Parser Json PHP

Question

    [{"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>

Answer 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']

Answer 2

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'];