Bash string regex replace with ".*"

Question

for a string

s='abc_somedef'

use regex replace (it works)

echo ${s//_some/}

use regex replace with .* (it NOT works)

echo ${s//^.*_some/}

I want the result to be def

how I write it with bash internally (not sed/awk) ? maybe some escape char ?

Maybe use non-greedy version of *: /^.*?_some/?

Kevin Ji
– Kevin Ji

2013-11-25 07:24:54 +00:00
Commented Nov 25, 2013 at 7:24 — Kevin Ji
– Kevin Ji, Commented Nov 25, 2013 at 7:24
Nope, that doesn't work either. See my answer below.

Chandranshu
– Chandranshu

2013-11-25 07:25:28 +00:00
Commented Nov 25, 2013 at 7:25 — Chandranshu
– Chandranshu, Commented Nov 25, 2013 at 7:25

Chandranshu · Accepted Answer · 2013-11-25 14:19:52Z

1

Use this:

echo ${s//*_some/}

Your version didn't match since it was trying to match a dot literally.

UPDATE:

Chepner has correctly explained below that * here is not part of a regex at all. It is simply being used as a globbing character.

edited Nov 25, 2013 at 14:19

answered Nov 25, 2013 at 7:27

Chandranshu

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2 Comments

devnull Over a year ago

You could have made this 30 character long by explaining why OPs version doesn't work.

chepner Over a year ago

"Unlike the regexes in most languages...": not true. Parameter expansion does not use regular expressions; it uses pattern matching. Where bash does use regular expressions (after the =~ operator), . has the standard meaning of "any single character".

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