求该结果的preg_replace的替换写法
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2022-05-03 12:59:32
...
$url=http://www.xxx.com/66/
preg_replace("#^\/(\d{2,4})\/?$#","name/index.php?t=1&id=$1",strtolower($url));
通过以上我能得到一个重写的网址:name/index.php?t=1&sid=66
我如何能直接通过一次正则替换,替换出一个数组:如下?
array(
'name'=>index.php,
'param'=>array('t'=>1,'id'=66)
);
(
[name] => index.php
[param] => Array
(
[t] => 1
[id] => 66
)
)
preg_replace("#^\/(\d{2,4})\/?$#","name/index.php?t=1&id=$1",strtolower($url));
通过以上我能得到一个重写的网址:name/index.php?t=1&sid=66
我如何能直接通过一次正则替换,替换出一个数组:如下?
array(
'name'=>index.php,
'param'=>array('t'=>1,'id'=66)
);
回复讨论(解决方案)
一步不行,需要自行分析形成数组。
$url = 'http://www.xxx.com/66/';@preg_replace("#/(\d{2,4})/?$#e","\$t=array('name'=>'index.php','param'=>array('t'=>1,'id'=>$1))",strtolower($url));print_r($t);66Array
(
[name] => index.php
[param] => Array
(
[t] => 1
[id] => 66
)
)
$url='http://www.xxx.com/66/';$s= preg_replace("#\/(\d{2,4})\/?$#","/name/index.php?t=1&id=$1",strtolower($url));$ar=parse_url($s);$arr['name']=substr($ar['path'],strrpos($ar['path'],'/')+1);parse_str($ar['query'],$t);$arr['param']=$t;print_r($arr);