XSLT将同名兄弟节点的值合并/连接成单个节点

Question

输入xml

<catalog>
    <product id="1">
        <name>abc</name>
        <category>aaa</category>
        <category>bbb</category>
        <category>ccc</category>
    </product>
    <product id="2">
        <name>cde</name>
        <category>aaa</category>
        <category>bbb</category>
    </product>
</catalog>

预期输出xml

<products>
    <product>
        <id>1</id>
        <name>abc</name>
        <category>aaa,bbb,ccc</category>
    </product>
    <product>
        <id>2</id>
        <name>cde</name>
        <category>aaa,bbb</category>
    </product>
</products>

用于转换的XSLT

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" indent="yes"/>
    <xsl:template match="/catalog">
        <products>
            <xsl:for-each select="product">
                <product>
                    <id><xsl:value-of select="@id"/></id>
                    <name><xsl:value-of select="name"/></name>
                    <category><xsl:value-of select="category" /></category>
                </product>
            </xsl:for-each>
        </products>
    </xsl:template>
</xsl:stylesheet>

实际输出xml :(

<products>
    <product>
        <id>1</id>
        <name>abc</name>
        <category>aaa</category>
    </product>
    <product>
        <id>2</id>
        <name>cde</name>
        <category>aaa</category>
    </product>
</products>

在每个'产品'下通过名称'category'循环遍历所有兄弟节点以及合并/连接到用逗号分隔的单个节点所需的代码.每个产品的"类别"数量各不相同,因此计数未知.

Answer 1

使用此处定义的这个方便的连接调用模板,这变得如此简单:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" indent="yes"/>
    <xsl:template match="/catalog">
        <products>
            <xsl:for-each select="product">
                <product>
                    <id>
                        <xsl:value-of select="@id"/>
                    </id>
                    <name>
                        <xsl:value-of select="name"/>
                    </name>
                    <category>
                        <xsl:call-template name="join">
                            <xsl:with-param name="list" select="category" />
                            <xsl:with-param name="separator" select="','" />
                        </xsl:call-template>
                    </category>
                </product>
            </xsl:for-each>
        </products>
    </xsl:template>

    <xsl:template name="join">
        <xsl:param name="list" />
        <xsl:param name="separator"/>

        <xsl:for-each select="$list">
            <xsl:value-of select="." />
            <xsl:if test="position() != last()">
                <xsl:value-of select="$separator" />
            </xsl:if>
        </xsl:for-each>
    </xsl:template>

</xsl:stylesheet>

输出:

<products>
  <product>
    <id>1</id>
    <name>abc</name>
    <category>aaa,bbb,ccc</category>
  </product>
  <product>
    <id>2</id>
    <name>cde</name>
    <category>aaa,bbb</category>
  </product>
</products>

Answer 2

在XSLT 2.0中,您只需对代码进行一次小的更改:

<category><xsl:value-of select="category" separator=","/></category>

请注意,如果您需要XSLT 1.0解决方案,最好这样说.某些环境中的某些人被卡在1.0上,但很多人都没有.