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XSLT 将两个兄弟元素内的元素分组

转载 作者:行者123 更新时间:2023-12-05 01:19:54 24 4
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我正在努力解决这个问题,我认为解释它的最简单方法就是在下面向您展示。我看过this但它并不总是适用,因为我在末尾也有独立的项目可以匹配。

看似棘手的部分是Whatever3到Whatever6,然后是Whatever7和Whatever8,最后是Whatever9的新位置——它们需要分组并保持原来的顺序。 (忽略我的命名,没有办法使用xsl:sort)

我考虑过 xsl:for-each 和 xsl:if 在里面,但问题是你不能保证有多少“组”和“非组”项。

谢谢!

XML

<root>
<item>
<position>1</position>
<label>Whatever1</label>
</item>
<item>
<position>2</position>
<label>Whatever2</label>
</item>
<item>
<position>3</position>
<label>Whatever3</label>
<marker id="unique1">start_group</marker>
</item>
<item>
<position>4</position>
<label>Whatever4</label>
</item>
<item>
<position>5</position>
<label>Whatever5</label>
</item>
<item>
<position>6</position>
<label>Whatever6</label>
<marker>last_in_group</marker>
</item>
<item>
<position>7</position>
<label>Whatever7</label>
<marker id="unique2">start_group</marker>
</item>
<item>
<position>8</position>
<label>Whatever8</label>
<marker>last_in_group</marker>
</item>
<item>
<position>9</position>
<label>Whatever9</label>
</item>
</root>

结果

<structure>
<item>
<position>1</position>
<label>Whatever1</label>
</item>
<item>
<position>2</position>
<label>Whatever2</label>
</item>
<group position="3" id="unique1">
<item>
<position>1</position>
<label>Whatever3</label>
</item>
<item>
<position>2</position>
<label>Whatever4</label>
</item>
<item>
<position>3</position>
<label>Whatever5</label>
</item>
<item>
<position>4</position>
<label>Whatever6</label>
</item>
</group>
<group position="4" id="uniqueid2">
<item>
<position>1</position>
<label>Whatever7</label>
</item>
<item>
<position>2</position>
<label>Whatever8</label>
</item>
</group>
<item>
<position>**5**</position>
<label>Whatever9</label>
</item>
</structure>

======================

这是我目前遇到的问题,我唯一的问题(除了困惑之外)是 Whatever4 和 Whatever5 出现在组外。

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>

<xsl:template match="root">
<structure>
<xsl:apply-templates select="item[not(marker)] | item[marker='start_group']"/>
</structure>
</xsl:template>

<xsl:template match="item[marker='start_group']">
<group>
<xsl:variable name="currentPosition" select="number(position/text())"/>
<xsl:variable name="lastGroup" select="count(following-sibling::*[local-name() = 'item' and marker='last_in_group'][1]/preceding-sibling::*) + 1"/>

<xsl:attribute name="position">
<xsl:value-of select="$currentPosition"/>
</xsl:attribute>
<xsl:attribute name="id">
<xsl:value-of select="marker/@id"/>
</xsl:attribute>

<item>
<position><xsl:value-of select="number(position/text()) - $currentPosition + 1"/></position>
<label><xsl:value-of select="label/text()"/></label>
</item>

<!-- position() gets reset in for-loop, so need to adjust with outer position -->
<xsl:for-each select="following-sibling::item[(position() + $currentPosition) &lt;= $lastGroup]">
<item>
<position><xsl:value-of select="number(position/text()) - $currentPosition + 1"/></position>
<label><xsl:value-of select="label/text()"/></label>
</item>
</xsl:for-each>
</group>
</xsl:template>

<xsl:template match="item[not(marker)]">
<item>
<position><xsl:value-of select="position/text()"/></position>
<label><xsl:value-of select="label/text()"/></label>
</item>
</xsl:template>

</xsl:stylesheet>

最佳答案

我。 XSLT 1.0 解决方案:

这个转换:

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>

<xsl:key name="kFollowing"
match="item[not(marker[. = 'start_group'])
and
preceding-sibling::*[marker][1]/marker = 'start_group'
]"
use="generate-id(preceding-sibling::*
[marker[. = 'start_group']]
[1])"/>

<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>

<xsl:template match="item[marker[. = 'start_group']]">
<group position="{1 +count(preceding-sibling::*[. = 'start_group'])}"
id="{marker/@id}">
<xsl:copy-of select=".|key('kFollowing', generate-id())"/>
</group>
</xsl:template>

<xsl:template match=
"item[not(marker[. = 'start_group'])
and
preceding-sibling::*[marker][1]/marker = 'start_group'
]"/>
</xsl:stylesheet>

应用于提供的 XML 文档时:

<root>
<item>
<position>1</position>
<label>Whatever1</label>
</item>
<item>
<position>2</position>
<label>Whatever2</label>
</item>
<item>
<position>3</position>
<label>Whatever3</label>
<marker id="unique1">start_group</marker>
</item>
<item>
<position>4</position>
<label>Whatever4</label>
</item>
<item>
<position>5</position>
<label>Whatever5</label>
</item>
<item>
<position>6</position>
<label>Whatever6</label>
<marker>last_in_group</marker>
</item>
<item>
<position>7</position>
<label>Whatever7</label>
<marker id="unique2">start_group</marker>
</item>
<item>
<position>8</position>
<label>Whatever8</label>
<marker>last_in_group</marker>
</item>
<item>
<position>9</position>
<label>Whatever9</label>
</item>
</root>

产生想要的、正确的结果:

<root>
<item>
<position>1</position>
<label>Whatever1</label>
</item>
<item>
<position>2</position>
<label>Whatever2</label>
</item>
<group position="1" id="unique1">
<item>
<position>3</position>
<label>Whatever3</label>
<marker id="unique1">start_group</marker>
</item>
<item>
<position>4</position>
<label>Whatever4</label>
</item>
<item>
<position>5</position>
<label>Whatever5</label>
</item>
<item>
<position>6</position>
<label>Whatever6</label>
<marker>last_in_group</marker>
</item>
</group>
<group position="1" id="unique2">
<item>
<position>7</position>
<label>Whatever7</label>
<marker id="unique2">start_group</marker>
</item>
<item>
<position>8</position>
<label>Whatever8</label>
<marker>last_in_group</marker>
</item>
</group>
<item>
<position>9</position>
<label>Whatever9</label>
</item>
</root>

二。 XSLT 2.0 解决方案:

<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema" exclude-result-prefixes="xs">
<xsl:output omit-xml-declaration="yes" indent="yes"/>

<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>

<xsl:template match="/*">
<root>
<xsl:for-each-group select="item" group-starting-with=
"*[marker eq 'start_group'
or
not(marker)
and
preceding-sibling::*[marker][1]/marker eq 'last_in_group'
]
">
<xsl:choose>
<xsl:when test="current-group()[1]/marker">
<group position=
"{1 +count(current-group()[1]
/preceding-sibling::*
[marker = 'start_group'])}"
id="{marker/@id}">
<xsl:apply-templates select="current-group()"/>
</group>
</xsl:when>
<xsl:otherwise>
<xsl:apply-templates select="current-group()"/>
</xsl:otherwise>
</xsl:choose>
</xsl:for-each-group>
</root>
</xsl:template>
</xsl:stylesheet>

当这个 XSLT 2.0 转换应用到同一个 XML 文档(上图)时,会产生相同的正确结果

关于XSLT 将两个兄弟元素内的元素分组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11145979/

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