Sub-string extraction using regexp and sed in bash script

Question

In a bash script, I would like to extract sub-string of the form key=[value], so that I can get the value in a variable by specifying the correspondingkey.

For instance, given this variable txt:

txt="something... key=[value] number=[0.42] ...other things... text=[foo] etc"

I would like to extract value for key, 0.42 for number, foo for text ... and empty string for missing keys.

I tried this command, to extract value:

echo "$txt" | sed 's/^*key=\[[*]\]*/\1/'

If I understood well, the command sed "s/regexp/replacement/" try to match here the following regexp:

^ the beginning of the line

* anything

key=\[ the beginning of what I want to find

[*^\[] match anything, except character [

\] the end of what I want to find

* anything

$ the end of the line

and replace it with what has been matched (due to \1).

But I am missing something since I get the following error message : sed: -e expression #1, char 27: invalid reference \1 on `s' command's RHS

I also tried this, without using \1:

echo "$txt" | sed 's/^*key=\[[*]\]*/TEST/'

But the regex failed to match and all the string of txt is returned...

You are mixing up globs and regexes. With a glob, * means any character, any time. With regexes, it is only a quantifier (any amount of time), but you still have to specify which pattern is repeated. The equivalent in regex is .*. This website is a good tool to test your regexes. — Aserre
– Aserre, Commented May 24, 2018 at 11:47
Also, there are lots of other basic errors in your regex. Consider trying the website I linked in my previous comment to learn the regex syntax. — Aserre
– Aserre, Commented May 24, 2018 at 11:53

choroba · Accepted Answer · 2018-05-24 11:56:49Z

1

* doesn't match any string. * is a quantifier which says "the previous could be repeated zero or more times". You need a regex for sed, not a wildcard pattern:

sed 's/.*key=\[\([^]]*\)\].*/\1/'

$...$ are needed to create a capture group, referenced as \1 (because it's the first such group)
[^]]* means "anything but ] zero or more times", so it matches the string inside square brackets

answered May 24, 2018 at 11:56

choroba

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