preg_replace - how to match anything that's NOT \*\*

I can't seem to figure out the regular expression to match any string that's in the format

**(anything that's not **)**

I tried doing this in php

$str = "** hello world * hello **";
$str = preg_replace('/\*\*(([^开发者_JAVA百科\*][^\*]+))\*\*/s','<strong>$1</strong>',$str);

but no string replacement was done.

You can use an assertion ?! paired with a character-wise . placeholder:

= preg_replace('/\*\*(((?!\*\*).)+)\*\*/s',

This basically means to match any number of anythings (.)+, but the . can never occupy the place of a \*\*

You could use lazy match

\*\*(.+?)\*\*
# "find the shortest string between ** and **

or a greedy one

\*\*((?:[^*]|\*[^*])+)\*\*
# "find the string between ** and **,
#  comprising of only non-*, or a * followed by a non-*"

This should work:

$result = preg_replace(
    '/\*\*      # Match **
    (           # Match and capture...
     (?:        # the following...
      (?!\*\*)  # (unless there is a ** right ahead)
     .          # any character
     )*         # zero or more times
    )           # End of capturing group
    \*\*        # Match **
    /sx', 
    '<strong>\1</strong>', $subject);

preg_replace( '/\*\*(.*?)\*\*/', '<strong>$1</strong>', $str );

Try with:

$str = "** hello world * hello **";
$str = preg_replace('/\*\*(.*)\*\*/s','<strong>$1</strong>',$str);