如何用preg_replace替换匹配的正则表达式来实现这一点？
例如，我想要字符串：
There was a movie made of Alan Turing's life in 2014 8
将被取代
There was a movie made of Alan Turing's life in 2014
而不替换年份的四位数或任何其他非一位数。
您可以在此处使用单词边界'b：
$s = "There was a movie made of Alan Turing's life in 2014 8";
$s = preg_replace('#'b'd'b#', '', $s);
//=> There was a movie made of Alan Turing's life in 2014
'b'd'b将匹配任何一位数字，但不会匹配超过一位的数字。