How can I get the first string from a Bash list?

Question

I do have a Bash list (space separated string) and I just want to extract the first string from it.

Example:

 VAR="aaa bbb ccc" -> I need "aaa"
 VAR="xxx" -> I need "xxx"

Is there another trick than using a for with break?

Possible canonical question (3 years prior, 13 answers, and 135 upvotes) : *[How can I retrieve the first word of the output of a command in Bash?](https://stackoverflow.com/questions/2440414)* — Peter Mortensen, Apr 25 '21 at 19:22

score 11 · Answer 1 · edited May 03 '21 at 22:34

11

Use cut:

echo $VAR | cut --delimiter " " --fields 1  # Number after fields is the 
                                            # index of pattern you are retrieving

edited May 03 '21 at 22:34

Peter Mortensen

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answered Aug 28 '13 at 11:23

Juto

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In some flavors of Linux it can be .... echo $VAR | cut -d " " -f 1 – blissweb May 21 '22 at 09:56

konsolebox · Answer 2 · 2013-08-28T11:31:42.853

7

Try this format:

echo "${VAR%% *}"

Another way is:

read FIRST __ <<< "$VAR"
echo "$FIRST"

edited Aug 28 '13 at 11:31

answered Aug 28 '13 at 11:16

konsolebox

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2

Did you mean to use ## ? Shouldn't that be %% instead? – IanM_Matrix1 Aug 28 '13 at 11:24
This (parameter expansion) is the cleanest solution. Except for using an array in the first place, of course. – Adrian Frühwirth Aug 28 '13 at 12:07

score 7 · Answer 3 · answered Aug 28 '13 at 11:35

7

If you want arrays, use arrays. ;)

VAR=(aaa bbb ccc)
echo ${VAR[0]} # -> aaa
echo ${VAR[1]} # -> bbb

answered Aug 28 '13 at 11:35

michas

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1

He asked for a "space separated string" – WPWoodJr Sep 02 '20 at 18:51
1

Substituting the first line with `VAR2="aaa bbb ccc"` and `VAR=($VAR2)` would work. – Peter Mortensen Apr 25 '21 at 17:13
`echo ${VAR[0]}` can be reduced to `echo ${VAR}` or `echo $VAR`. – Peter Mortensen Apr 25 '21 at 17:15
In the general case, it will fail if e.g. "aaa" is instead "*" (an [asterisk](https://en.wiktionary.org/wiki/asterisk#Noun)). It expands to the list of all files and all directories in the current directory... – Peter Mortensen Apr 25 '21 at 19:31

score 6 · Answer 4 · answered Aug 28 '13 at 11:32

6

I'm not sure how standard this is, but this works in Bash 4.1.11

NewVAR=($VAR)
echo $NewVAR

answered Aug 28 '13 at 11:32

IanM_Matrix1

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It does indeed work. And it is much simpler (and faster, especially compared to invoking external processes like [cut](https://en.wikipedia.org/wiki/Cut_(Unix)), [AWK](https://en.wikipedia.org/wiki/AWK), or [Perl](https://en.wikipedia.org/wiki/Perl)). – Peter Mortensen Apr 25 '21 at 17:21
But it will fail if the first "word" is "*" (an [asterisk](https://en.wiktionary.org/wiki/asterisk#Noun)) - it expands to the list of all files and all directories in the current directory... – Peter Mortensen Apr 25 '21 at 19:27

score 2 · Answer 5 · edited May 03 '21 at 22:35

2

At this moment the only solution that worked, on both Linux and OS X was:

 IP="1 2 3"
 for IP in $IP:
 do
   break
 done

edited May 03 '21 at 22:35

Peter Mortensen

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answered Aug 28 '13 at 11:30

sorin

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Just checked if cut was in Mac OS, found [this](http://stackoverflow.com/questions/1211943/extract-text-from-hostname) – Juto Aug 28 '13 at 11:39
`cut` exists but it is not the same, quite an old one in fact. – sorin Aug 28 '13 at 11:40
I am not sure by what you mean by quite an old one, does it still work? I guess if it works, programs like cut, don't need to be updated every month? – Juto Aug 29 '13 at 08:07
So old that it doesn't have the parameters used by you :p – sorin Aug 29 '13 at 18:46
echo $VAR | cut -d " " -f 1 (works on OSX) – blissweb May 21 '22 at 09:57

How can I get the first string from a Bash list?

5 Answers5