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When I 'echo *' I get the following output:

file1 file2 file3 ...

What I want is to pick out the first word. How can I proceed?

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Assuming this wasn't just an example, to get the first file in a directory, use ls | head -1. – mattdm Feb 24 '13 at 15:41
@mattdm Using ls won't work if one of the filenames contains a blank. – helpermethod Feb 25 '13 at 17:07
How do you define word? If the first file is Sunset on a beach.jpg, should it be Sunset or the whole file name? What about Sea, sex and sun.ogg? Sea, Sea, or the whole file name? – Stéphane Chazelas Mar 21 at 12:46

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.

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The would return the first word of every line, not the first word. – Stéphane Chazelas Feb 24 '13 at 20:19
i wonder how many lines echo * generates – Bananguin Feb 25 '13 at 14:10
That depends how many files have newline characters in their name or depending on the environment or how bash was compiled or whether some file called -e or -ee... appears in the list, how many time \n appears in a file name. If there's a file called -n, it might not even return any line at all... – Stéphane Chazelas Feb 25 '13 at 15:05
Well, we're still talking bash and 1. usually bash will not pass a string with newlines as one string with new lines 2. it's very hard and unusual (impossible?) to have a newline in a filename 3. a \n in a filename will show up as a \n 4. 3 holds for filenames starting with - 5. even when called with -n or -e echo will open stdout and close it when it's done so of course it will return a line, and at least one string, for that matter 6. i edited my advice to at least take care of the multiline problem – Bananguin Feb 25 '13 at 16:34
All 5 points are false. Try in an empty dir: touch '$a\nb' 'a\nb'; env BASHOPTS=xpg_echo bash -c 'echo * | wc -l' (xpg_echo is enabled wherever bash is required to be Unix conformant). And in another empty directory: touch ./-n; bash -c 'echo * | wc -l'. A line is a sequence of characters terminated by a newline character. If echo doesn't output a newline character, it doesn't output any line. Behavior of text utilities like cut, awk or sed is unspecified if the input has extra characters after the last newline character and behavior varies across implementations. – Stéphane Chazelas Feb 25 '13 at 17:52

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
printf '%s\n' "${files[0]}"
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First word would be easy, too: files=($(echo *)) – Hauke Laging Feb 24 '13 at 12:54
It would break on "-n", "-e", "-ne", "-Enenene"..., and depending on how bash was compiled or the environment, possibly on backslash characters, though. – Stéphane Chazelas Feb 24 '13 at 20:18
@HaukeLaging, you'd need to disable globbing. Like: text=$(echo *); set -f; files=($text), otherwise more wildcards could be expanded. – Stéphane Chazelas Feb 24 '13 at 20:23
and files=$(echo *); echo ${files%% *} – 발렌탕 Feb 27 '13 at 19:41
@memnoch_proxy It will break on filenames containing whitespace, use it with caution. – Chris Down May 22 '13 at 2:22

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.

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You can use the positional parameters

set -- *
echo "$1"
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Be aware that this will destroy any other arguments to your script, unless run in an auxiliary scope. It will also expand to * if there are no files in the directory. – Chris Down Feb 24 '13 at 12:47
shopt -s nullglob would handle that – glenn jackman Feb 24 '13 at 21:13

Check one of the following alternatives:

$ FILE=($(echo *))
$ FILE=$(echo * | grep -o "^\S*")
$ FILE=$(echo * | grep -o "[^ ]*")
$ FILE=$(find . -type f -print -quit)

Then you can print it via echo $FILE.

See also: grep the only first word from output?

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This works:

echo * | grep -o "^\w*\b"

Creds to http://unix.stackexchange.com/a/57879/3920

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Getting the whole first file name:

shopt -s nullglob
printf '%s\000' * | grep -z -m 1 '^..*$'
printf '%s\000' * | ( IFS="" read -r -d "" var; printf '%s\n' "$var" )
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