Understanding a sed example
bash, regex, sed
Solution
Here is sed command:
sed -En '/^password: / s,^password: "(.*)"$,\1,p'
Commas are used as regex delimiter it can very well be another delimiter like `#`:
sed -En '/^password: / s#^password: "(.*)"$#\1#p'`
`/^password: /` finds an input line that starts with `password:`
- `s#^password: "(.*)"$#\1#p` finds and captures double-quoted string after `password:` and replaces the entire line with the captured string `\1` ( so all that remains is the password )
Problem
I found a solution for extracting the password from a Mac OS X Keychain item. It uses `sed` to get the password from the `security` command: ``` security 2>&1 >/dev/null find-generic-password -ga $USER | \ sed -En '/^password: / s,^password: "(.*)"$,\1,p' ``` The code is here in a comment by 'sr105'. The part before the `|` evaluates to `password: "secret"`. I'm trying to figure out exactly how the `sed` command works. Here are some thoughts: I understand the flags `-En`, but what are the commas doing in this example? In the sed docs it says a comma separates an address range, but there's 3 commas. The first 'address' `/^password: /` has a trailing `s`; in the docs `s` is only mentioned as the replace command like `s/pattern/replacement/`. Not the case here. The `^password: "(.*)"$` part looks like the Regex for isolating `secret`, but it's not delimited. I can understand the end part where the back-reference `\1` is printed out, but again, what are the commas doing there?? Note that I'm not interested in an easier alternative to this sed example. This will only be part of a larger bash script which will include some more sed parsing in an .htaccess file, so I'd really like to learn the syntax even if it is obscure. Thanks for your help!