的preg_replace - 如何匹配任何非\ * \ *

Question

我似乎無法找出正則表達式匹配的格式的任何字符串

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

我試圖在PHP這樣做

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

但沒有字符串替換完成。

Answer 1

您可以使用assertion?!一個字符明智配對.佔位符：

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

這基本上意味着匹配任意數量的anythings (.)+，但.永遠不能佔據的\*\*

Answer 2

你可以用懶惰匹配

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

或貪婪的一個

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

Answer 3

這應該工作：

$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); 

Answer 4

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

Answer 5

的地方嘗試使用：

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