Bash: How to tokenize a string variable?

Question

If I have a string variable who's value is "john is 17 years old" how do I tokenize this using spaces as the delimeter? Would I use awk?

Diego Torres Milano · Answer 1 · 2016-10-13T20:33:13.747

70

$ string="john is 17 years old"
$ tokens=( $string )
$ echo ${tokens[*]}

For other delimiters, like ';'

$ string="john;is;17;years;old"
$ IFS=';' tokens=( $string )
$ echo ${tokens[*]}

edited Oct 13 '16 at 20:33

answered Mar 22 '11 at 07:22

Diego Torres Milano

57,580
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Very nice, feels much more like an array. – Adam Eberlin Dec 21 '13 at 21:35
echo ${tokens[*]} doesn't work for me I get 'bash: ${tokens[*}: bad substitution ' error. – JPM Mar 11 '20 at 16:04
you are missing the `*`: ```$ tokens=( a ); $ echo ${tokens[]}; -bash: ${tokens[]}: bad substitution $ echo ${tokens[*]}; a``` – Diego Torres Milano Mar 11 '20 at 21:15

score 62 · Accepted Answer · answered Mar 21 '11 at 19:50

Use the shell's automatic tokenization of unquoted variables:

$ string="john is 17 years old"
$ for word in $string; do echo "$word"; done
john
is
17
years
old

If you want to change the delimiter you can set the $IFS variable, which stands for internal field separator. The default value of $IFS is " \t\n" (space, tab, newline).

$ string="john_is_17_years_old"
$ (IFS='_'; for word in $string; do echo "$word"; done)
john
is
17
years
old

(Note that in this second example I added parentheses around the second line. This creates a sub-shell so that the change to $IFS doesn't persist. You generally don't want to permanently change $IFS as it can wreak havoc on unsuspecting shell commands.)

for your examples, how would you re-use the third token (17) for example? use the for loop and count tokens? — kurumi, Mar 22 '11 at 07:31
@Allen, then i can do this `IFS="_";set -- $string; echo $2.` or directly set it to an array like what `dtmilano` did. There is no need to use a for loop isn't it? — kurumi, Mar 24 '11 at 05:40

score 14 · Answer 3 · answered Mar 22 '11 at 07:33

14

$ string="john is 17 years old"
$ set -- $string
$ echo $1
john
$ echo $2
is
$ echo $3
17

answered Mar 22 '11 at 07:33

kurumi

23,171
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score 3 · Answer 4 · answered Mar 21 '11 at 19:46

3

you can try something like this :

#!/bin/bash
n=0
a=/home/file.txt
for i in `cat ${a} | tr ' ' '\n'` ; do
   str=${str},${i}
   let n=$n+1
   var=`echo "var${n}"`
   echo $var is ... ${i}
done

answered Mar 21 '11 at 19:46

harshit

7,659
22
66
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The use of `tr` makes this the best solution. Your exemple code could be much simpler : `echo john is 17 years old | tr ' ' '\n'` – Titou May 11 '17 at 08:49

score 1 · Answer 5 · answered Dec 21 '18 at 09:36

with POSIX extended regex:

$ str='a b     c d'
$ echo "$str" | sed -E 's/\W+/\n/g' | hexdump -C
00000000  61 0a 62 0a 63 0a 64 0a                           |a.b.c.d.|
00000008

this is like python's re.split(r'\W+', str)

\W matches a non-word character,
including space, tab, newline, return, [like the bash for tokenizer]
but also including symbols like quotes, brackets, signs, ...

... except the underscore sign _,
so snake_case is one word, but kebab-case are two words.

leading and trailing space will create an empty line.

Bash: How to tokenize a string variable?

5 Answers5

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