这是来自Instagram JSON的片段:
"images":{
"standard_resolution":
{ "url":"http:\/\/scontent-a.cdninstagram.com\/hphotos-xfa1\/t51.2885-15\/10593467_370803786404217_1595289732_n.jpg","width":640,"height":640}
}
我想从['images']['standard_resolution']['url']
剥离协议。我尝试过:
.parse_url($value['images']['standard_resolution']['url'], PHP_URL_HOST) . parse_url($value['images']['standard_resolution']['url'], PHP_URL_PATH).
但什么也没做!我认为这与在JSON(/
)中完成的http:\/\/
转义有关。因为如果我尝试
.parse_url("http://www.google.com", PHP_URL_HOST) . parse_url("http://www.google.com", PHP_URL_PATH).
..它运作正常。我想保持简单......不要使用正则表达式。 parse_url
会很完美。
答案 0 :(得分:1)
为什么你甚至需要替换或正则表达式呢?
如果您使用json_decode
,则转义的斜杠是未转义的。
这样做:
<?php
$foo = '{"images":{"standard_resolution":{"url":"http:\/\/scontent-a.cdninstagram.com\/hphotos-xfa1\/t51.2885-15\/10593467_370803786404217_1595289732_n.jpg","width":640,"height":640}}}';
$bar = json_decode($foo, true);
$baz =
parse_url($bar['images']['standard_resolution']['url'], PHP_URL_HOST) .
parse_url($bar['images']['standard_resolution']['url'], PHP_URL_PATH);
echo $baz;
输出:
scontent-a.cdninstagram.com/hphotos-xfa1/t51.2885-15/10593467_370803786404217_1595289732_n.jpg
请参阅: