85

来自 Twitter 搜索 API 的 JSON 格式的趋势数据。

使用以下方法获取文件：

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

如何处理来自该对象的数据。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
)
4

3 回答 3

145

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";
}
于 2008-11-04T20:59:07.717 回答
35

如果您使用json_decode($string, true)，您将不会获得任何对象，而是将所有内容作为关联数组或数字索引数组。更容易处理，因为 PHP 提供的 stdObject 只不过是一个具有公共属性的愚蠢容器，无法使用您自己的功能进行扩展。

$array = json_decode($string, true);

echo $array['trends'][0]['name'];
于 2010-11-03T12:58:47.963 回答
8

就像你定义的对象一样使用它。IE

$trends = $json_output->trends;
于 2008-11-04T21:03:52.857 回答