如何在php中解析url以获取数字和字符串

Question

我有网址： http://example.com/list4

我想在url名称中解析“list4”以获取分隔的字符串和数字。

最终的网址将是：

<?php  echo  'http://example.com/'.$string.$number; ?>

我现在能够做到这一点，但是：

当我尝试用$url = 'http://'.$_SERVER['HTTP_HOST'].$_SERVER['PHP_SELF'];替换$ url时，无效。我试过$url = basename($_SERVER['REQUEST_URI']);但是也没有用......为什么？

Answer 1

试一试：

$url = "http://example.com/list4";
preg_match("#\/(\w+)(\d+)$#", $url, $matches);
echo "String: " . $matches[1] . "<br>";
echo "Number: " . $matches[2] . "<br>";
// print out the URL you wanted above
echo "http://example.com/" . $matches[1] . $matches[2];

---

修改

为了满足问题的需要，您需要在网址末尾说明.php。这将起作用：

$url = $_SERVER['HTTP_HOST'] . $_SERVER['PHP_SELF'];
preg_match("#\/(\w+)(\d+)\.?\w*$#", $url, $matches); // updated
echo "String: " . $matches[1] . "<br>";
echo "Number: " . $matches[2] . "<br>";
// print out the URL you wanted above
echo "http://example.com/" . $matches[1] . $matches[2];

Answer 2

<?php

$网址= $ _ SERVER [ 'HTTP_HOST'] $ _ SERVER [ 'PHP_SELF'];

回声“

”;

$第一= preg_match_all（ '/ [/] +（[A-Z] ）/？'，$网址，$结果）; $字符串= $结果[1] [0];

$第二= preg_match_all（ '/ [/] + [A-Z] （[0-9] *）/？'，$网址，$结果）; $数= $结果[1] [0];

echo $ _SERVER ['HTTP_HOST']。$ string。$ number ？＆GT;