Question

I have bash function which run python (which return all finded regex from stdin)

function find-all() {
    python -c "import re
import sys
print '\n'.join(re.findall('$1', sys.stdin.read()))"
}

When I use this regex find-all 'href="([^"]*)"' < index.html it should return first group from the regex (value of href attribute from file index.html)

How can I write this in sed or awk?

Answer 1

I suggest you use grep -o.

-o, --only-matching
       Show only the part of a matching line that matches PATTERN.

E.g.:

$ cat > foo
test test test
test
bar
baz test
$ grep -o test foo
test
test
test
test
test

---

Update

If you were extracting href attributes from html files, using a command like:

$ grep -o -E 'href="([^"]*)"' /usr/share/vlc/http/index.html
href="style.css"
href="iehacks.css"
href="old/"

You could extract the values by using cut and sed like this:

$ grep -o -E 'href="([^"]*)"' /usr/share/vlc/http/index.html| cut -f2 -d'=' | sed -e 's/"//g'
style.css
iehacks.css
old/

But you’d be better off using html/xml parsers for reliability.