Java 中带参数的 XSL 转换
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XSL Transformation in Java with parameters
提问by Anirudh
I have a xsl file where i need to use parameters from an external source. I am using Java and my code looks something like this:
我有一个 xsl 文件，我需要在其中使用来自外部源的参数。我正在使用 Java，我的代码如下所示：
TransformerFactory transformerFactory = TransformerFactory.newInstance();
Transformer xsltTransformer = transformerFactory.newTransformer(xsltSource);
xsltTransformer.setParameter(parameterName, parameterValue);
However, an exception is thrown at the 2nd line - Variable or parameter 'variable_name' is undefined.I realize that XSL is compiled and is probably compiled when the transformer is created.
但是，在第 2 行抛出异常 -变量或参数“variable_name”未定义。我意识到 XSL 已编译，并且可能在创建转换器时编译。
So, how do i pass parameters to my transformation? How is the setParameter method supposed to be used?
那么，我如何将参数传递给我的转换？应该如何使用 setParameter 方法？
采纳答案by rsp
If you pass a parameter like:
如果传递如下参数：
transformer.setParameter("render_id", "1234");
the parameter can be picked up by the transform:
参数可以通过变换获取：
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >
<xsl:output method="xml" encoding="UTF-8" omit-xml-declaration="yes"/>
<!-- Receives the id of the menu being rendered. -->
<xsl:param name="render_id" />
回答by TinyRacoon
rsp's answer was spot on. Thanks. Just want to add that you cannot pass a parameter to a variable in the same way (I am setting parameters via Java's TransformerFactory).
rsp 的回答是正确的。谢谢。只是想补充一点，您不能以相同的方式将参数传递给变量（我正在通过 Java 的 TransformerFactory 设置参数）。
I made the mistake of thinking variables and params were interchangeable :)
我错误地认为变量和参数是可以互换的:)
