[{"data":{"attr":{"href":"http://www.google.com/","title":"http://www.google.com/"},"title":"Google"},"attr":{},"metadata":{}},
{"data":{"attr":{"href":"http://www.yandex.com/","title":"http://www.yandex.com/"},"title":"Яндекс"},"attr":{"class":""},"metadata":{}}]
帮助从Json获取链接。获取给定形式的Json
的链接<a href="http://www.google.com/" title="http://www.google.com/">Google</a>
<a href="http://www.yandex.com/" title="http://www.yandex.com/">Яндекс</a>
答案 0 :(得分:1)
因为这似乎是正确的答案,而且我只是设定了我可以用接受的答案代表做的赏金:
google或RTM - &gt; $parsed[0]['data']['attr']['href']
答案 1 :(得分:0)
我认为你要求从PHP中的json获取链接:
$links = json_decode($putYourJSONinsideThisVariable, true);
echo $links[0]['data']['attr']['href'];