頂端回答這個問題介紹了一種技術,在Java中實現高效的XSLT管道:高效XSLT管道,使用參數,在Java
Efficient XSLT pipeline in Java (or redirecting Results to Sources)
不幸的是,當變壓器似乎暴露了一個API設置XSLT參數,這似乎沒有任何影響。例如,我有以下代碼:
Transformer.java
import javax.xml.transform.sax.SAXTransformerFactory;
import javax.xml.transform.Templates;
import javax.xml.transform.sax.TransformerHandler;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.stream.StreamResult;
import javax.xml.transform.stream.StreamSource;
import javax.xml.transform.sax.SAXResult;
import javax.xml.transform.Transformer;
import java.io.File;
public class MyTransformer {
public static void main(String[] args) throws javax.xml.transform.TransformerConfigurationException, javax.xml.transform.TransformerException{
SAXTransformerFactory stf = (SAXTransformerFactory)TransformerFactory.newInstance();
// These templates objects could be reused and obtained from elsewhere.
Templates templates1 = stf.newTemplates(new StreamSource(new File("MyStylesheet1.xslt")));
Templates templates2 = stf.newTemplates(new StreamSource(new File("MyStylesheet2.xslt")));
TransformerHandler th1 = stf.newTransformerHandler(templates1);
TransformerHandler th2 = stf.newTransformerHandler(templates2);
th1.setResult(new SAXResult(th2));
th2.setResult(new StreamResult(System.out));
Transformer t = stf.newTransformer();
//SETTING PARAMETERS HERE
t.setParameter("foo","this is from param 1");
t.setParameter("bar","this is from param 2");
t.transform(new StreamSource(new File("in.xml")), new SAXResult(th1));
// th1 feeds th2, which in turn feeds System.out.
}
}
MyStylesheet1.xslt
<?xml version="1.0"?>
<stylesheet xmlns="http://www.w3.org/1999/XSL/Transform" xmlns:foo="urn:foo" version="1.0">
<output method="xml"/>
<param name="foo"/>
<template match="@*|node()">
<copy>
<apply-templates select="@*|node()"/>
</copy>
</template>
<template match="foo:my/foo:hello">
<copy>
<foo:world>
foo is : <value-of select="$foo"/>
</foo:world>
</copy>
</template>
</stylesheet>
MyStylesheet2.xslt
<?xml version="1.0"?>
<stylesheet xmlns="http://www.w3.org/1999/XSL/Transform" xmlns:foo="urn:foo" version="1.0">
<output method="xml"/>
<param name="bar"/>
<template match="@*|node()">
<copy>
<apply-templates select="@*|node()"/>
</copy>
</template>
<template match="foo:my/foo:hello/foo:world">
<copy>
<apply-templates select="@*|node()"/>
<attribute name="attr">
<value-of select="$bar"/>
</attribute>
</copy>
</template>
</stylesheet>
in.xml
<my xmlns="urn:foo">
<hello/>
</my>
這給了我下面的輸出:
<?xml version="1.0" encoding="UTF-8"?><my xmlns="urn:foo">
<hello><foo:world xmlns:foo="urn:foo">foo is : </foo:world></hello>
</my>
正如你可以看到富:世界/ @ attr爲空,foo的文本內容:大家說: 「foo是:」。預期的行爲是應該使用傳遞給setParameter方法的參數填充它們。
有沒有辦法使用這種技術設置XSL轉換參數。如果沒有,任何人都可以推薦一種替代技術來在Java中高效地轉換樣式表,這樣XSLT參數也可以被設置?