使用sed替换空格分隔的字符串

Question

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'

Answer 1

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=[^ ]*//'

Answer 2

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.