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Java中高效的XSLT管道,带参数

xslt pipeline sax dom java

jbeard4 · 技术社区 · 14 年前

这个问题的最高答案描述了一种在Java中实现高效XSLT管道的技术:

Efficient XSLT pipeline in Java (or redirecting Results to Sources)

变压器.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.
    }
}

<?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>

<my xmlns="urn:foo">
    <hello/>
</my>

它提供了以下输出:

<?xml version="1.0" encoding="UTF-8"?><my xmlns="urn:foo">
        <hello><foo:world xmlns:foo="urn:foo">foo is : </foo:world></hello>
</my>

如你所见foo:world/@属性为空,并且福:世界说“foo is:”。预期的行为是,应该用传递到setParameter方法中的参数填充它们。

有没有一种方法可以使用这种技术设置XSL转换参数。如果没有,有人能推荐一种替代技术来在Java中高效地转换样式表吗,比如可以设置XSLT参数?

2 回复 | 直到 7 年前

Jon Freedman 14 年前

//SETTING PARAMETERS HERE
th1.getTransformer().setParameter("foo","this is from param 1");
th2.getTransformer().setParameter("bar","this is from param 2");

tornike 11 年前

与最后一个音符有关。在th1.getTransformer()上调用transform(),结果再次指向th1是不正确的。它将被处理两次。使用新的Transformer()是正确的方法。

t.transform(new StreamSource(new File("in.xml")), new SAXResult(th1));