How to get the first word of a string?

Question

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?

@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 '16 at 12:46
@mattdm `ls | head -1` gives me random things like `a.patch p.py` which is not the first word and not even files in alphabetical order. — phuclv, Feb 13 '17 at 09:10

score 149 · Answer 1 · edited Mar 21 '16 at 12:25

149

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.

edited Mar 21 '16 at 12:25

Willi Mentzel

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answered Feb 24 '13 at 19:08

Bananguin

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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
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i wonder how many lines `echo *` generates – Bananguin Feb 25 '13 at 14:10
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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
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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
`head` not needed with `awk 'NR==1{print $1}` or `awk '{print $1;exit}'` and `sed -n '1s/\s.*//p'` (don't need $ because greedy) – dave_thompson_085 Apr 04 '22 at 01:22
I found that `sed` includes the whole line and `awk` includes `\u00a0` in the end of an extracted word. If you have `node` installed, here is a clever command you can use to extract the first word: `echo * | xargs node -e 'console.log(process.argv[1])'` ([0] being node exec itself). Maybe it helps someone, who hacks some cli tools. – Viktor M Sep 28 '22 at 00:20

score 56 · Answer 2 · answered May 06 '15 at 10:55

56

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 06 '15 at 10:55

starfry

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Nice one. But I liked the simple and more non-hacky other version with `head` and `cut` – Anwar Jul 22 '17 at 18:24
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IMO, head, cut & sed are less simple. Why spawn another tool when builtin parameter substitution does the job efficiently. This answer is most efficient and most portable way to pick off the first word of a list of space-separated words (which is all the OP asked for). – Juan Dec 06 '18 at 14:00
@Juan , first how is builtin substitution efficient - not intended as rhetorical, but who knows what the actual complexity is. Second, unlikely, but if `echo *` were a stream or some # of lines approaching infinity, wouldn't this be a particularly fun way to OOM your box? I don't if you can OOM of an assignment like this, as in does any shell provide constraints (like buffer + block writes to memory), but my guess is that there arent and that you definitely can write to all available memory on a casual, nothing to see here, assignment. – christian elsee Apr 03 '21 at 23:08

Chris Down · Answer 3 · 2016-04-04T15:19:14.397

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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 '%s\n' "${files[0]}"

edited Apr 04 '16 at 15:19

answered Feb 24 '13 at 12:07

Chris Down

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First word would be easy, too: `files=($(echo *))` – Hauke Laging Feb 24 '13 at 12:54
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It would break on "-n", "-e", "-ne", "-Enenene"..., and depending on how `bash` was compiled or the environment, possibly on backslash characters, though. – Stéphane Chazelas Feb 24 '13 at 20:18
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@HaukeLaging, you'd need to disable globbing. Like: `text=$(echo *); set -f; files=($text)`, otherwise more wildcards could be expanded. – Stéphane Chazelas Feb 24 '13 at 20:23
I also find that easy to use `echo * | cut -d ' ' -f 1` is that good practice ? – vdegenne Feb 27 '13 at 19:40
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and `files=$(echo *); echo ${files%% *}` – vdegenne Feb 27 '13 at 19:41
@Oddant - the ${a%% *} is a CHARM, thank you! – memnoch_proxy May 21 '13 at 23:49
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@memnoch_proxy It will break on filenames containing whitespace, use it with caution. – Chris Down May 22 '13 at 02:22
@ChrisDown good point, but otherwise it's very useful for pulling a date string name off string output, like I could turn "2013:15:21 01:05:13" into 2013-15-21 using b=${a%% *}; c=${b/:/-} – memnoch_proxy May 22 '13 at 17:12
@Oddant: ${files%% *} saved the day! Thanks! – neu242 Sep 24 '13 at 07:24
Portability note: this method is full of bash-isms. – Juan Dec 06 '18 at 14:26
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@Juan This post is explicitly *about* bash, it has the "bash" tag – Chris Down Dec 06 '18 at 16:19
Or just `$files` because bash automatically uses element [0] – dave_thompson_085 Apr 04 '22 at 01:17

score 8 · Answer 4 · answered Feb 24 '13 at 12:39

8

You can use the positional parameters

set -- *
echo "$1"

answered Feb 24 '13 at 12:39

glenn jackman

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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

score 5 · Answer 5 · edited Apr 13 '17 at 12:36

5

This works:

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

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

edited Apr 13 '17 at 12:36

Community

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answered Oct 16 '15 at 13:40

Alexander

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Unless the first filename is something like `-n` , `-e`, `-ne`, `-en`, etc. On some platforms. – roaima Apr 04 '22 at 08:20
I like this one, which can catch all rows with the first word. I am using NAME="Red Hat Enterprise Linux Server" VERSION="7.9 (Maipo)" – dave Mar 28 '23 at 19:11

score 3 · Answer 6 · edited Apr 13 '17 at 12:36

3

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?

edited Apr 13 '17 at 12:36

Community

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answered Oct 16 '15 at 13:14

kenorb

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score 1 · Answer 7 · answered Mar 21 '16 at 14:27

1

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" )

answered Mar 21 '16 at 14:27

markle

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score 0 · Answer 8 · answered Apr 03 '22 at 21:59

0

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 03 '22 at 21:59

Safayet Ahmed

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Already in Chris Down's answer and equivalent to kenorb's but you actually need only `$STRARRAY` because bash automatically selects [0] – dave_thompson_085 Apr 04 '22 at 01:19

How to get the first word of a string?

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