将URL作为参数传递给XSL [英] Pass URL as a parameter to XSL

问题描述

我想将当前页面的URL作为属性传递给XSL模板.据我了解，它应该作为参数传递，然后用作属性.

I'd like to pass current page URL as an attribute to XSL template. As far as I understood it should be passed as a parameter, and then used as an attribute.

我使用PHP加载XML& XSL文件:

I use PHP to load XML & XSL files:

<?php
$xml = new DOMDocument;
$xml->load('main.xml');

$xsl = new DOMDocument;
$xsl->load('blocks/common.xsl');

$proc = new XSLTProcessor;

$proc->importStyleSheet($xsl);

echo $proc->transformToXML($xml);
?> 

例如，应如何更改此代码以将URL作为名为"current-url"的参数传递?

How should this code be altered to pass URL as a parameter named "current-url", for example?

我在这里看到过很多类似的问题，但使用了不同的解决方案，但到目前为止，没有一个问题对我有用.预先谢谢你.

I've seen a lot of similar questions here with different solutions, but none has worked for me so far. Thank you in advance.

推荐答案

也许您已经尝试过这种方法，但是如果没有，请尝试以下方法:

Maybe you already tried this approach, but in case if not:

<?php

  $params = array('current-url' => $_SERVER['REQUEST_URI']);

  $xml = new DOMDocument;
  $xml->load('main.xml');

  $xsl = new DOMDocument;
  $xsl->load('blocks/common.xsl');

  $proc = new XSLTProcessor;
  $proc -> registerPHPFunctions();
  $proc->importStyleSheet($xsl);

  foreach ($params as $key => $val)
    $proc->setParameter('', $key, $val);

  echo $proc->transformToXML($xml);
  ?>

在xsl中，在模板上方添加

In the xsl, add above the templates

<xsl:param name="current-url" />

在模板中，您可以使用

<xsl:value-of select="$current-url" />

如果还不存在，则必须在xsl:stylesheet声明中添加xmlns:php="http://php.net/xsl".
供参考: registerPHPFunctions()以及您可能已经在SO上检查过的解决方案: 将变量传递到XSLT

If not already there, you have to add xmlns:php="http://php.net/xsl" into the xsl:stylesheet declaration.
For reference: registerPHPFunctions() and a solution you maybe already checked on SO: Passing variables to XSLT

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