无法理解如何做这个preg_replace,没试过,因为不知道该尝试什么,太难理解..
index-D.html where d is a digit from 0-99999
如何替换该字符串的出现,index-D.html
为空
答案 0 :(得分:0)
manual非常清楚并提供了示例:
$string = "index-D.html where d is a digit from 0-99999";
$pattern = "index-D.html";
$new_string = preg_replace($pattern, "", $string);