Bash variable substitution

Question

I have a variable containing four numbers separated by a space, such as for instance:

a="12.3 423.4 11.0033 14.02"

But sometimes, I have a trailing whitespace:

a="12.3 423.4 11.0033 14.02 "

I want to replace the spaces with " & ", and for that, I do:

echo ${a// / & }

Which gives me:

12.3 & 423.4 & 11.0033 & 14.02

or if I have a trailing whitespace:

12.3 & 423.4 & 11.0033 & 14.02 &

My problem is that I don't know if I'll have a space at the end of my string and I don't want that extra "&" in any case. What would be the most elegant way to avoid this extra character? Is there a way to say "replace if a space and the next character a digit"?

Edit: I knew I could use sed, but since there is a mechanism of variable substitution in bash, I would like to know how could I use it to do what I want. I don't know how to write "not end of line" or "is a digit" in the bash substitution.

possible duplicate of [How to trim whitespace from bash variable?](http://stackoverflow.com/questions/369758/how-to-trim-whitespace-from-bash-variable) — el_tenedor, Sep 27 '13 at 13:47
@danadam I want the result to be "12.3 & 423.4 & 11.0033 & 14.02" with or without the trailing space. — Maxime Chéramy, Sep 27 '13 at 13:49
@el_tenedor it is not a duplicate but it might be a possible solution. However, I am looking for the most elegant solution. — Maxime Chéramy, Sep 27 '13 at 13:50

score 9 · Accepted Answer · answered Sep 27 '13 at 13:55

9

This will remove trailing space if there is any:

a=${a% }

Then you can do your replace:

a=${a// / & }

answered Sep 27 '13 at 13:55

danadam

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Thank you, I will use `${${a% }// / & }` unless someone comes with a better idea. I'll pick the best answer later. *Edit:* hum, bash says bad substitution while it's fine for zsh... I need to do it in 2 steps as you did. – Maxime Chéramy Sep 27 '13 at 14:05

ben_re · Answer 2 · 2013-09-27T14:17:31.377

1

You could use sed and capture the match.

$ echo 12.3 423.4 11.0033 14.02 | sed 's/ [0-9]/ \&&/g'
12.3 & 423.4 & 11.0033 & 14.02

So with the / [0-9]/ you're only replacing where there is a space followed by a number.

If you didn't want to use sed you could check that the next character isn't EOL.

${a// [^$]/ & }

edited Sep 27 '13 at 14:17

answered Sep 27 '13 at 13:50

ben_re

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Is there a way to do it directly with bash variable substitution (without sed)? – Maxime Chéramy Sep 27 '13 at 13:53
I don't believe so, as you can only substitute with bash substitution. You cannot use regex and I don't think you can capture the match. Sed is the way forward! :) – ben_re Sep 27 '13 at 14:01
So apparently, you can use regex! – ben_re Sep 27 '13 at 14:17
Sorry but your solution doesn't work actually: you are removing the characters that match [^$]. – Maxime Chéramy Sep 27 '13 at 14:28
Hmm... It worked when i tested it. Have you definitely got the space before the `[^$]`. (i.e. `/ [^$]/`)? – ben_re Sep 27 '13 at 14:33
On the example, I have: `12.3 & 23.4 & 1.0033 & 4.02` (your solution using sed works however). – Maxime Chéramy Sep 27 '13 at 14:38
@lifesaspanner `bash` parameter substitution uses glob patterns, not regular expressions. – chepner Sep 27 '13 at 14:50
@chepner Aaah. That would be I can't do a negative lookahead! – ben_re Sep 27 '13 at 15:06
@Maxime You're quite right, it does replace the character. It seems the only options are removing the trailing space, or use sed. – ben_re Sep 27 '13 at 15:07

score 1 · Answer 3 · answered Sep 27 '13 at 14:24

shopt -s extglob            # enable extended globbing patterns
b=${a/%+([[:blank:]])/}     # remove trailing whitespace
b=${b/#+([[:blank:]])/}     # remove leading whitespace
b=${b//+([[:blank:]])/ & }  # globally replace whitespace by " & "
echo ">$a<"
echo ">$b<"

>12.3 423.4 11.0033 14.02 <
>12.3 & 423.4 & 11.0033 & 14.02<

references:

http://www.gnu.org/software/bash/manual/bashref.html#Pattern-Matching
http://www.gnu.org/software/bash/manual/bashref.html#Shell-Parameter-Expansion

iruvar · Answer 4 · 2013-09-27T15:00:57.903

1

This is probably a contrived approach, but FWIW. If you are ok with multiple consecutive spaces being replaced with a single &, you could use bash read and printf built-ins to split the string into a array (assuming default whitespace IFS for the split) and then join back using a temporary IFS of &, see below.

a="12.3 423.4 11.0033 14.02 "
read -r -a x <<< "${a}"
IFS='&' eval 'printf -v a "%s" "${x[*]}"'
echo $a
12.3&423.4&11.0033&14.02

edited Sep 27 '13 at 15:00

answered Sep 27 '13 at 14:31

iruvar

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I like your definition of "elegant" :). Definitively not mine but +1 for the creativity. – Maxime Chéramy Sep 27 '13 at 14:35

score 0 · Answer 5 · answered Sep 27 '13 at 13:44

0

Maybe you could just do

a=`echo "12.3 423.4 11.0033 14.02 " | sed 's/ *$//'`
echo ${a// / & }

That way you're removing the trailing whitespaces.

answered Sep 27 '13 at 13:44

jlhonora

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Bash variable substitution

5 Answers5