How to match string in single or double quoted using regex

Question

I'm trying to write a regex that matches strings as the following:

translate("some text here")

and

translate('some text here')

I've done that:

preg_match ('/translate\("(.*?)"\)*/', $line, $m) 

but how to add if there are single quotes, not double. It should match as single, as double quotes.

Answer 1

You could go for:

translate\( # translate( literally
(['"])      # capture a single/double quote to group 1
.+?         # match anything except a newline lazily
\1          # up to the formerly captured quote
\)          # and a closing parenthesis

See a demo for this approach on regex101.com.

---

In PHP this would be:

<?php

$regex = '~
            translate\( # translate( literally
            ([\'"])     # capture a single/double quote to group 1
            .+?         # match anything except a newline lazily
            \1          # up to the formerly captured quote
            \)          # and a closing parenthesis
         ~x';

if (preg_match($regex, $string)) {
    // do sth. here
}
?>

Note that you do not need to escape both of the quotes in square brackets ([]), I have only done it for the Stackoverflow prettifier.
Bear in mind though, that this is rather error-prone (what about whitespaces, escaped quotes ?).

---

In the comments the discussion came up that you cannot say anything BUT the first captured group. Well, yes, you can (thanks to Obama here), the technique is called a tempered greedy token which can be achieved via lookarounds. Consider the following code:

translate\(
(['"])
(?:(?!\1).)*
\1
\)

It opens a non-capturing group with a negative lookahead that makes sure not to match the formerly captured group (a quote in this example).
This eradicates matches like translate("a"b"c"d") (see a demo here).

---

The final expression to match all given examples is:

translate\(
(['"])
(?:
   .*?(?=\1\))
)
\1
\)