PHP-使用YQL选择特定条目 [英] PHP - selecting a particular entry with YQL

问题描述

所以我正在使用 YQL 来获取youtube用户的订阅者人数。

so I'm using YQL to get a youtube user's subscriber count.

要执行此操作，我正在使用如下网址进行查询：

to do this I'm making a query using a URL like this:

$query = 'https://query.yahooapis.com/v1/public/yql?q='.urlencode("SELECT * FROM xml WHERE url='http://gdata.youtube.com/feeds/api/users/{$id}'").'&format=json&callback=';

然后我解码响应并找到我想要的数组项：

then I decode the response and find the array entry I'm looking for like this:

  $response = json_decode($response, true);
  $subscriber_count = $response['query']['results']['entry']['statistics']['subscriberCount'];

工作正常，除了我不喜欢这样做的方式:)
是我可以直接从上面的 $ query URL获取 subscriberCount 值的方法吗？我不需要整个XML，只需要一个条目。

it works fine, except I don't like the way I'm doing this :) I mean, is there a way I can get the subscriberCount value directly from the $query URL above? I don't need the entire XML, just that one entry.

推荐答案

您可以做的是限制数据量YQL通过执行如下操作将YQL返回到您真正感兴趣的数据：

What you could do is to limit the amount of data that is returned by YQL to the data you are really interested in by doing sth like this:

SELECT statistics.subscriberCount FROM xml WHERE ...

仍然会在您感兴趣的数字周围保留一些结构化XML / JSON元素但至少要少一些（见下文）。不确定这是否是您想要的吗？

Still you would have some structural XML/JSON elements around the number that you are interested in but at least it is less (see below). Not sure if that is what you wanted?

<query xmlns:yahoo="http://www.yahooapis.com/v1/base.rng"
    yahoo:count="1" yahoo:created="2011-01-06T23:21:32Z" yahoo:lang="en-US">
    <results>
        <entry xmlns="http://www.w3.org/2005/Atom">
            <yt:statistics
                xmlns:yt="http://gdata.youtube.com/schemas/2007" subscriberCount="7"/>
        </entry>
    </results>
</query>

这篇关于PHP-使用YQL选择特定条目的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！