读取XML并输出RSS提要。 [英] reading an XML and outputing an RSS feed.

问题描述

您好，


我已经在这方面工作了一天而感到沮丧。基本上我想读取一个引用许多其他xml文件的主xml文件，并使用此信息输出RSS Feed。


要引用的文件。这个文件通常非常大，并且有很多类别，所以它需要能够读取每个类别并输出一个频道，然后输出文章中每个文章ID的项目内容。

展开 | 选择 < span class =codeDivider> | Wrap | 行号

解决方案

我可以看到你的xslt和xpath有几个问题。首先：

1. For-each context problem

展开 | 选择 | Wrap < span class =codeDivider> | 行号

我可以看到你的xslt和xpath有几个问题。首先：

1. For-each context problem

展开 | 选择 | Wrap < span class =codeDivider> | 行号

好的，可以这个会让你更好地理解我需要的东西，


我必须阅读以下所有信息并导出一个RSS文件。


所以它将是


< rss>


< channel> （将每个类别ID视为一个频道）


内容

< item>

内容

< / item>

< / channel>


< channel>

目录

< item>

内容

< / item>

< / channel>

ect ect。

< / rss>

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Hello,

I''ve been working on this for a days and getting frustrated. Basically i want read a master xml file that references many other xml files and use this information to output an RSS Feed.

File to reference. this file is usally quite large and has many categories so it needs to be able to read each category and output a channel then item contents for each article id in the catecory.

Expand|Select|Wrap|Line Numbers

解决方案

I can see several problems with your xslt and xpath. To start with:
1. For-each context problem

Expand|Select|Wrap|Line Numbers

I can see several problems with your xslt and xpath. To start with:
1. For-each context problem

Expand|Select|Wrap|Line Numbers

Ok, mayeb this will give you a better understanding of what i need,

I have to read all the below information and export a single RSS file.

So it would be

<rss>

<channel> (treat each category id as a channel)

contents
<item>
contents
</item>
</channel>

<channel>
Contents
<item>
contents
</item>
</channel>
ect ect ect.
</rss>

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