我是xsl转换的新手,循环子元素时遇到问题。我正在尝试从IN.A.Number1获取元素值树。我的xsd看起来像:
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" version="1.0">
<xs:element name="Document">
<xs:complexType>
<xs:all>
<xs:element name="IN">
<xs:complexType>
<xs:all>
<xs:element name="A" minOccurs="0">
<xs:complexType>
<xs:all>
<xs:element name="Number1" type="Integer" minOccurs="0"/>
<xs:element name="Number2" type="Integer" minOccurs="0"/>
<xs:element name="Number3" type="Integer" minOccurs="0"/>
</xs:all>
</xs:complexType>
</xs:element>
<xs:element name="B" minOccurs="0">
<xs:complexType>
<xs:all>
<xs:element name="Date1" type="Date" minOccurs="0"/>
<xs:element name="String1" type="String" minOccurs="0"/>
</xs:all>
</xs:complexType>
</xs:element>
<xs:element name="Arraydate" minOccurs="0">
<xs:complexType>
<xs:sequence>
<xs:element name="data_type" type="xs:string" fixed="date"/>
<xs:element ref="date_format"/>
<xs:element name="element" type="String Array" maxOccurs="10" minOccurs="0"/>
</xs:sequence>
</xs:complexType>
</xs:element>
</xs:all>
</xs:complexType>
</xs:element>
</xs:all>
</xs:complexType>
</xs:element>
</xs:schema>
输出应如下所示:
<?xml version="1.0"?>
<mapping name="IN_A_Number1" original name= "B.Date1"/>
.
.
<mapping name="IN_B_Date1 original name= "B.Date1"" />
<mapping name="IN_B_Date1 original name= "B.Date1""/>
.
.
.
</mappings>
到目前为止,我已经获得了第一级的元素,但是不知道如何编写元素的子元素。
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="xs"
version="1.0">
<xsl:template match="/">
<xsl:copy>
<xsl:for-each select="*">
<mappings>
<xsl:variable name="pom" select="."></xsl:variable>
<xsl:element name="mapping">
<xsl:attribute name="name">
<xsl:value-of select="$pom" />
</xsl:attribute>
<xsl:apply-templates
select="//xs:element[@name = 'IN']/ xs:complexType/xs:all/xs:element"/>
</xsl:element>
</mappings>
</xsl:for-each>
</xsl:copy>
</xsl:template>
<xsl:template match="xs:element">
<mappings>
<xsl:element name="mapping">
<xsl:attribute name="name">
<xsl:call-template name = "xmlToJava">
<xsl:with-param name="xmlName" select = "concat(./@name,'.',//xs:element[@name = 'IN']/ xs:complexType/xs:all/xs:element[@name='.'])"/>
</xsl:call-template>
</xsl:attribute>
</xsl:element>
</mappings>
</xsl:template>
<xsl:template name="xmlToJava">
<xsl:param name="xmlName" />
<xsl:value-of select="translate($xmlName, '.','_')" />
</xsl:template>
</xsl:stylesheet>
谢谢您的帮助:)
答案 0 :(得分:0)
您可以使用此XSLT-1.0模板创建所有xs:element
叶节点的列表:
<xsl:template match="/xs:schema" >
<xsl:element name="mappings">
<xsl:for-each select=".//xs:element[not(descendant::xs:element)]">
<mapping>
<xsl:attribute name="name">
<xsl:for-each select="ancestor-or-self::xs:element">
<xsl:value-of select="@name" />
<xsl:if test="descendant::xs:element">
<xsl:text>_</xsl:text>
</xsl:if>
</xsl:for-each>
</xsl:attribute>
<xsl:attribute name="originalName">
<xsl:for-each select="ancestor-or-self::xs:element[3 > position()]">
<xsl:value-of select="@name" />
<xsl:if test="descendant::xs:element">
<xsl:text>.</xsl:text>
</xsl:if>
</xsl:for-each>
</xsl:attribute>
</mapping>
</xsl:for-each>
</xsl:element>
</xsl:template>
其输出是:
<?xml version="1.0"?>
<mappings xmlns:xs="http://www.w3.org/2001/XMLSchema">
<mapping name="Document_IN_A_Number1" originalName="A.Number1"/>
<mapping name="Document_IN_A_Number2" originalName="A.Number2"/>
<mapping name="Document_IN_A_Number3" originalName="A.Number3"/>
<mapping name="Document_IN_B_Date1" originalName="B.Date1"/>
<mapping name="Document_IN_B_String1" originalName="B.String1"/>
<mapping name="Document_IN_Arraydate_data_type" originalName="Arraydate.data_type"/>
<mapping name="Document_IN_Arraydate_" originalName="Arraydate."/>
<mapping name="Document_IN_Arraydate_element" originalName="Arraydate.element"/>
</mappings>