나는 다음과 같은 XSLT가 :XSLT는 템플릿의 요소를 무시합니까?
<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:html="http://www.w3.org/TR/REC-html40"
xmlns:fo="http://www.w3.org/1999/XSL/Format"
xmlns:o="urn:schemas-microsoft-com:office:office"
xmlns:x="urn:schemas-microsoft-com:office:excel"
xmlns:ss="urn:schemas-microsoft-com:office:spreadsheet">
<xsl:output method="xml" indent="yes" />
<xsl:strip-space elements="*" />
<xsl:template match="ss:Workbook/o:DocumentProperties/o:*"/>
<xsl:template match="ss:Workbook/x:ExcelWorkbook/x:*"/>
<xsl:template match="ss:Workbook/x:ExcelWorkbook/x:*"/>
<xsl:template match="ss:Workbook/ss:Worksheet/x:WorksheetOptions/x:*"/>
<xsl:template match="ss:Workbook/ss:DocumentProperties/ss:*"/>
<xsl:template match='ss:Workbook/ss:Worksheet/ss:Table'>
<grade-dist>
<xsl:apply-templates select='ss:Workbook/ss:Worksheet/ss:Table'/>
</grade-dist>
</xsl:template>
<xsl:template match='ss:Workbook/ss:Worksheet/ss:Table'>
....
내 XML 출력 벌금을,하지만 난 해달라고 왜 <grade-dist>
하고 모두 완전히 무시처럼의 </grade-dist>
는 것, 어떤 생각을?
감사합니다,
좋은 질문, +1. 설명과 두 가지 해결책에 대한 내 대답을보십시오. :) –