来自JSON中Twitter搜索API的趋势数据。
使用以下命令抓取文件:
$jsonurl = "http://search.twitter.com/trends.json";
$json = file_get_contents($jsonurl,0,null,null);
$json_output = json_decode($json);
如何处理此对象中的数据。作为一个数组?只需要从Name值中提取数据。
JSON对象包含:
stdClass Object
(
[trends] => Array
(
[0] => stdClass Object
(
[name] => Vote
[url] => http://search.twitter.com/search?q=Vote
)
[1] => stdClass Object
(
[name] => Halloween
[url] => http://search.twitter.com/search?q=Halloween
)
[2] => stdClass Object
(
[name] => Starbucks
[url] => http://search.twitter.com/search?q=Starbucks
)
[3] => stdClass Object
(
[name] => #flylady
[url] => http://search.twitter.com/search?q=%23flylady
)
[4] => stdClass Object
(
[name] => #votereport
[url] => http://search.twitter.com/search?q=%23votereport
)
[5] => stdClass Object
(
[name] => Election Day
[url] => http://search.twitter.com/search?q=%22Election+Day%22
)
[6] => stdClass Object
(
[name] => #PubCon
[url] => http://search.twitter.com/search?q=%23PubCon
)
[7] => stdClass Object
(
[name] => #defrag08
[url] => http://search.twitter.com/search?q=%23defrag08
)
[8] => stdClass Object
(
[name] => Melbourne Cup
[url] => http://search.twitter.com/search?q=%22Melbourne+Cup%22
)
[9] => stdClass Object
(
[name] => Cheney
[url] => http://search.twitter.com/search?q=Cheney
)
)
[as_of] => Mon, 03 Nov 2008 21:49:36 +0000
)
发布于 2008-11-04 20:59:08
你是说像这样的东西?
<?php
$jsonurl = "http://search.twitter.com/trends.json";
$json = file_get_contents($jsonurl,0,null,null);
$json_output = json_decode($json);
foreach ( $json_output->trends as $trend )
{
echo "{$trend->name}\n";
}
发布于 2010-11-03 20:58:48
如果您使用json_decode($string, true)
,您将不会得到任何对象,但所有内容都是一个关联或数字索引数组。更容易处理,因为PHP提供的stdObject只是一个带有公共属性的哑容器,不能用你自己的功能来扩展。
$array = json_decode($string, true);
echo $array['trends'][0]['name'];
发布于 2008-11-04 21:03:53
就像你定义的对象一样使用它。即
$trends = $json_output->trends;
https://stackoverflow.com/questions/263392
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