264

I've been looking for a solution and found similar questions, only they were attempting to split sentences with spaces between them, and the answers do not work for my situation.

Currently a variable is being set to something a string like this:
ABCDE-123456
and I would like to split that into 2 variables, while eliminating the "-". i.e.:
var1=ABCDE
var2=123456

How is it possible to accomplish this?

This is the solution that worked for me:
var1=$(echo $STR | cut -f1 -d-)
var2=$(echo $STR | cut -f2 -d-)

Is it possible to use the cut command that will work without a delimiter (each character gets set as a variable)?

var1=$(echo $STR | cut -f1 -d?)
var2=$(echo $STR | cut -f1 -d?)
var3=$(echo $STR | cut -f1 -d?)
etc.

Improve this question
3
  • For your second question, see @mkb's comment to my answer below - that's definitely the way to go!
    – Rob I
    Commented May 9, 2012 at 19:22
  • 1
    See my edited answer for one way to read individual characters into an array.
    – Dennis Williamson
    Commented Jul 4, 2012 at 16:14
  • 1
    Here is the same thing in a more concise form: var1=$(cut -f1 -d- <<<$STR)
    – Nick Weedon
    Commented Dec 31, 2015 at 11:04

5 Answers 5

Reset to default
320

To split a string separated by -, you can use read with IFS:

$ IFS=- read -r var1 var2 <<< ABCDE-123456
$ echo "$var1"
ABCDE
$ echo "$var2"
123456

Edit:

Here is how you can read each individual character into array elements:

$ read -ra foo <<<"$(echo "ABCDE-123456" | sed 's/./& /g')"

Dump the array:

$ declare -p foo
declare -a foo='([0]="A" [1]="B" [2]="C" [3]="D" [4]="E" [5]="-" [6]="1" [7]="2" [8]="3" [9]="4" [10]="5" [11]="6")'

If there are spaces in the string:

$ IFS=$'\v' read -ra foo <<<"$(echo "ABCDE 123456" | sed $'s/./&\v/g')"
$ declare -p foo
declare -a foo='([0]="A" [1]="B" [2]="C" [3]="D" [4]="E" [5]=" " [6]="1" [7]="2" [8]="3" [9]="4" [10]="5" [11]="6")'
Improve this answer
6
  • 2
    this solution also has the benefit that if delimiter is not present, the var2 will be empty
    – Martin Serrano
    Commented Jan 11, 2018 at 4:34
  • The read does not work inside loops with input redirects. read will pick a wrong file descriptor to read from.
    – akwky
    Commented Feb 24, 2020 at 10:59
  • I initially gave this answer a plus as an elegant solution, but now figured out it works differently on Bash v3 and v4, thereby doesn't work on macos with pre-installed bash v3. Unfortunatelly I can't downvote the answer now since the vote is locked :(
    – Jerry Green
    Commented Nov 11, 2020 at 14:30
  • A more general, correct, way: IFS=- read -r -d '' var1 var2 < <(printf %s "ABCDE-123456"). The -r -d '' and <(printf %s ...) are important
    – Fravadona
    Commented Jan 21, 2022 at 10:22
  • 1
    @akwky: Use an alternate file descriptor. while read -r line <&3; do ssh_or_something "$line"; done 3<file
    – Dennis Williamson
    Commented Jun 24, 2022 at 0:14
286

If you know it's going to be just two fields, you can skip the extra subprocesses. Like this:

STR="ABCDE-12345" 
var1=${STR%-*} # ABCDE
var2=${STR#*-} # 12345

Explanation:

  • ${STR%-*} deletes the shortest substring of $STR that matches the pattern -* (deletes - and anything after it). It starts from the end of the string.
  • ${STR#*-} deletes the shortest substring of $STR that matches the pattern *- (deletes - and anything before it). It starts from the beginning of the string.

They each have counterparts %% and ## which find the longest anchored pattern match. To memorize, use this mnemonic, shared by @DS:

"#" is to the left of "%" on a standard keyboard, so "#" removes a prefix (on the left), and "%" removes a suffix (on the right).

See the bash documentation for more information.

Improve this answer
9
  • 22
    Plus 1 For knowing your POSIX shell features, avoiding expensive forks and pipes, and the absence of bashisms.
    – Jens
    Commented Jan 30, 2015 at 15:17
  • 3
    Dunno about "absence of bashisms" considering that this is already moderately cryptic .... if your delimiter is a newline instead of a hyphen, then it becomes even more cryptic. On the other hand, it works with newlines, so there's that.
    – Steven Lu
    Commented May 1, 2015 at 20:19
  • 5
    I've finally found documentation for it: Shell-Parameter-Expansion
    – Marek Podyma
    Commented Aug 9, 2016 at 15:58
  • 31
    Mnemonic: "#" is to the left of "%" on a standard keyboard, so "#" removes a prefix (on the left), and "%" removes a suffix (on the right).
    – DS.
    Commented Jan 13, 2017 at 19:56
  • 4
    Another mnemonic, since your keyboard may be different (and some just "feel" the layout, rather than know it): the % symbol is typically encountered after a number, e.g. 90%, hence it is a suffix. The # symbol is typically leading comments or even just the first char in hashtags, so it's a common prefix. The purpose of both modifiers is to remove, one just removes a prefix (#), the other removes the suffix (%).
    – Oliver W.
    Commented Nov 8, 2022 at 21:49
198

If your solution doesn't have to be general, i.e. only needs to work for strings like your example, you could do:

var1=$(echo $STR | cut -f1 -d-)
var2=$(echo $STR | cut -f2 -d-)

I chose cut here because you could simply extend the code for a few more variables...

Improve this answer
5
  • Can you look at my post again and see if you have a solution for the followup question? thanks!
    – crunchybutternut
    Commented May 9, 2012 at 17:40
  • You can use cut to cut characters too! cut -c1 for example.
    – Matt K
    Commented May 9, 2012 at 17:59
  • 1
    Although this is very simple to read and write, is a very slow solution because forces you to read twice the same data ($STR) ... if you care of your script performace, the @anubhava solution is much better
    – FSp
    Commented Nov 27, 2012 at 10:26
  • 1
    Apart from being an ugly last-resort solution, this has a bug: You should absolutely use double quotes in echo "$STR" unless you specifically want the shell to expand any wildcards in the string as a side effect. See also stackoverflow.com/questions/10067266/…
    – tripleee
    Commented Jan 25, 2016 at 6:47
  • 1
    You're right about double quotes of course, though I did point out this solution wasn't general. However I think your assessment is a bit unfair - for some people this solution may be more readable (and hence extensible etc) than some others, and doesn't completely rely on arcane bash feature that wouldn't translate to other shells. I suspect that's why my solution, though less elegant, continues to get votes periodically...
    – Rob I
    Commented Feb 10, 2016 at 13:57
56

Sounds like a job for set with a custom IFS.

IFS=-
set $STR
var1=$1
var2=$2

(You will want to do this in a function with a local IFS so you don't mess up other parts of your script where you require IFS to be what you expect.)

Improve this answer
9
  • Nice - I knew about $IFS but hadn't seen how it could be used.
    – Rob I
    Commented May 9, 2012 at 19:20
  • I used triplee's example and it worked exactly as advertised! Just change last two lines to <pre> myvar1=echo $1 && myvar2=echo $2 </pre> if you need to store them throughout a script with several "thrown" variables.
    – Sigg3.net
    Commented Jun 19, 2013 at 8:08
  • 1
    No, don't use a useless echo in backticks.
    – tripleee
    Commented Jun 19, 2013 at 13:25
  • 4
    This is a really sweet solution if we need to write something that is not Bash specific. To handle IFS troubles, one can add OLDIFS=$IFS at the beginning before overwriting it, and then add IFS=$OLDIFS just after the set line.
    – Daniel Andersson
    Commented Mar 27, 2015 at 6:46
  • 3
    Maybe add a set -f to disable pathname expansion before set -- $STR, or it will capture paths files names if $STR contains patterns.
    – Léa Gris
    Commented Oct 11, 2019 at 16:56
39

Using bash regex capabilities:

re="^([^-]+)-(.*)$"
[[ "ABCDE-123456" =~ $re ]] && var1="${BASH_REMATCH[1]}" && var2="${BASH_REMATCH[2]}"
echo $var1
echo $var2

OUTPUT

ABCDE
123456
Improve this answer
1
  • 2
    Love pre-defining the re for later use(s)!
    – Cometsong
    Commented Oct 21, 2016 at 13:29

Not the answer you're looking for? Browse other questions tagged or ask your own question.