处理PHP JSON对象中的数据。

Trends data from Twitter Search API in JSON.

来自JSON的Twitter搜索API的趋势数据。

Grabbing the file using:

把文件使用:

$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.

如何处理来自这个对象的数据。作为一个数组?只需要从[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
)

4 个解决方案

#1

147

You mean something like this?

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";
}

#2

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，而是所有对象都作为一个关联数组或数字索引数组。很容易处理，因为PHP提供的stdObject只是一个具有公共属性的哑容器，不能使用自己的功能对其进行扩展。

$array = json_decode($string, true);

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

#3

Just use it like it was an object you defined. i.e.

就像你定义的对象一样使用它。即。

$trends = $json_output->trends;

#4

-2

The clean way would be:

清洁的方法是:

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

foreach ($trends as $trend) {
    your_func($trend['name']);
}

#1

147