Question

I’m trying to match the following:
A pattern begins with /* and ends with */\n(override\s+)public, tried doing something like this:

sed -n '/\/\*/,/\*\/\n(override\s+)public/p' file

And here I am asking you how to write that properly 🙂

If it’s possible to do with grep or awk – that’s an option, if it’s a one-liner in perl, that’s fine too.

PS. I’m not sure of sed regex syntax, but \s is meant to be a whitespace, ideally [\ \t\r\n].

EDIT:

Suppose you have:

/**
 * Some comment goes here
 */
override public function ...

/**
 * A comment for another function
 */
public function ...

The expression would give you these two results:

/**
 * Some comment goes here
 */
override public

and

/**
 * A comment for another function
 */
public

EDIT:
Even better, if someone could translate the below regex into sed, that would do the job.

/\/\*(!?\*\/)*\*\/\s*(overrides\s+)?public/gm

Note, that m flag means the regex is multiline, that is the negative lookahead group will match line ends too. But if that’s not possible, but matching line ends is still possible, then the expression would look like this:

/\/\*((!?\*\/)*$)*\*\/[\r\n\t\ ]*(overrides[\r\n\t\ ]+)?public/g

Answer 1

I think this will do what you want:

perl -0777 -n -e '
    for (m/(\/\*.*?\*\/\s+(override\s+)?public)/s) { 
        print "$1\n"; 
    }
' file

The -0777 is the perl convention for “read all of file in at once”, rather than the default of line-by-line.

The -n tells perl to not print the input file (as opposed to -p which does print it).

The -e tells perl to read the program from the command line.

The regex is largely as you wrote it, though I removed the comma bit since it wasn’t clear what you were trying to do. I used the reluctant *? quantifier to match the shortest possible match, along with the s flag which treats the entire file as a ‘single line’ and which means that .*? matches across line boundaries when it otherwise would not.

The m in the if clause causes the expression to evaluate to true if the regex matches. The s at the end of the expression treats the input as a single line. Inside the if clause, we’re printing $1 which is the first group matched by the regex. If you wanted to print the second group, you’d use $2.

Updated: Changed the groups so it will print out the override and public, made override optional per your edit, changed if to for so it matches multiple instances of the pattern.