我有这样的输入文件
<items> <item> <id>5</id> <name>Harry</name> <age>18</age> <color>blue</color> </item> <item> <id>5</id> <name>Harry</name> <age>18</age> <fav_food>pizza</fav_food> </item> <item> <id>5</id> <name>Harry</name> <age>18</age> <is_single>true</is_single> </item> </items>我想将元素分组，以便文件看起来像这样
<items> <item> <id>5</id> <name>Harry</name> <age>18</age> <color>blue</color> <fav_food>pizza</fav_food> <is_single>true</is_single> </item> </items>编辑 - 在此链接之后进行XSLT转换（ 当ID相同时，XSLT合并数据 ）但这不会打印任何内容。
这是我的转换（使用XSLT 2.0） -
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output indent="yes" omit-xml-declaration="yes" /> <xsl:strip-space elements="*" /> <xsl:template match="items"> <xsl:apply-templates select="@*|node()"/> </xsl:template> <xsl:template match="item"> <xsl:apply-templates select="@*" /> <xsl:for-each-group select="item" group-by="id"> <item> <id><xsl:value-of select="current-grouping-key()"/></id> <xsl:apply-templates select="current-group()" /> </item> </xsl:for-each-group> </xsl:template> <xsl:template match="@*|node()[not(self::*)]"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> <xsl:template match="*"> <xsl:element name="{local-name()}"> <xsl:apply-templates select="@*|node()"/> </xsl:element> </xsl:template> </xsl:stylesheet>I have an input file like this
<items> <item> <id>5</id> <name>Harry</name> <age>18</age> <color>blue</color> </item> <item> <id>5</id> <name>Harry</name> <age>18</age> <fav_food>pizza</fav_food> </item> <item> <id>5</id> <name>Harry</name> <age>18</age> <is_single>true</is_single> </item> </items>I'd like to group up the elements so that the file looks like this
<items> <item> <id>5</id> <name>Harry</name> <age>18</age> <color>blue</color> <fav_food>pizza</fav_food> <is_single>true</is_single> </item> </items>EDIT - Made an XSLT transform following this link (XSLT Consolidating data when ID is the same) but this just doesn't print anything.
Here is my transform (using XSLT 2.0) -
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output indent="yes" omit-xml-declaration="yes" /> <xsl:strip-space elements="*" /> <xsl:template match="items"> <xsl:apply-templates select="@*|node()"/> </xsl:template> <xsl:template match="item"> <xsl:apply-templates select="@*" /> <xsl:for-each-group select="item" group-by="id"> <item> <id><xsl:value-of select="current-grouping-key()"/></id> <xsl:apply-templates select="current-group()" /> </item> </xsl:for-each-group> </xsl:template> <xsl:template match="@*|node()[not(self::*)]"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> <xsl:template match="*"> <xsl:element name="{local-name()}"> <xsl:apply-templates select="@*|node()"/> </xsl:element> </xsl:template> </xsl:stylesheet>最满意答案
您遇到的主要问题是您有一个匹配item元素的模板：
<xsl:template match="item">但在此范围内，您有一个xsl：for-each-group ，它也会查找item元素
<xsl:for-each-group select="item" group-by="id">这意味着您正在寻找属于其他项元素的子元素的项元素，其中XML中没有。 我认为您需要将前两个模板合并为一个：
<xsl:template match="items"> <xsl:for-each-group select="item" group-by="id"> <item> ....有一点不清楚， 项目元素中的元素将被重复。 它会永远是身份，姓名和年龄吗？ 尽管这里最好是灵活的，并假设只有id才是常用元素。 我还假设如果重复两个元素（如名称 ），它将始终具有相同的值。
因此，要获得所有其他不同的非id元素，您可以对元素名称进行更多分组
<xsl:for-each-group select="current-group()/*[not(self::id)]" group-by="local-name()">然后，在这个循环中你可以输出元素。
试试这个XSLT
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output indent="yes" omit-xml-declaration="yes" /> <xsl:strip-space elements="*" /> <xsl:template match="items"> <xsl:for-each-group select="item" group-by="id"> <item> <id><xsl:value-of select="current-grouping-key()"/></id> <xsl:for-each-group select="current-group()/*[not(self::id)]" group-by="local-name()"> <xsl:apply-templates select="self::*" /> </xsl:for-each-group> </item> </xsl:for-each-group> </xsl:template> <xsl:template match="@*|node()[not(self::*)]"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> <xsl:template match="*"> <xsl:element name="{local-name()}"> <xsl:apply-templates select="@*|node()"/> </xsl:element> </xsl:template> </xsl:stylesheet>应用于XML时，将输出以下内容
<item> <id>5</id> <name>Harry</name> <age>18</age> <color>blue</color> <fav_food>pizza</fav_food> <is_single>true</is_single> </item>The main problem you have is you have a template matching the item element:
<xsl:template match="item">But within this, you have an xsl:for-each-group that also looks for item elements
<xsl:for-each-group select="item" group-by="id">What this means is that you are looking for item elements that are children of other item elements, of which there are none in the XML. I think you need to combine the first two templates into one here:
<xsl:template match="items"> <xsl:for-each-group select="item" group-by="id"> <item> ....One thing that is not clear is what elements within item elements will be repeated. Will it always be the id, name and age? It's probably best to be flexible here though, and assume only id will be the common element. I will also assume if two elements are repeated (like name) it will always have the same value.
So, to get all the other distinct non-id elements, you could some more grouping on element name
<xsl:for-each-group select="current-group()/*[not(self::id)]" group-by="local-name()">Then, within this loop you can just output the element.
Try this XSLT
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output indent="yes" omit-xml-declaration="yes" /> <xsl:strip-space elements="*" /> <xsl:template match="items"> <xsl:for-each-group select="item" group-by="id"> <item> <id><xsl:value-of select="current-grouping-key()"/></id> <xsl:for-each-group select="current-group()/*[not(self::id)]" group-by="local-name()"> <xsl:apply-templates select="self::*" /> </xsl:for-each-group> </item> </xsl:for-each-group> </xsl:template> <xsl:template match="@*|node()[not(self::*)]"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> <xsl:template match="*"> <xsl:element name="{local-name()}"> <xsl:apply-templates select="@*|node()"/> </xsl:element> </xsl:template> </xsl:stylesheet>When applied to your XML, the following is output
<item> <id>5</id> <name>Harry</name> <age>18</age> <color>blue</color> <fav_food>pizza</fav_food> <is_single>true</is_single> </item>更多推荐
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