PHP get URL after jump link

$url =
http://s.click.taobao.com/t?e=m%3D2%26s%3DNUvtFDcIwOocQipKwQzePOeEDrYVVa64LKpWJ% 2bin0xjradhuf14fmr8wzheppgyi8svium61dt3fihka92r7zcnjyd38oaemguafy6v5pipjbfxotquef4wnvsvnnsb8
Jump to
$result =
http://item.taobao.com/item.htm?id=23547888655&ali_trackid=2:mm_41992893_0_0:1385292898_6k3_1991321817

$url is from Ali mother to get the promotion link, his page content is JS, he jumped to the $result through JS, I want to get $result by using the text, PHP can encapsulate this function?

Reply to discussion (solution)

Then you have to look at its JS jump code, after using PHP simulation.

Then you have to look at its JS jump code, after using PHP simulation.

I tried this method, but I didn't get it.

Is there any other way, for example, curl, Fsockopen, Get_headers, File_get_contents, others, with what can be achieved?

Then you have to look at its JS jump code, after using PHP simulation.

I tried this method, but I didn't get it.

Is there any other way, for example, curl, Fsockopen, Get_headers, File_get_contents, others, with what can be achieved?

Then try this, http://blog.segmentfault.com/pcp/1190000000341558.

PHP Curl
Set curl_setopt ($ch, Curlopt_header, true);
Return information in the regular access location: part
Preg_match ('/^location: (? P . *?) $/m ', $HMTL, $match)
Location: The following address is the post-jump address.

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