处理 PHP JSON 对象中的数据
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Handling data in a PHP JSON Object
提问by Peter Bailey
Trends data from Twitter Search API in JSON.
来自 Twitter 搜索 API 的 JSON 格式的趋势数据。
Grabbing the file using:
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);
How do I work with data from this object. As an array? Only really need to extract data from the [name] values.
我如何处理来自这个对象的数据。作为数组？只有真正需要从 [name] 值中提取数据。
JSON object contains:
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
)
回答by Peter Bailey
You mean something like this?
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";
}
回答by Sven
If you use json_decode($string, true), you will get no objects, but everything as an associative or number indexed array. Way easier to handle, as the stdObject provided by PHP is nothing but a dumb container with public properties, which cannot be extended with your own functionality.
如果您使用json_decode($string, true)，您将不会得到任何对象，而是将所有内容作为关联或数字索引数组。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'];
回答by Zak
Just use it like it was an object you defined. i.e.
就像使用您定义的对象一样使用它。IE
$trends = $json_output->trends;
