1
我有xslt2.0,我降級到1.0並使用xalan,但得到以下異常。如何在xslt 2.0中降級後在xslt1.0中寫入函數
可恢復的錯誤:第9行:不支持的XSL元素「函數」。
xslt的部分如下。
<xsl:function name="nav:adjustDate">
<xsl:param name="dateStr" />
<xsl:param name="age" />
<xsl:variable name="minutes">
<xsl:choose>
<xsl:when test="$age = 1">
<xsl:value-of select="0" />
</xsl:when>
<xsl:when test="$age = 2">
<xsl:value-of select="-10" />
</xsl:when>
<xsl:when test="$age = 3">
<xsl:value-of select="-20" />
</xsl:when>
<xsl:when test="$age = 4">
<xsl:value-of select="-30" />
</xsl:when>
<xsl:when test="$age = 5">
<xsl:value-of select="-40" />
</xsl:when>
<xsl:when test="$age = 6">
<xsl:value-of select="-50" />
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="-60" />
</xsl:otherwise>
</xsl:choose>
</xsl:variable>
<xsl:variable name="dateFormatterStr">
<xsl:text>yyyy-MM-dd'T'HH:mm:ss.SSSZ</xsl:text>
</xsl:variable>
<!-- output date format should match the input date format of the job file -->
<xsl:variable name="outDateFormatterStr">
<xsl:text>yyyy-MM-dd'T'HH:mm:ssZ</xsl:text>
</xsl:variable>
<xsl:variable name="bo" select="bool:new('FALSE')" />
<xsl:variable name="dateFormatter" select="dateFormat:new($dateFormatterStr)" />
<xsl:variable name="outDateFormatter" select="dateFormat:new($outDateFormatterStr)" />
<xsl:value-of select="dateFormat:setLenient($dateFormatter,$bo)" />
<!-- Have to remove the colon in the timezone offset(eg. +05:00) otherwise date formatter wont work correctly -->
<xsl:variable name="testDate"
select="dateFormat:parse($dateFormatter,concat(substring($dateStr,1,string-length($dateStr)-3),'00'))" />
<xsl:variable name="cal" select="gregorianCal:new()" />
<xsl:value-of select="gregorianCal:setTime($cal,$testDate)" />
<!-- xslt version 2 does not accept contants 12 represents the value for java.util.Calendar.MINUTE
Follow section of code will subtract the number of minutes-->
<xsl:value-of select="gregorianCal:add($cal,12,$minutes)" />
<xsl:variable name="outputDate" select="gregorianCal:getTime($cal)" />
<xsl:sequence select="dateFormat:format($outDateFormatter,$outputDate)" />
</xsl:function>
也想知道XSL替代:在xslt1.0
序列有人能指導我這個?如何進行 ? 對於xslt來說相當新穎。