CODEWITHSUNDEEP
regexbashsed
Microsoft Linux TM
Microsoft Linux TM

Reputation: 412

Using sed to replace space delimited strings

echo 'bar=start "bar=second CONFIG="$CONFIG bar=s buz=zar bar=g bar=ggg bar=f bar=foo bar=zoo really?=yes bar=z bar=yes bar=y bar=one bar=o que=idn"' | sed -e 's/^\|\([ "]\)bar=[^ ]*[ ]*/\1/g'

Actual output:

CONFIG="$CONFIG buz=zar bar=ggg bar=foo really?=yes bar=yes bar=one que=idn"

Expected output:

CONFIG="$CONFIG buz=zar really?=yes que=idn"

What I'm missing in my regex?

Edit:

This works as expected (with GNU sed):

's/\(^\|\(['\''" ]\)\)bar=[^ ]*/\2/g; s/[ ][ ]\+/ /g; s/[ ]*\(['\''"]\+\)[ ]*/\1/g'

Upvotes: 0

Views: 2049

Answers (2)

Zhro
Zhro

Reputation: 2614

If you want to remove all instances of bar=something then you can simplify your regex as such:

\sbar=\w+

This matches all bar= plus all whole words. The bar= must be preceded by a whitespace character.

Demonstration: https://regex101.com/r/xbBhJZ/3

As sed:

s/\sbar=\w\+//g

This correctly accounts for foobar=bar.

Like Waxrat's answer, you have to insert a space at the beginning for it to properly match as it's now matching against a preceding whitespace character before the bar=. This can be easily done since you're quoting your string explicitly.

Upvotes: 1

Waxrat
Waxrat

Reputation: 2185

sed regular expressions are pretty limited. They don't include \w as a synonym for [a-zA-Z0-9_], for example. They also don't include \b which means the zero-length string at the beginning or end of a word (which you really want in this situation...).

s/ bar=[^ ]* *//

is close, but the problem is the trailing * removes the space that might precede the next bar=. So, in ... bar=aaa bar=bbb ... the first match is bar=aaa leaving bar=bbb ... to try for the second match but it won't match because you already consumed the space before bar.

s/ bar=[^ ]*//

is better -- don't consume the trailing spaces, leave them for the next match attempt. If you want to match bar=something even if it's at the beginning of the string, insert a space at the beginning first:

sed 's/^bar=/ bar=/; s/ bar=[^ ]*//'

Upvotes: 2

Related Questions