YUI3 YQL同时使用限制和排序

发布于 2024-10-09 11:22:11 字数 521 浏览 3 评论 0原文

我有一个 YQL 查询，它组合了一堆 RSS 提要，然后按日期对它们进行排序。它工作得很好，但我希望能够对结果进行分页。

这是我的查询：

select channel.item.title, channel.item.link, channel.item.pubDate, channel.item.description(0) from xml where url in(... urls go here ...) | unique(field='channel.item.link') | sort(field='channel.item.pubDate', descending='true')

问题是 yql 在排序过滤器之前执行 LIMIT 和 OFFSET。因此，如果我有 LIMIT 5，我最终只会得到列表中第一个 RSS feed 的前 5 项……而不是所有组合 feed 的前 5 项。

有没有办法链接查询，以便我可以获得对所有结果进行排序的所有查询，并调用限制我的结果的查询。

感谢您的帮助。

原文

I have a YQL query that combines a bunch of RSS feeds and then sorts them by date. It's working great, but I would like to be able to paginate the results.

Here is my query:

select channel.item.title, channel.item.link, channel.item.pubDate, channel.item.description(0) from xml where url in(... urls go here ...) | unique(field='channel.item.link') | sort(field='channel.item.pubDate', descending='true')

The problem is that yql executes LIMIT and OFFSET before the sort filter. So if I have a LIMIT 5 I end up with only the first 5 items from the first RSS feed in the list ... not the first 5 items of all my combined feeds.

Is there a way to chain queries, so I can get all my query that sorts all the results and the call a query that limits my results.

Thanks for the help.

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傲娇萝莉攻 2024-10-16 11:22:11

使用 truncate(count=5) 获取过滤后的 feed 的前 5 项。

select channel.item.title, channel.item.link, channel.item.pubDate, channel.item.description(0) from xml where url in(... urls go here ...) | unique(field='channel.item.link') | sort(field='channel.item.pubDate', descending='true') | truncate(count=5)

或者具有反向排序的 tail() 方法（上面发布的）也可以工作。

请参阅http://developer.yahoo.com/yql/guide/sorting.html

Use the truncate(count=5) to get the first 5 items of the filtered feed.

select channel.item.title, channel.item.link, channel.item.pubDate, channel.item.description(0) from xml where url in(... urls go here ...) | unique(field='channel.item.link') | sort(field='channel.item.pubDate', descending='true') | truncate(count=5)

Or the tail() method with a reversed sort (posted above) would work as well.

See http://developer.yahoo.com/yql/guide/sorting.html

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红焚 2024-10-16 11:22:11

在另一个方向上使用 tail() 和另外的 sort() 怎么样？

像这样的事情：

select channel.item.title, channel.item.link, channel.item.pubDate, channel.item.description(0) from xml where url in(... urls go here ...) | unique(field='channel.item.link') | sort(field='channel.item.pubDate', descending='false') | tail(count=5)

我还没有尝试过这个，我不确定你需要哪种排序顺序，但你肯定可以使用tail()。显然没有等效的 head() 函数，因此请确保最新的项目是排序后的末尾。

欲了解更多信息，请参阅
http://developer.yahoo.com/yql/guide/sorting.html

What about using tail() and additionally sort() in the other direction?

Something like this:

select channel.item.title, channel.item.link, channel.item.pubDate, channel.item.description(0) from xml where url in(... urls go here ...) | unique(field='channel.item.link') | sort(field='channel.item.pubDate', descending='false') | tail(count=5)

I have not tried this and I am not sure which sort order you need but you can for sure use tail(). There is no equivalent head() function apparently, so make sure the newest items are are the end after your sorting.

For further information see
http://developer.yahoo.com/yql/guide/sorting.html

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