如何提取出/2013/userup/100/131a2z34-800.doc
回复讨论(解决方案) $str = {user:link islocal='1'} /2013/userup/100/131a2z34-800.doc {/user:link};$preg = #\{.*\}(.*)\{.*\}#iuus;preg_match_all($preg, $str, $matches);print_r($matches[1]);
正则 #}\s*(\s+)\s*{#
或者用split()依据空格或花括号切分也能得到