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找到特定的字符串想在PHP这里应用的preg_match以获得链接是我的代码我如何通过正则表达式的preg_match
$string = playlist: [{
image: "http://77.81.98.55/i/01/00012/d6ntku2p3tnn.jpg",
provider: "http://static.vidzi.tv/nplayer/vidzi.swf",
sources: [{
file: "http://vidzi.tv/6c90fqha0k9i-4b4a926f75863405c12bfaeb331b331c.m3u8?embed=",
type: 'hls'
},{
file: "http://77.81.98.55/gzuqiu4h342qedz7nknr5h3sli74im5vv7f2ae7z2pxjxhqnqxk4625rsdpq/v.mp4"
}]
,tracks: [{file: "http://vidzi.tv/6c90fqha0k9i-4b4a926f75863405c12bfaeb331b331c.vtt", kind: "thumbnails"}]
}]
preg_match('/file: "(.*?)",/i', $result, $dl_link);
echo '<pre>';
print_r($dl_link);
echo '</pre>';
,输出是
Array
(
[0] => file: "http://vidzi.tv/6c90fqha0k9i-4b4a926f75863405c12bfaeb331b331c.m3u8?embed=",
[1] => http://vidzi.tv/6c90fqha0k9i-4b4a926f75863405c12bfaeb331b331c.m3u8?embed=
)
从上面的preg_match我得到第一
file: "http://vidzi.tv/6c90fqha0k9i-4b4a926f75863405c12bfaeb331b331c.m3u8?embed="
,但我希望得到这个
file: "http://77.81.98.55/gzuqiu4h342qedz7nknr5h3sli74im5vv7f2ae7z2pxjxhqnqxk4625rsdpq/v.mp4"
帮我感谢
尝试使用'(文件:\“HTTP:\/\/\ d {2}。*?“)' –
使用'preg_match_all'并从结果数组中取得第二个匹配 – anubhava
分配给变量$ string的数据看起来应该是json - 是吗? – RamRaider