preg_replace_callback正则表达式问题，与（。*？）匹配返回数组

Question

Given the string {{esc}}"Content"{{/esc}} ... {{esc}}"More content"{{/esc}} I would like to output \"Content\" ... \"More content\" e.g., I am trying to escape the quotes inside a string. (This is a contrived example, though, so an answer with something like 'just use this library to do it' would be unhelpful.)

Here is my current solution:

return preg_replace_callback(
  '/{{esc}}(.*?){{\/esc}}/', 
  function($m) {
    return str_replace('"', '\\"', $m[1]);
  },
  $text
);

As you can see, I need to say $m[1], because a print_r reveals that $m looks like this:

Array 
(
    [0] => {{esc}}"Content"{{/esc}}
    [1] => "Content"
)

or, for the second match,

Array 
(
    [0] => {{esc}}"More content"{{/esc}}
    [1] => "More content"
)

My question is: why does my regex cause $m to be an array? Is there any way I can get the result of $m[1] as just a single variable $m?

Answer 1

With the help of Jack, I answered my own question here since srain did not make this point clear: The second element of the array is the result captured by the parenthesized subexpression (.*?), per the PHP manual. Indeed, there does not appear to be a convenient way to extract the string matched by this subexpression otherwise.