How to get the leading whitespaces of a bash variable

Question

for example, the variable is like:

ANY='    ab1 de$%   '

The middle part 'ab1 de$%' can be any combination of characters.

I want to get the substring of leading whitespaces, to indent other lines.

What do you want to trim off? Just the graph characters (e.g. ab1de$%)? or also the trailing whitespace? — Travis Clarke
– Travis Clarke, Commented May 17, 2017 at 5:01

Community · Accepted Answer · 2017-05-23 11:47:22Z

4

You can use search-and-replace parameter substitution:

leading_whitespace="${ANY/[^[:space:]]*/}"

EDIT:

Or also substring-removal parameter substitution as mentioned by David C. Rankin below:

leading_whitespace="${ANY%%[![:space:]]*}"

edited May 23, 2017 at 11:47

CommunityBot

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answered May 17, 2017 at 5:03

Travis Clarke

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3 Comments

rici Over a year ago

[^[:space:]]* might be closer to the problem specification.

David C. Rankin Over a year ago

parameter-expansion w/substring removal also works lws="${any%%[![:space:]]*}"

Travis Clarke Over a year ago

@DavidC.Rankin – that is an elegant alternative as well.

Otheus · Accepted Answer · 2025-10-07 09:56:02Z

2

You can use parameter expansion of ${parameter%%word}.

leading="${ANY%%[[:graph:]]*}"

(Edited) or more restrictively:

leading="${ANY%%[^[:space:]]*}"

Depending on circumstances, trimming ONLY whitespace characters might be preferred, whereas in other circumstances, trimming all non-graphical characters might be better. Consider the BEL, BS, and DEL characters. See https://www.regular-expressions.info/posixbrackets.html for the subtle difference.

edited Oct 7 at 9:56

Otheus

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answered May 17, 2017 at 5:16

thatseeyou

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2 Comments

Travis Clarke Over a year ago

I would favor the solution by David C. Rankin – lws="${ANY%%[![:space:]]*}" for the reason mentioned by rici above – in that is more closely matches the problem description.

Otheus Oct 6 at 8:06

I edited this answer, because yours was honestly a bit muddled, and saved by Rici. 1/2 dozen 6 of the other in most cases.

codeforester · Accepted Answer · 2017-05-17 04:59:49Z

1

You can use sed:

leading=$(sed -E 's/^([[:space:]]+).*/\1/' <<< "$string")

answered May 17, 2017 at 4:59

codeforester

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