我是 XSLT 的新手。我想复制一个特定节点的所有前面的节点,我尝试使用复制和前面,但输出不是我所期望的..
源示例 XML:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<MetaData>
<Owner>adad</Owner>
<date>2013-1-12</date>
</MetaData>
<Orders>
<Order name="123">
<OrderName>Order1</OrderName>
<OrderNo>1</OrderNo>
</Order>
<Order name="1234">
<OrderName>Order2</OrderName>
<OrderNo>2</OrderNo>
</Order>
<Order>
<OrderName>Order3</OrderName>
<OrderNo>3</OrderNo>
</Order>
</Orders>
<tail>1111</tail>
</root>
Xsl:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:output method="xml" indent="yes"/>
<xsl:template match = "Orders" name="split">
<xsl:for-each select="Order">
<xsl:if test="position() = 2">
<xsl:copy-of select="preceding::node()"/>
<xsl:copy-of select="."/>
</xsl:if>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
输出:
<?xml version="1.0" encoding="UTF-8"?>
adad
2013-1-12
<MetaData>
<Owner>adad</Owner>
<date>2013-1-12</date>
</MetaData>
<Owner>adad</Owner>adad
<date>2013-1-12</date>2013-1-12
<Order name="123">
<OrderName>Order1</OrderName>
<OrderNo>1</OrderNo>
</Order>
<OrderName>Order1</OrderName>Order1
<OrderNo>1</OrderNo>1
<Order name="1234">
<OrderName>Order2</OrderName>
<OrderNo>2</OrderNo>
</Order>
1111
预期的 XML:
<?xml version="1.0" encoding="UTF-8"?>
<root>
<MetaData>
<Owner>adad</Owner>
<date>2013-1-12</date>
</MetaData>
<Orders>
<Order name="123">
<OrderName>Order1</OrderName>
<OrderNo>1</OrderNo>
</Order>
<Order name="1234">
<OrderName>Order2</OrderName>
<OrderNo>2</OrderNo>
</Order>
<Orders>
</root>
<MetaData>
看起来像和的后代<Order>
被复制了三遍。为什么<tail>
节点丢失了元素名称?
谁能帮我解决这个问题?谢谢