这是一个来自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']
中删除协议。我尝试过:
.parse_url($value['images']['standard_resolution']['url'], PHP_URL_HOST) . parse_url($value['images']['standard_resolution']['url'], PHP_URL_PATH).
但什么也不做!我认为这与JSON中的/
转义(http:'/'/
)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).
它工作得很好。我想保持轻松。。并且不要使用regex。CCD_ 4将是完美的。
为什么您甚至需要执行替换或regex?
如果使用json_decode
,则转义斜杠将取消转义。
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;
将输出:
scontent-a.cdninstagram.com/hphotos-xfa1/t51.2885-15/10593467_370803786404217_1595289732_n.jpg
参见:
http://ideone.com/F0c4m9