Bash get first word

AWK is a good option if you have to deal with trailing whitespace because it’ll take care of it for you:

echo "   word1  word2 " | awk '{print $1;}' # Prints "word1"

cut won’t take care of this though:

echo "  word1  word2 " | cut -f 1 -d " " # Prints nothing/whitespace

‘cut’ here prints nothing/whitespace, because the first thing before a space was another space.

There isn’t any need to use external commands. Bash itself can do the job. Assuming «word1 word2» you got from somewhere and stored in a variable, for example,

$ string="word1 word2"
$ set -- $string
$ echo $1
word1
$ echo $2
word2

Now you can assign $1, $2, etc. to another variable if you like.

I think one efficient way is the use of Bash arrays:

array=( $string ) # Do not use quotes in order to allow word expansion
echo ${array[0]}  # You can retrieve any word. Index runs from 0 to length-1

Also, you can directly read arrays in a pipe-line:

echo "word1 word2" | while read -a array; do echo "${array[0]}" ; done

echo "word1 word2 word3" | { read first rest ; echo $first ; }

This has the advantage that is not using external commands and leaves the $1, $2, etc. variables intact.

answered May 20, 2014 at 2:33

Serge StroobandtSerge Stroobandt

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answered Mar 13, 2010 at 23:21

answered Nov 13, 2014 at 14:00

I wondered how several of the top answers measured up in terms of speed. I tested the following:

1 @mattbh’s

echo "..." | awk '{print $1;}'

2 @ghostdog74’s

string="..."; set -- $string; echo $1

3 @boontawee-home’s

echo "..." | { read -a array ; echo ${array[0]} ; }

and 4 @boontawee-home’s

echo "..." | { read first _ ; echo $first ; }

I measured them with Python’s timeit in a Bash script in a zsh terminal on macOS, using a test string with 215 5-letter words. I did each measurement five times (the results were all for 100 loops, best of 3), and averaged the results:

Method       Time
--------------------------------
1. awk       9.2 ms
2. set       11.6 ms (1.26 * "1")
3. read -a   11.7 ms (1.27 * "1")
4. read      13.6 ms (1.48 * "1")

answered Mar 14, 2018 at 17:49

echo "word1 word2" | cut -f 1 -d " "

cut cuts the first field (-f 1) from a list of fields delimited by the string » » (-d " ").

gniourf_gniourfgniourf_gniourf

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As Perl incorporates AWK’s functionality, this can be solved with Perl too:

echo " word1 word2" | perl -lane 'print $F[0]'

Klas MellbournKlas Mellbourn

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answered Jan 5, 2018 at 10:49

vdegennevdegenne

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You can pipe it through awk and make it echo the first word

echo * | head -n1 | awk '{print $1;}'

or you cut the string up and select the first word:

echo *  | head -n1 | cut -d " " -f1

or you pipe it thorugh sed and have it remove everything but the first word

echo * | head -n1 | sed -e 's/s.*$//'

Added the | head -n1 to satisfy nitpickers. In case your string contains newlines | head -n1 will select the first line first before the important commands select the first word from the string passed to it.

Assuming a posixy shell (/bin/sh or /bin/bash can do this)

all=$(echo *)
first=${all%% *}

The construct ${all%% *} is an example of substring removal. The %% means delete the longest match of * (a space followed by anything) from the right-hand end of the variable all. You can read more about string manipulation here.

This solution assumes that the separator is a space. If you’re doing this with file names then any with spaces will break it.

answered May 6, 2015 at 10:55

Assuming that you really want the first filename and not the first word, here’s a way that doesn’t break on whitespace:

shopt -s nullglob
files=(*)
printf '%sn' "${files[0]}"

answered Feb 24, 2013 at 12:07

You can use the positional parameters

set -- *
echo "$1"

answered Feb 24, 2013 at 12:39

answered Oct 16, 2015 at 13:14

Another approach is to list all the file names as an array, and then index the array for the first element:

STRARRAY=($(echo *))
FIRST=${STRARRAY[0]}

answered Apr 3, 2022 at 21:59

muru

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murumuru

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heemaylheemayl

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