PHP的preg_replace函数的问题
$string = "april 15, 2003";
$pattern = "/(/w+) (/d+), (/d+)/i";
$replacement = "/${1}1,/$3";
print preg_replace($pattern, $replacement, $string);
/* output
======
april1,2003
*/
$replacement = "/${1}1,/$3";
"/${1}1,/$3"这个是什么意思？
$patterns = array ("/(19|20)(/d{2})-(/d{1,2})-(/d{1,2})/",
"/^/s*{(/w+)}/s*=/");
$replace = array ("//3///4///1//2", "$//1 =");
print preg_replace ($patterns, $replace, "{startdate} = 1999-5-27");
$replace = array ("//3///4///1//2", "$//1 =");这个也是什么意思？
------解决方案--------------------
双引号中 php 会试图将 $ 解释为变量的前导
字符串 "${1}" 将引起变量未定义的错误
所以要转义掉，这样才会把 ${1} 交给 prea_replace 去处理
