删除前导空格在 Bash 中不起作用

Question

I have a string in Bash which may or may not start with any number of leading spaces, eg

"  foo bar baz"
" foo bar baz"
"foo bar baz"

I want to delete the first instance of "foo" from the string, and any leading spaces (there may not be any).

Following the advice from this question , I have tried the following:

str=" foo bar baz"
regex="[[:space:]]*foo"
echo "${str#$regex}"
echo "${str#[[:space:]]*foo}"

If str has one or more leading spaces, then it will return the result I want, which is _bar baz (underscore = leading space). If the string has no leading spaces, it won't do anything and will return foo bar baz . Both 'echoes' return the same results here.

My understanding is that using * after [[:space:]] should match zero or more instances of [[:space:]], not one or more. What am I missing or doing wrong here?

EDITS

@Raman - I've tried the following, and they also don't work:

echo "${str#[[:space:]]?foo}"
echo "${str#?([[:space:]])foo}"
echo "${str#*([[:space:]])foo}"

All three solutions will not delete 'foo' whether or not there is a trailing space. The only solution that kind of works is the one I posted with the asterisk - it will delete 'foo' when there is a trailing space, but not when there isn't.

Answer 1

The best thing to do is to use parameter expansions (with extended globs) as follows:

# Make sure extglob is enabled
shopt -s extglob

str=" foo bar baz"
echo "${str##*([[:space:]])}"

This uses the extended glob *([[:space:]]) , and the ## parameter expansion (greedy match).

Edit. Since your pattern has the suffix foo , you don't need to use greedy match:

echo "${str#*([[:space:]])foo}"

is enough.

Note. you can put foo in a variable too, but just be careful, you'll have to quote it:

pattern=foo
echo "${str#*([[:space:]])"$pattern"}"

will work. You have to quote it in case the expansion of pattern contains glob characters. For example when pattern="foo[1]" .

Answer 2

My understanding is that using * after [[:space:]] should match zero or more instances of [[:space:]], not one or more

That's wrong.

What am I missing

That glob is not regex . In regex * matches zero or more preceding characters or groups. In glob * matches anything. It's the same as for filename expansion, think along ls [[:space:]]*foo .

You can use extended bash glob and do:

shopt -s extglob
str=' foo bar baz'
echo "${str#*([[:space:]])foo}"

To do anything more complicated, actually use a regex.

str=' foo bar baz';
[[ $str =~ ^[[:space:]]*foo(.*) ]];
echo "${BASH_REMATCH[1]}"

Answer 3

If what you want is a real regex match, you should be using a real regex match:

$: [[ "$str" =~ [[:space:]]*(.*) ]]
$: echo "[${BASH_REMATCH[1]}]"
[foo  bar       baz]

A more pedestrian approach would be to skip the quotes.

$: echo "[$str]"
[ foo bar baz]
$: new=$( echo $str )
$: echo "[$new]"
[foo bar baz]

Be aware that this opens you up to all sorts of messes in any more complex situations. It breaks if you wanted to preserve more than a single consecutive space between values, or a tab instead of just a quote, etc.

$: str=' foo  bar'$'\t''baz';
$: echo "[$str]"
[ foo  bar      baz]
$: new=$( echo $str )
$: echo "[$new]"
[foo bar baz]

It can cause other sorts of havoc too, but it's good to know the trick for the cases when it's appropriate.