preg_replace怀疑

Question

我做错了什么？

echo preg_replace('#\d{3}\d{3}\d{3}\d{2}#', '$1.$2.$3-$4', '12345678901');

输出为： 123.456.789-01 。还没有形成字符串!!

Answer 1

<?php
echo preg_replace('#(\d{3})(\d{3})(\d{3})(\d{2})#', '$1.$2.$3-$4', '12345678901');

是正确的，因为美元符号+整数是指（）括号（分组）中的内容

demo

Answer 2

您没有正确分组（缺少括号）：

echo preg_replace('#(\d{3})(\d{3})(\d{3})(\d{2})#', '$1.$2.$3-$4', '12345678901');