如果我在GMT中获得格式为dateTime
的格式,我该如何在XSLT中转换dateTime
XML输入:
<items>
<item>
<lastname>Lisa</lastname>
<firstname>Rimpell</firstname>
<checkintime>2017-02-05T05:40:00+03:00</checkintime>
<chekouttime>2017-02-05T10:40:00+03:00</chekouttime>
<address></address>
</item>
</items>
XSLT是:
<xsl:template match="/">
<Response>
<Data>
<xsl:call-template name="Buildusers" />
</Data>
</Response>
</xsl:template>
<xsl:template name="Buildusers">
<Rows>
<xsl:for-each select="//items/item">
<Row Action="ADD">
<xsl:value-of select="lastname" />
|<xsl:value-of select="firstname" />
|<xsl:value-of select="checkintime" />
|<xsl:value-of select="chekouttime" />
|<xsl:value-of select="address" />
</Row>
</xsl:for-each>
</Rows>
</xsl:template>
我需要使用“dateTime
的转换来构建如下所示的行。那是当我得到dateTime
的值时,它应该转换并构建行
预期产出:
Lisa|Rimpell|2017-02-05 02:40:00|2017-02-05 07:40:00||
答案 0 :(得分:0)
在XSLT 2.0中,您可以执行以下操作:
brew doctor
接收:
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xmlns:my="http://www.example.com/my"
exclude-result-prefixes="xs my">
<xsl:output method="xml" indent="yes" version="1.0" encoding="UTF-8" />
<xsl:strip-space elements="*" />
<xsl:function name="my:adjust-dateTime">
<xsl:param name="dateTime"/>
<xsl:variable name="UTC" select="adjust-dateTime-to-timezone($dateTime, xs:dayTimeDuration('PT0H'))"/>
<xsl:sequence select="format-dateTime($UTC, '[Y0001]-[M01]-[D01] [H01]:[m01]:[s01]')" />
</xsl:function>
<xsl:template match="/items">
<Response>
<Data>
<Rows>
<xsl:for-each select="item">
<Row Action="ADD">
<xsl:value-of select="lastname, firstname, my:adjust-dateTime(checkintime), my:adjust-dateTime(chekouttime), address" separator="|" />
</Row>
</xsl:for-each>
</Rows>
</Data>
</Response>
</xsl:template>
</xsl:stylesheet>