我有一个带图像链接的字符串.
$image_link_raw = 'http://website.com/files/2012/10/image001.png';
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现在我想要删除http://website.com,只是得到/files/2012/10/image001.png:
$image_link_raw = '/files/2012/10/image001.png';
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有没有办法在PHP中这样做?
Bra*_*tie 11
$path = parse_url($url, PHP_URL_PATH);
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parse_url 效果很好(并且可以获得其他详细信息,如主机,端口,协议等.)
<?php
$url = 'http://website.com/files/2012/10/image001.png';
echo 'URL Parts:' . PHP_EOL;
var_dump(parse_url($url));
echo PHP_EOL . 'And specific to path:' . PHP_EOL;
echo parse_url($url, PHP_URL_PATH);
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结果输出:
URL Parts:
array(3) {
["scheme"]=>
string(4) "http"
["host"]=>
string(11) "website.com"
["path"]=>
string(27) "/files/2012/10/image001.png"
}
And specific to path:
/files/2012/10/image001.png
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小智 8
我想我们只是给出答案呢?
<?php
$image_link_raw = 'http://website.com/files/2012/10/image001.png';
$p=parse_url($image_link_raw);
//print_r($p);
echo $p['path'];
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