我是XSLT的新手,所以我希望我能在这里得到一些帮助。
我试图转换以下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>
到
<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>
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(.,$separator)">
<ArticleNo><xsl:value-of select="../SKU"/>.<xsl:value-of select="position()" /></ArticleNo>
</xsl:for-each>
</xsl:template>
<xsl:template match="SKU" />
</xsl:stylesheet>
如何正确访问SKU?
答案 0 :(得分:2)
使用<item>
<layer-list>
<item>
<shape>
<gradient android:angle="90" android:endColor="#ffffff" android:startColor="#ffffff" android:type="linear" />
<stroke android:width="0.33dp" android:color="#0fb1fa" />
<corners android:radius="0dp" />
<padding android:bottom="3dp" android:left="3dp" android:right="3dp" android:top="3dp" />
</shape>
</item>
<item android:right="5dp">
<bitmap android:gravity="center_vertical|right" android:src="@drawable/arrow_down_gray" />
</item>
</layer-list>
</item>
可以通过在共享XSL中进行小调整来实现输出。 XSLT 2.0
值可以添加到变量中,用于格式化所需的输出。
<SKU>
输出
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="Meta" name="tokenize">
<xsl:param name="separator" select="', '" />
<xsl:variable name="skuValue" select="../SKU" />
<xsl:for-each select="tokenize(.,$separator)">
<ArticleNo>
<xsl:value-of select="$skuValue" />
.
<xsl:value-of select="position()" />
</ArticleNo>
</xsl:for-each>
</xsl:template>
<xsl:template match="SKU" />
</xsl:stylesheet>