获取正确的图像网址

Question

我有以下图片链接:

http://ws.assoc-amazon.com/widgets/q?_encoding=UTF8&ASIN=B008EYEYBA&Format=_SL110_&ID=AsinImage&MarketPlace=US&ServiceVersion=20070822&WS=1&tag=mytwitterpage-20 

但是如果你点击它并在浏览器中查看它,图像文件的实际URL是这样的:

http://ecx.images-amazon.com/images/I/418lsVTc0aL._SL110_.jpg

任何想法我如何解析上面的图像链接,以获得使用PHP的实际jpg文件？

Answer 1

<?php

function get_url($url) {
    $ch = curl_init($url);
    curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);

    curl_exec($ch);

    if (!curl_errno($ch)) {
        $url = curl_getinfo($ch, CURLINFO_EFFECTIVE_URL);
    }

    curl_close($ch);

    return $url;
}

echo get_url("http://ws.assoc-amazon.com/widgets/q?_encoding=UTF8&ASIN=B008EYEYBA&Format=_SL110_&ID=AsinImage&MarketPlace=US&ServiceVersion=20070822&WS=1&tag=mytwitterpage-20");

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