带有 sed 的 Bash 正则表达式

Question

Trying to get only words that come before "/", I wrote:

T='He/She is a very handsome/beautiful man/woman indeed.'
echo "$T" | sed -E 's#\b(.*)/(.*)\b#\1#g'

However, I only got to make it work in the last occurrence (although I'm using "g" in sed sentence):

He/She is a very handsome/beautiful man.

My desired output is: "He is a very handsome man indeed."

Any ideas? Thank you.

Answer 1

A POSIX compliant one:

T='He/She is a very handsome/beautiful man/woman indeed.'
echo "$T" | sed 's:/[^ ]*::g'

Answer 2

To remove non-space characters after every slash you can use:

$ sed 's:/\S*::g' <<<'He/She is a very handsome/beautiful man/woman indeed.'
He is a very handsome man indeed.

The pattern, :/\\S*: , matches a slash followed by zero or more non-space characters. The replacement string, :: , is empty and it is applied globally g . The<<< is a here-string that passes input to sed.