处理PHP JSON对象中的数据

Question

来自JSON中的Twitter Search API的趋势数据.

使用以下方法抓取文件:

$jsonurl = "http://search.twitter.com/trends.json";
$json = file_get_contents($jsonurl,0,null,null);
$json_output = json_decode($json);

如何处理来自此对象的数据.作为阵列？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
)

Answer 1

你的意思是这样的？

<?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";
}

Answer 2

如果使用json_decode($string, true),则不会获得任何对象,而是将所有对象作为关联或数字索引数组.更容易处理的方式,因为PHP提供的stdObject只是一个具有公共属性的哑容器,无法使用您自己的功能进行扩展.

$array = json_decode($string, true);

echo $array['trends'][0]['name'];

Answer 3

只需使用它就像你定义的对象一样.即

$trends = $json_output->trends;