XSLT1.0:分组和删除重复项
我有一个xml分组挑战,我需要对其分组并删除重复项,如下所示:XSLT1.0:分组和删除重复项,xslt,muenchian-grouping,Xslt,Muenchian Grouping,我有一个xml分组挑战,我需要对其分组并删除重复项,如下所示: <Person> <name>John</name> <date>June12</date> <workTime taskID=1>34</workTime> <workTime taskID=1>35</workTime> <workTime taskID=2>12</workTime> </P
<Person>
<name>John</name>
<date>June12</date>
<workTime taskID=1>34</workTime>
<workTime taskID=1>35</workTime>
<workTime taskID=2>12</workTime>
</Person>
<Person>
<name>John</name>
<date>June13</date>
<workTime taskID=1>21</workTime>
<workTime taskID=2>11</workTime>
<workTime taskID=2>14</workTime>
</Person>
??
看来我需要两级钥匙 此转换:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:key name="kwrkTimeByNameTask" match="workTime"
use="concat(../name, '+', @taskID)"/>
<xsl:key name="kDateByName" match="date"
use="../name"/>
<xsl:key name="kwrkTimeByNameTaskDate" match="workTime"
use="concat(../name, '+', @taskID, '+', ../date)"/>
<xsl:template match="/">
<xsl:for-each select=
"*/*/workTime
[generate-id()
=
generate-id(key('kwrkTimeByNameTask',
concat(../name, '+', @taskID)
)[1]
)
]
">
<xsl:sort select="../name"/>
<xsl:sort select="@taskID" data-type="number"/>
<xsl:variable name="vcurTaskId" select="@taskID"/>
<Person>
<name><xsl:value-of select="../name"/></name>
<taskID><xsl:value-of select="@taskID"/></taskID>
<xsl:for-each select=
"key('kDateByName', ../name)
[key('kwrkTimeByNameTaskDate',
concat(../name, '+', current()/@taskID, '+', .)
)
]
">
<workTime>
<date><xsl:value-of select="."/></date>
<time>
<xsl:value-of select=
"key('kwrkTimeByNameTaskDate',
concat(../name, '+', $vcurTaskId, '+', .)
)"/>
</time>
</workTime>
</xsl:for-each>
</Person>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
<Person>
<name>John</name>
<taskID>1</taskID>
<workTime>
<date>June12</date>
<time>34</time>
</workTime>
<workTime>
<date>June13</date>
<time>21</time>
</workTime>
</Person>
<Person>
<name>John</name>
<taskID>2</taskID>
<workTime>
<date>June12</date>
<time>12</time>
</workTime>
<workTime>
<date>June13</date>
<time>11</time>
</workTime>
</Person>
是:
对于该
名称
的每个日期
,例如该名称
的工作时间
,日期
和@taskID
(外部
的当前工作时间
)存在,执行本
指令正文中的任何操作 XSLT中的分组通常使用一种称为Muenchian方法的方法来完成。在这里找到更多数据:只是为了好玩,另一个解决方案有两个键。此样式表:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:key name="kWorkTimeByName-TaskID" match="workTime"
use="concat(../name,'++',@taskID)"/>
<xsl:key name="kWorkTimeByName-Date-TaskID" match="workTime"
use="concat(../name,'++',../date,'++',@taskID)"/>
<xsl:template match="/">
<xsl:variable name="vAllWorkTime" select="*/*/workTime"/>
<result>
<xsl:for-each select="$vAllWorkTime
[count(.|key('kWorkTimeByName-TaskID',
concat(../name,'++',@taskID))[1])=1]">
<xsl:sort select="../name"/>
<xsl:sort select="@taskID" data-type="number"/>
<Person>
<xsl:copy-of select="../name"/>
<taskID>
<xsl:value-of select="@taskID"/>
</taskID>
<xsl:for-each select="$vAllWorkTime
[count(.|key('kWorkTimeByName-Date-TaskID',
concat(current()/../name,'++',
../date,'++',current()/@taskID))[1])=1]">
<xsl:sort select="../date"/>
<xsl:copy>
<xsl:copy-of select="../date"/>
<time>
<xsl:value-of select="."/>
</time>
</xsl:copy>
</xsl:for-each>
</Person>
</xsl:for-each>
</result>
</xsl:template>
</xsl:stylesheet>
输出:
<result>
<Person>
<name>John</name>
<taskID>1</taskID>
<workTime>
<date>June12</date>
<time>34</time>
</workTime>
<workTime>
<date>June13</date>
<time>21</time>
</workTime>
</Person>
<Person>
<name>John</name>
<taskID>2</taskID>
<workTime>
<date>June12</date>
<time>12</time>
</workTime>
<workTime>
<date>June13</date>
<time>11</time>
</workTime>
</Person>
</result>
约翰
1.
6月12日
34
6月13日
21
约翰
2.
6月12日
12
6月13日
11
这个问题既有趣又困难(+1)。请参阅我的答案,以获得一个高效且简短的解决方案。您能否解释一下您的解决方案的设计。它看起来又短又漂亮,但我想从中尽可能多地学习。Thanks@Daniel:我添加了一个解释。我想知道使用一个简单的慕尼黑分组,然后检查前面的兄弟姐妹是否重复是否更好。这是一个好的解决方案吗?@Daniel:如果我们有钥匙的力量,那么为什么要恢复到兄弟姐妹比较?我想知道使用简单的Muenchian分组,然后检查前面的兄弟姐妹是否有重复是否更好。这是一个好的解决方案吗?concat中的“++”、“+”或“无”之间有什么区别?@Daniel:关于分隔符字符串:它必须是一个不能在任何一个键中的字符串,所以请把Dimitre comment大部分当作一个玩笑;)关于分组:按名称和任务分组,然后按日期分组(因此键变成了名称、任务和日期);如果您使用最后一个当前组的所有节点或仅第一个组的所有节点,则算法逻辑没有差异。
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:key name="kwrkTimeByNameTask" match="workTime"
use="concat(../name, '+', @taskID)"/>
<xsl:key name="kDateByName" match="date"
use="../name"/>
<xsl:key name="kwrkTimeByNameTaskDate" match="workTime"
use="concat(../name, '+', @taskID, '+', ../date)"/>
<xsl:template match="/">
<xsl:for-each select=
"*/*/workTime
[generate-id()
=
generate-id(key('kwrkTimeByNameTask',
concat(../name, '+', @taskID)
)[1]
)
]
">
<xsl:sort select="../name"/>
<xsl:sort select="@taskID" data-type="number"/>
<xsl:variable name="vcurTaskId" select="@taskID"/>
<Person>
<name><xsl:value-of select="../name"/></name>
<taskID><xsl:value-of select="@taskID"/></taskID>
<xsl:for-each select=
"key('kDateByName', ../name)
[key('kwrkTimeByNameTaskDate',
concat(../name, '+', current()/@taskID, '+', .)
)
]
">
<workTime>
<date><xsl:value-of select="."/></date>
<time>
<xsl:value-of select=
"key('kwrkTimeByNameTaskDate',
concat(../name, '+', $vcurTaskId, '+', .)
)"/>
</time>
</workTime>
</xsl:for-each>
</Person>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
<t>
<Person>
<name>John</name>
<date>June12</date>
<workTime taskID="1">34</workTime>
<workTime taskID="1">35</workTime>
<workTime taskID="2">12</workTime>
</Person>
<Person>
<name>John</name>
<date>June13</date>
<workTime taskID="1">21</workTime>
<workTime taskID="2">11</workTime>
<workTime taskID="2">14</workTime>
</Person>
</t>
<Person>
<name>John</name>
<taskID>1</taskID>
<workTime>
<date>June12</date>
<time>34</time>
</workTime>
<workTime>
<date>June13</date>
<time>21</time>
</workTime>
</Person>
<Person>
<name>John</name>
<taskID>2</taskID>
<workTime>
<date>June12</date>
<time>12</time>
</workTime>
<workTime>
<date>June13</date>
<time>11</time>
</workTime>
</Person>
<xsl:for-each select=
"key('kDateByName', ../name)
[key('kwrkTimeByNameTaskDate',
concat(../name, '+', current()/@taskID, '+', .)
)
]
">
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:key name="kWorkTimeByName-TaskID" match="workTime"
use="concat(../name,'++',@taskID)"/>
<xsl:key name="kWorkTimeByName-Date-TaskID" match="workTime"
use="concat(../name,'++',../date,'++',@taskID)"/>
<xsl:template match="/">
<xsl:variable name="vAllWorkTime" select="*/*/workTime"/>
<result>
<xsl:for-each select="$vAllWorkTime
[count(.|key('kWorkTimeByName-TaskID',
concat(../name,'++',@taskID))[1])=1]">
<xsl:sort select="../name"/>
<xsl:sort select="@taskID" data-type="number"/>
<Person>
<xsl:copy-of select="../name"/>
<taskID>
<xsl:value-of select="@taskID"/>
</taskID>
<xsl:for-each select="$vAllWorkTime
[count(.|key('kWorkTimeByName-Date-TaskID',
concat(current()/../name,'++',
../date,'++',current()/@taskID))[1])=1]">
<xsl:sort select="../date"/>
<xsl:copy>
<xsl:copy-of select="../date"/>
<time>
<xsl:value-of select="."/>
</time>
</xsl:copy>
</xsl:for-each>
</Person>
</xsl:for-each>
</result>
</xsl:template>
</xsl:stylesheet>
<result>
<Person>
<name>John</name>
<taskID>1</taskID>
<workTime>
<date>June12</date>
<time>34</time>
</workTime>
<workTime>
<date>June13</date>
<time>21</time>
</workTime>
</Person>
<Person>
<name>John</name>
<taskID>2</taskID>
<workTime>
<date>June12</date>
<time>12</time>
</workTime>
<workTime>
<date>June13</date>
<time>11</time>
</workTime>
</Person>
</result>