我需要一个相当复杂的正则表达式来完成以下任务：
> replace numbers in a string, i.e. 700, 12.43 by a label (format: {NUMBER:xx})
> ignore: when number is between {braces}, i.e. {7}, {7th}
> ignore: when any character is attached to number, i.e. G3, 7x, 1/2
> except: when
> preceded by $, i.e. $840
> followed by .!?:, i.e. 33! 45.65? 4...
综合来看：
Buy 4 {5} G3 Mac computers for 80% at $600 or 2 for 1/2 price: 200...
dollar. Twice - 2x - as cheap!
期望输出：
Buy {NUMBER:4} {5} G3 Mac computers for 80% at
$ {NUMBER:600} or {NUMBER:2} for 1/2 price:
{$NUMBER:200} dollar. Twice - 2x - as cheap!
我现在有了这个：
preg_replace("/(?<!{)(?>[0-9]+(?:'.[0-9]+)?)(?!})/", " {NUMBER:$0} ", $string);
输出：
Buy {NUMBER:4} {5} G {NUMBER:3} Mac computers for {NUMBER:80} % at
$ {NUMBER:600} or {NUMBER:2} for {NUMBER:1} / {NUMBER:2} price:
{NUMBER:200} ... dollar. Twice - {NUMBER:2} x - as cheap!
换句话说：忽略异常还不起作用，我不知道如何正确地实现它。谁能帮助我？
这适用于您的测试用例，并遵循您的规则，假设大括号正确匹配且未被测试：
$result = preg_replace(
'/(?<!'{) # Assert no preceding {
(?<![^'s$]) # Assert no preceding non-whitespace except $
'b # Match start of number
('d+(?:'.'d+)?+) # Match number (optional decimal part)
'b # Match end of number
(?![^{}]*'}) # Assert that next brace is not a closing brace
(?![^'s.!?,]) # Assert no following non-whitespace except .!?,
/x',
'{NUMBER:'1}', $string);
$string="Buy 4 {5} G3 Mac computers for 80% at '$600 or 2 for 1/2 price: 200... 'ndollar. Twice - 2x - as cheap!";
$pattern='/['s|^|'$]([0-9]+('.'s+)*)['s|$|'.|'!|'?|':|',]/';
//$count=preg_match_all($pattern, $string, $matches);
//echo "$count'n";
//print_r($matches[1]);
echo preg_replace($pattern,"{NUMBER:'$1}",$string);