带/不带空格的Bash字符串操作

Question

Consider having a String in bash holding a list of items:

# BLUBB="item foobar blubb bar"

Further consider having an exclude list of items:

# EXCLUDE="item bar blubb"

What would be the easiest way to resolve this to a list holding BLUBB w/o EXCLUDE. My first approach was:

# for i in $EXCLUDE; do BLUBB=${BLUBB//$i/}; done

But this additionally removes the bar from foobar . So it seems, one has to look for whitespace or nothing before and after the $i . How is the syntax for this?

Answer 1

Something like this can be done:

#/bin/bash
BLUBB="item foobar blubb bar"
EXCLUDE="item bar blubb"
for word in ${BLUBB} ${EXCLUDE} ${EXCLUDE}; do
    echo ${word}
done | sort | uniq -u

Answer 2

I'd do this in bash with an associative array:

$ BLUBB="item foobar blubb bar"
$ EXCLUDE="item bar blubb"
$ declare -A h
$ for word in $BLUBB; do h[$word]=1; done
$ declare -p h
declare -A h='([blubb]="1" [bar]="1" [foobar]="1" [item]="1" )'
$ for word in $EXCLUDE; do unset "h[$word]"; done
$ declare -p h
declare -A h='([foobar]="1" )'
$ words="${!h[*]}"
$ echo "$words"
foobar

If any of the "words" in BLUBB or EXCLUDE contain whitespace, you have to use indexed arrays to hold them:

$ BLUBB=(item foobar blubb bar "word with spaces")
$ EXCLUDE=(item bar blubb with)
$ declare -A h
$ for elem in "${BLUBB[@]}"; do h["$elem"]=1; done
$ for elem in "${EXCLUDE[@]}"; do unset "h[$elem]"; done
$ declare -p h
declare -A h='(["word with spaces"]="1" [foobar]="1" )'