I'm sure there are many ways to do this: how can I count the number of lines in a text file?

$ <cmd> file.txt
1020 lines

8 Answers 8


The standard way is with wc, which takes arguments to specify what it should count (bytes, chars, words, etc.); -l is for lines:

$ wc -l file.txt
1020 file.txt
  • How do I count the lines in a file if I want to ignore comments? Specifically, I want to not count lines that begin with a +, some white space (could be no white space) and then a %, which is the way comment lines appear in a git diff of a MATLAB file. I tried doing this with grep, but couldn't figure out the correct regular expression.
    – Gdalya
    Commented Jul 11, 2013 at 1:36
  • @Gdalya I hope the following pipeline will do this (no tests were perfomed): cat matlab.git.diff | sed -e '/^\+[ ]*.*\%$/d' | wc -l. /regexp/d deletes a line if it matches regexp, and -e turns on an adequate (IMNSHO) syntax for regexp.
    – dbanet
    Commented Nov 6, 2013 at 21:29
  • 2
    Why not simply grep -v '^+ *%' matlab.git.diff | wc -l?
    – celtschk
    Commented Jul 6, 2014 at 19:51
  • @celtschk , as long as this is usual in comment lines: is it possible to modify your grep command in order to consider as comment cases like " + Hello" (note the space(s) before the +)? Commented Jan 18, 2015 at 20:46
  • 1
    @SopalajodeArrierez: Of course it is possible: grep -v '^ *+' matlab.git.diff | wc -l (I'm assuming the quote signs were not actually meant to be part of the line; I also assume that both lines with and without spaces in front of the + are meant to be comments; if at least one space is mandatory, either replace the star * with \+, or just add another space in front of the star). Probably instead of matching only spaces, you'd want to match arbitrary whitespace; for this replace the space with [[:space:]]. Note that I've also removed matching the % since it's not in your example.
    – celtschk
    Commented Feb 14, 2015 at 15:02

Steven D forgot GNU sed:

sed -n '$=' file.txt

Also, if you want the count without outputting the filename and you're using wc:

wc -l < file.txt

Just for the heck of it:

cat -n file.txt | tail -n 1 | cut -f1
  • 2
    Or grep -c '', or tr -dc '\n' | wc -c, or nl -ba -nln | tail -n 1 |sed -e 's/[^0-9].*//'... Is any of these useful in itself (as opposed to things to build upon to make a program that does more than counting lines), other than wc -l and pure (ba)sh? Commented Dec 3, 2010 at 1:58
  • 1
    @Gilles: I think the phrase "many ways" in the question triggered a challenge that Steve and I rose to. Commented Dec 3, 2010 at 2:03
  • 1
    @Gilles: sed 's/.*//' file.txt | uniq -c Commented Dec 3, 2010 at 2:30
  • 2
    @Gilles: Oh, you meant first. uniq -c -w 0 file.txt and you can cut -c -7 to keep only the number. Or, more POSIXly: uniq -c file.txt | awk '{c+=$1}END{print c}'. How about dc (even though it's not POSIX)? uniq -c file.txt | cut -c -7 | sed '$alax' | dc -e '[pq]sb[+z1=blax]sa' -. bc is POSIX: uniq -c file.txt | cut -c -7 | sed -n ':a;${s/\n/ + /gp;b};N;ba' | bc. The easy answer if you assume a limited line length: uniq -c -f 100000 file.txt. Commented Dec 3, 2010 at 16:21
  • 1
    @JosipRodin: Quotes added Commented Oct 18, 2015 at 0:09

As Michael said, wc -l is the way to go. But, just in case you inexplicably have bash, perl, or awk but not wc, here are a few more solutions:


$ LINECT=0; while read -r LINE; do (( LINECT++ )); done < file.txt; echo $LINECT

Perl Solutions

$ perl -lne 'END { print $. }' file.txt

and the far less readable:

$ perl -lne '}{ print $.' file.txt

Awk Solution

$  awk 'END {print NR}' file.txt

Word of warning when using

wc -l

because wc -l functions by counting \n, if the last line in your file doesn't end in a newline effectively the line count will be off by 1. (hence the old convention leaving newline at the end of your file)

Since I can never be sure if any given file follows the convention of ending the last line with a newline or not, I recommend using any of these alternate commands which will include the last line in the count regardless of newline or not.

sed -n $= filename
perl -lne 'END { print $. }' filename
awk 'END {print NR}' filename
grep -c '' filename
  • nice summary. And welcome to unix & linux
    – Sebastian
    Commented Sep 18, 2014 at 15:29
  • Hm is the last piece really line?
    – gena2x
    Commented Sep 18, 2014 at 19:48
  • 1
    I'm sure it depends on everyone's usecase; for the 'last piece' is usually a line of text that someone didn't cap off with a newline. The usecase I most often encounter is a file with a single string of text that does not end in a newline. wc -l would count this as "0", when I would otherwise expect a count of "1". Commented Sep 23, 2014 at 16:22
  • It is not the count of wc that will be off by one, but your count: While in Windows (and DOS before it), the CR/LF sequence is a line separator, on Unix the LF character is a line terminator. That is, without a newline at the end, you don't have a line (and strictly speaking, not a valid text file).
    – celtschk
    Commented Feb 7, 2020 at 9:15

You can always use the command grep as follows:

grep -c "^" file.txt

It will count all the actual rows of file.txt, whether or not its last row contains a LF character at the end.


In case you only have bash and absolutely no external tools available, you could also do the following:

while read
done <file.txt
echo $count

Explanation: the loop reads standard input line by line (read; since we do nothing with the read input anyway, no variable is provided to store it in), and increases the variable count each time. Due to redirection (<file.txt after done), standard input for the loop is from file.txt.

  • This is a very inefficient way to do it. Remember, bash reads are slow. Commented Feb 7, 2020 at 1:51
  • @codeforester: That's true, but (a) it was a solution for when you have no other tool available, and (b) slow doesn't mean unawaitable. I just tried with a text file of 125MB (taking an actual text file and concatenating it a thousand times) and more than 2.6 million lines, and it took slightly less than 14 seconds. Not nothing — the tools do it in a fraction of a second — but certainly awaitable.
    – celtschk
    Commented Feb 7, 2020 at 8:58
  • 1
    This would miscount if any line ended with a backslash.
    – Kusalananda
    Commented Apr 11, 2021 at 17:54

If you're looking to count smaller files a simple wc -l file.txt could work.

Looking for an answer to this question myself, working with large files that are several gigs, I found the following tool:


Also, depending on your system configuration--if you're using an older version of wc you might be better off piping larger chunks with dd like so:

dd if={file_path} bs=128M | wc -l


grep -c $ is very simple and works great.

I even saved it as an alias since I use it a lot (lc stand for line count):

alias lc="grep -c $"

It can be used either this way:

lc myFile

Or that way:

cat myFile | lc

Note that this will not count the last line if it is empty. For my uses that is almost always OK though.

  • If your grep implementation does not count the last line if it's empty, it's severely broken. On the contrary there are some grep implementations that count the extra bytes found after the last newline in non-text files as an extra line. For instance, printf foo | grep -c $ outputs 1 with GNU grep even though printf outputs no line. printf foo | wc -l correctly outputs 0. Commented Oct 30, 2022 at 10:38
  • @StéphaneChazelas But wc -l will be off by 1 for files without a newline at the end of the file (in those cases it will not count the last line, even if it has text). Isn't that sort of more broken? For my personal use, I'd rather skip counting the newline at EOF than skip counting a line with actual stuff in it. But I guess everybody counts lines for different purposes, and wc -l might be better for some! :)
    – pitamer
    Commented Oct 30, 2022 at 13:10
  • That's the point. A line has to delimited by a newline character. The characters after the last newline if any don't form part of a line. If there are such characters, by definition the file is not a text file. The output of printf foo does not form a text file. Commented Oct 30, 2022 at 17:21

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