Remove characters to right of first space in Bash

Question

2492  some string continues here

I would like to convert this to

in Bash. How would I go about that?

This feels close, but is not working:

var="2492  some string continues here  "
echo ${var%[[:space:]]*}

If you are asking how to do it in bash you should probably not be using bash to do it. Use Perl or something. — Ben, Apr 13 '18 at 09:34

glenn jackman · Answer 1 · 2018-04-12T21:07:09.847

15

Because there are multiple spaces you want to use

${var%%[[:space:]]*}
# ...^^

to remove the longest trailing substring that starts with a space

With just a single % you're removing the shortest sequence of a space followed by zero or more characters, which is just the last space in the string.

$ echo ">$var<"; echo ">${var%[[:space:]]*}<"; echo ">${var%%[[:space:]]*}<"
>2492  some string continues here  <
>2492  some string continues here <
>2492<

If you're just looking for the first word, you can do this:

read -r word rest_of_string <<<"$var"
echo "I have: $word"

That will take care of leading whitespace, assuming you have not altered the IFS variable.

edited Apr 12 '18 at 21:07

answered Apr 12 '18 at 20:35

glenn jackman

85,964

How about if there is a leading space in the string? 2492 some string continues here – Joshua Soileau Apr 12 '18 at 20:56
That will remove all the characters. – glenn jackman Apr 12 '18 at 21:05
[:space:] character class includes space, tab, CR, NL, etc. So it's a catch-all for any whitespace. – glenn jackman Apr 12 '18 at 22:36
1

a=($var); echo "${a[0]}" also works for accessing the first word in the case of leading whitespace. – Digital Trauma Apr 12 '18 at 22:52
You expose the value to filename expansion as well as word splitting unless you do set -f first – glenn jackman Apr 12 '18 at 22:58

score 4 · Answer 2 · edited Jun 11 '20 at 14:16

There is the simple solution of using %% (${var%% *})instead of % (${var% *}). That will remove everything (*) after an initial space.

$ var='2492  some string continues here'
$ echo "${var%% *}"
2492

But that will fail if the string in var has any leading spaces. It is possible to remove the leading spaces with:

$ var=$' \t 2492  some string continues here  '
$ var="${var#"${var%%[![:space:]]*}"}"
$ echo "$var"
2492  some string continues here  
$ echo "${var%%[[:space:]]*}"
2492

That works even if the white-spaces are spaces tabs NL or CR.

Regex

Maybe a more robust solution is to use a regex:

$ var=$' \t 2492  some string continues here  '
$ regex='^[[:space:]]*([^[:space:]]+)'
$ [[ $var =~ $regex ]] && var=${BASH_REMATCH[1]}
$ echo "$var"
2492

I think that's the first time I've seen anyone refer to regular expressions as "more robust" ;-) — Kusalananda, Apr 13 '18 at 07:06

score 2 · Answer 3 · answered Apr 13 '18 at 08:29

2

You can also use the simple tool cut, who cuts strings based on a delimiter :

echo "$mystring" | cut -d' ' -f 1

Where :

-d' ' sets the delimiter to a space
-f 1 gives the first field (based on the delimiter)

answered Apr 13 '18 at 08:29

Ikaros

882

I like this one. No regular expressions, no programming languages. – gronostaj Apr 13 '18 at 15:07

score 1 · Answer 4 · answered Apr 13 '18 at 09:22

You can use native shell string manipulation:

TEST="test  1234 foo"
SPLIT_VAR=${TEST/ */ }

It will replace the first pattern matching " *" (one space then anything) and replace it with " " (one space). So you keep the first word and the first space.

You can see http://www.tldp.org/LDP/abs/html/string-manipulation.html for more usage of string manipulation.

And as a side note, it's also works less evolved shell (tested on busybox's ash implementation).

anonymous at work · Answer 5 · 2018-04-13T08:18:42.650

0

If you don't mind an external call, use awk or sed:

$ string="first second third fourth"
$ echo "${string}" | awk '{print $1}' # will work even with tabs
first
$ echo "${string}" | sed -e "s/ .*$//" # will fail on leading whitespace
first

edited Apr 13 '18 at 08:18

answered Apr 13 '18 at 05:55

anonymous at work

1

Related: https://unix.stackexchange.com/questions/171346/security-implications-of-forgetting-to-quote-a-variable-in-bash-posix-shells – Kusalananda Apr 13 '18 at 07:13
Quoted my strings, thanks for catching that @Kusalananda – anonymous at work Apr 13 '18 at 08:19

Robert Riedl · Answer 6 · 2018-04-13T07:44:14.417

0

To use only bash and builtin commands, you could use Internal Field Seperator(IFS) and arrays

set -f                # To prevent filename globbing when creating the array, as pointed out by Kusalananda in the comments
array=(2492 some string continues here)
set +f                # enable it again
IFS=" "
echo "${array[0]}"
unset IFS

edited Apr 13 '18 at 07:44

answered Apr 13 '18 at 07:01

Robert Riedl

173

Try that with a string that starts with e.g. *. – Kusalananda Apr 13 '18 at 07:17
@Kusalananda tried it with array=(*2492* *some * *string continues* here)- still works – Robert Riedl Apr 13 '18 at 07:21
Now add a file in the current directory that contains 2492 in its name, for example myfile-2492.txt, and run again. – Kusalananda Apr 13 '18 at 07:23
1

@Kusalananda oh wow, why does this happen ? – Robert Riedl Apr 13 '18 at 07:31
Filename globbing happens when you create the array of unquoted strings. – Kusalananda Apr 13 '18 at 07:32
@Kusalananda TIL – Robert Riedl Apr 13 '18 at 07:35
1

@Kusalananda, I guess I'll have to start with set -f and end with set +f ? – Robert Riedl Apr 13 '18 at 07:37
That would be one solution to it, yes. You should also quote the ${array[0]}. – Kusalananda Apr 13 '18 at 07:39
Could you please explain the purpose of IFS here? echo "${array[0]}", the only command between the assignment and deletion of IFS, doesn't include anything that would be subject to tokenization with the default setting but not with IFS=" " or the other way around. – David Foerster Apr 13 '18 at 11:05

Remove characters to right of first space in Bash

6 Answers6

Regex