Proper Perl syntax for complex substitution

I've got a large number of PHP files and lines that need to be altered from a standard echo "string goes here"; syntax to: custom_echo("string goes here");

This is the line I'm trying to punch into Perl to accomplish this: perl -pi -e 's/echo \Q开发者_JAVA技巧(.?*)\E;/custom_echo($1);/g' test.php Unfortunately, I'm making some minor syntax error, and it's not altering "test.php" in the least. Can anyone tell me how to fix it?

Why not just do something like:

perl -pi -e 's|echo (\".*?\");|custom_echo($1);|g' file.php

I don't think \Q and \E are doing what you think they're doing. They're not beginning and end of quotes. They're in case you put in a special regex character (like .) -- if you surround it by \Q ... \E then the special regex character doesn't get interpreted.

In other words, your regular expression is trying to match the literal string (.?*), which you probably don't have, and thus substitutions don't get made.

You also had your ? and * backwards -- I assume you want to match non-greedily, in which case you need to put the ? as a non-greedy modifier to the .* characters.

Edit: I also strongly suggest doing:

perl -pi.bak -e ... file.php

This will create a "backup" file that the original file gets copied to. In my above example, it'll create a file named file.php.bak that contains the original, pre-substitution contents. This is incredibly useful during testing until you're certain that you've built your regex properly. Hell, disk is cheap, I'd suggest always using the -pi.bak command-line operator.

You put your grouping parentheses inside the metaquoting expression (\Q(pattern)\E) instead of outside ((\Qpattern\E)), so your parentheses also get escaped and your regex is not capturing anything.