处理 xslt 包含中的非法 URI 字符
Handle illegal URI characters in xslt inclusion
在 xsl 转换中,我有一个包含一些其他 xslt 的 xslt 文件。问题是这些 xslt 的 URI 包含非法字符,特别是“##”。 xslt 看起来像这样:
<xsl:include href="/appdm/tomcat/webapps/sentys##1.0.0/WEB-INF/classes/xslt/release_java/xslt/gen.xslt" />
当我尝试实例化一个 java Transformer
时,我得到了错误:
javax.xml.transform.TransformerConfigurationException: javax.xml.transform.TransformerConfigurationException: javax.xml.transform.TransformerException: org.xml.sax.SAXException: org.apache.xml.utils.URI$MalformedURIException: Fragment contains invalid character:#
这是 java 代码:
public String xslTransform2String(String sXml, String sXslt) throws Exception {
String sResult = null;
try {
Source oStrSource = createStringSource(sXml);
DocumentBuilderFactory oDocFactory = DocumentBuilderFactory.newInstance();
oDocFactory.setNamespaceAware(true);
//sXslt is the xslt content with the inclusions
//<xsl:include href="/appdm/tomcat/webapps/sentys##1.0.0/WEB-INF/classes/xslt/release_java/xslt/gen.xslt" />"
Document oDocXslt = oDocFactory.newDocumentBuilder().parse(new InputSource(new StringReader(sXslt)));
Source oXsltSource = new DOMSource(oDocXslt);
StringWriter oStrOut = new StringWriter();
Result oTransRes = createStringResult(oStrOut);
Transformer oTrans = createXsltTransformer(oXsltSource);
oTrans.transform(oStrSource, oTransRes);
sResult = oStrOut.toString();
} catch (Exception oEx) {
throw new BddException(oEx, XmlProvider.ERR_XSLT, null);
}
return sResult;
}
private Transformer createXsltTransformer(Source oXsltSource) throws Exception {
Transformer transformer = getXsltTransformerFactory().newTransformer(
oXsltSource);
ErrorListener errorListener = new DefaultErrorListener();
transformer.setErrorListener(errorListener);
return transformer;
}
有什么方法可以使用相对路径而不是绝对路径吗?
谢谢
为避免 MalformedURIException,请将第二个或两个 #
替换为 %23
。
见
在 xsl 转换中,我有一个包含一些其他 xslt 的 xslt 文件。问题是这些 xslt 的 URI 包含非法字符,特别是“##”。 xslt 看起来像这样:
<xsl:include href="/appdm/tomcat/webapps/sentys##1.0.0/WEB-INF/classes/xslt/release_java/xslt/gen.xslt" />
当我尝试实例化一个 java Transformer
时,我得到了错误:
javax.xml.transform.TransformerConfigurationException: javax.xml.transform.TransformerConfigurationException: javax.xml.transform.TransformerException: org.xml.sax.SAXException: org.apache.xml.utils.URI$MalformedURIException: Fragment contains invalid character:#
这是 java 代码:
public String xslTransform2String(String sXml, String sXslt) throws Exception {
String sResult = null;
try {
Source oStrSource = createStringSource(sXml);
DocumentBuilderFactory oDocFactory = DocumentBuilderFactory.newInstance();
oDocFactory.setNamespaceAware(true);
//sXslt is the xslt content with the inclusions
//<xsl:include href="/appdm/tomcat/webapps/sentys##1.0.0/WEB-INF/classes/xslt/release_java/xslt/gen.xslt" />"
Document oDocXslt = oDocFactory.newDocumentBuilder().parse(new InputSource(new StringReader(sXslt)));
Source oXsltSource = new DOMSource(oDocXslt);
StringWriter oStrOut = new StringWriter();
Result oTransRes = createStringResult(oStrOut);
Transformer oTrans = createXsltTransformer(oXsltSource);
oTrans.transform(oStrSource, oTransRes);
sResult = oStrOut.toString();
} catch (Exception oEx) {
throw new BddException(oEx, XmlProvider.ERR_XSLT, null);
}
return sResult;
}
private Transformer createXsltTransformer(Source oXsltSource) throws Exception {
Transformer transformer = getXsltTransformerFactory().newTransformer(
oXsltSource);
ErrorListener errorListener = new DefaultErrorListener();
transformer.setErrorListener(errorListener);
return transformer;
}
有什么方法可以使用相对路径而不是绝对路径吗?
谢谢
为避免 MalformedURIException,请将第二个或两个 #
替换为 %23
。
见