How to grep the first string delimited by ''

Question

How to grep the first string within '' in a text file? in:

path: '', redirectTo: '/login', pathMatch: 'full' },
path: 'login', component: LoginComponent },
path: 'registration',
path: 'employees',
path: 'recruitment',
path: 'work',

out:

login
registration
employees
recruitment
work

I tried:

grep -Po "path: '\K.*'" in

Answer 1

You might use gnu-awk with a regex and a capture group, capturing optional chars other than ' between 2 ' chars

awk 'match($0, /path: \047([^\047]*)\047/, a) {print a[1]}' file

Output

[empty line] <--
login
registration
employees
recruitment
work

If you are using grep -P for a Perl compatible regular expression with a negated character class [^']* and a lookahead , but that does not print the empty line:

grep -Po "path: '\K[^']*(?=')" file

Output

login
registration
employees
recruitment
work