XSLT：獲取文本標記模板中的先前節點值

Question

我是XSLT的新手，所以我希望這里能有所幫助。

我正在嘗試轉換以下XML

<?xml version="1.0"?><?xml-stylesheet type="text/xsl"?>
    <OrderLineItems>
    <OrderLineItem>
        <SKU>60</SKU>
        <Meta>Topic: one, Topic: two, Topic: three, Topic: four</Meta>
    </OrderLineItem>
    <OrderLineItem>
        <SKU>70</SKU>
        <Meta>Topic: one, Topic: two, Topic: three, Topic: four</Meta>
    </OrderLineItem>
    </OrderLineItems>

至

<ArticleNo>60.1</ArticleNo>
<ArticleNo>60.2</ArticleNo>
<ArticleNo>60.3</ArticleNo>
<ArticleNo>60.4</ArticleNo>

<ArticleNo>70.1</ArticleNo>
<ArticleNo>70.2</ArticleNo>
<ArticleNo>70.3</ArticleNo>
<ArticleNo>70.4</ArticleNo>

與以下xslt不起作用

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:template match="Meta" name="tokenize">
        <xsl:param name="separator" select="', '" />
        <xsl:for-each select="tokenize(.,$separator)">
                            <ArticleNo><xsl:value-of select="../SKU"/>.<xsl:value-of select="position()" /></ArticleNo>
        </xsl:for-each>
    </xsl:template>
  <xsl:template match="SKU" />

</xsl:stylesheet>

如何正確訪問SKU？

Answer 1

使用XSLT 2.0 ，可以通過對共享XSL進行小的調整來實現輸出。 可以將<SKU>值添加到變量中，並用於格式化所需的輸出。

<xsl:stylesheet version="2.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:template match="Meta" name="tokenize">
        <xsl:param name="separator" select="', '" />
        <xsl:variable name="skuValue" select="../SKU" />
        <xsl:for-each select="tokenize(.,$separator)">
            <ArticleNo>
                <xsl:value-of select="$skuValue" />
                .
                <xsl:value-of select="position()" />
            </ArticleNo>
        </xsl:for-each>
    </xsl:template>
    <xsl:template match="SKU" />
</xsl:stylesheet>

輸出量

<ArticleNo>60.1</ArticleNo>
<ArticleNo>60.2</ArticleNo>
<ArticleNo>60.3</ArticleNo>
<ArticleNo>60.4</ArticleNo>
<ArticleNo>70.1</ArticleNo>
<ArticleNo>70.2</ArticleNo>
<ArticleNo>70.3</ArticleNo>
<ArticleNo>70.4</ArticleNo>