taken from Handy one-liners for sed). There are many other ways to do it, but this way is short and sweet.

randyding · 12-09-2004, 08:00 PM

I'm a little unsure what you're trying so forgive me if this isn't what you need.

To pick line 4 out of a file foo.txt, you can do it this way

line4="`sed -n 4p foo.txt`"

bendeco13 · 12-09-2004, 09:05 PM

Thanks!!

That Works

figure20012 · 08-31-2012, 12:23 PM

Quote:

Originally Posted by randyding

I'm a little unsure what you're trying so forgive me if this isn't what you need.

To pick line 4 out of a file foo.txt, you can do it this way

line4="`sed -n 4p foo.txt`"

instead of 4 can i pass a variable let say $linenum
please help

SecretCode · 08-31-2012, 12:49 PM

Of course, yes. The only catch is you have to surround the variable name with { } to distinguish it from the alphanumeric character immediately following.

Code:

sed -n ${linenum}p foo.txt

firstfire · 08-31-2012, 12:54 PM

Hi.

Code:

line = $(sed -n "$linenum{p;q;}")

The q here tells sed to quit immediately after the print command -- we do not need to iterate through the rest of the file.

David the H. · 08-31-2012, 03:49 PM

There are any number of ways to print out a specific line number from a file. The sed way above is one. Here are a few more:

Code:

inputfile=yourfile.txt
lineno=10

awk "NR==$lineno" "$inputfile"

echo "${lineno}p" | ed -s "$inputfile"

cut -d $'\n' -f "$lineno" <"$inputfile"

head -n "$lineno" "$inputfile" | tail -n 1

mapfile -t -O 1 lines <"$inputfile"
echo "${lines[lineno]}"

n=1
while read line; do
	case $n in
		$lineno) echo "$line" && break ;;
		      *) : $(( n++ )) && continue ;;
	esac
done <"$inputfile"

The last two are pure shell versions. mapfile requires bash v.4+ and is not recommended for large files (since the contents get stored in RAM), but is otherwise an efficient solution.

The cut solution uses ansi-c style quoting to set the delimiter to newline. This is only available in modern shells like bash or ksh. You can use a literal newline instead if portability is a concern.