php中类型说明符的正则表达式

Question

I need a regular expression which extract type specifier ( like %d,%s) from a string.

   <?
     $price = 999.88;
     $str = "string";
     $string = "this is a %f sample %'-20s,<br> this string is mixed with type    specifier like   (number:%d's)";        
    //output 1 :echo sprintf($string,$price,$str,500);
    //output 2 should be $string replaced by [#]
   ?>

output 1

this is a 999.880000 sample --------------string,
this string is mixed with type specifier like (number:500's)

i want to replace all these type specifiers with [#]. how do i write an regular expression for that type specifiers.

what i need is

output 2

this is a [#] sample --------------[#],
this string is mixed with type specifier like (number:[#]'s)

Answer 1

Pascal beat me to it, but here is my (equally not very well tested regex).

'/%(?:(?<swap_position>[0-9]+)\$)?(?<sign>-|\+)?(?<padding>\'.|0|[[:space:]])?(?<alignment>-?)(?<width>[0-9]+)?(?:\.(?<precision>[0-9]+))?(?<type>[%bcdeufFosxX])?/m';

I would be concerned that by converting everything to # you lose the information where someone has swapped the order using 2$, 1$ etc.

I wanted to make it so that if you captured a custom padding character prefixed by a quote, that quote wouldn't be captured, but I couldn't work it out.