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在XSLT中与可选节点匹配并合并

[英]match and merge in XSLT with optional nodes

 原文  2016-08-17 10:18:08  6   2    xml/ xslt/ xslt-1.0

问题描述

I have a sample XML message which contains multiple parent node. 我有一个示例XML消息，其中包含多个父节点。 The requirement is if the two parent node are same, merge the child node. 要求是，如果两个父节点相同，则合并子节点。 This works fine when all the nodes are present but doesn't work when the optional node is absent 当所有节点都存在时，此方法工作正常，但在不存在可选节点时，则无效

Sample Message:1 With Optional Nodes present 样本消息：1带有可选节点 

<document>
<body>
    <party>
        <gtin>1000909090</gtin>
        <pos>
            <attrGroupMany name="temperatureInformation">
                <row>
                    <gtin>1000909090</gtin>
                    <attr name="temperatureCode">STORAGE</attr>
                    <attrQualMany name="temperature">
                        <value qual="FAH">10</value>
                        <value qual="CC">20</value>
                    </attrQualMany>
                    <attrGroupMany name="temperatureStats">   <!--  optional group -->
                        <row>
                            <attr name="StatsCode">CODE1</attr>
                        </row>
                        <row>
                            <attr name="StatsCode">CODE2</attr>
                        </row>
                    </attrGroupMany>
                </row>
                <row>
                    <attr name="temperatureCode">STORAGE</attr>
                    <attrQualMany name="temperature">
                        <value qual="FAH">10</value>
                        <value qual="CC">20</value>
                    </attrQualMany>
                    <attrGroupMany name="temperatureStats">    <!--  optional group -->
                        <row>
                            <attr name="StatsCode">CODE3</attr>
                        </row>
                        <row>
                            <attr name="StatsCode">CODE4</attr>
                        </row>
                    </attrGroupMany>
                </row>
                <row>
                    <attr name="temperatureCode">HANDLING</attr>
                    <attrQualMany name="temperature">
                        <value qual="FAH">10</value>                    
                    </attrQualMany>
                    <attrGroupMany name="temperatureStats">    <!--  optional group -->
                        <row>
                            <attr name="StatsCode">CODE5</attr>
                        </row>
                        <row>
                            <attr name="StatsCode">CODE6</attr>
                        </row>
                    </attrGroupMany>
                </row>
                <row>
                    <attr name="temperatureCode">HANDLING</attr>
                    <attrGroupMany name="temperatureStats">      <!--  optional group -->
                        <row>
                            <attr name="StatsCode">CODE7</attr>
                        </row>
                        <row>
                            <attr name="StatsCode">CODE8</attr>
                        </row>
                    </attrGroupMany>
                </row>
            </attrGroupMany>
        </pos>
    </party>    
</body>
</document>

The below sample XSLT works fine to remove the duplicate from 'attrGroupMany name="temperatureInformation" ' 下面的示例XSLT可以很好地从'attrGroupMany name =“ temperatureInformation”'中删除重复项

XSLT used 使用XSLT 

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">



<xsl:key name="group" match="party/pos/attrGroupMany[@name = 'temperatureInformation']/row"
    use="concat(generate-id(ancestor::pos), '|', attr[@name = 'temperatureCode'], '|', attrQualMany[@name = 'temperature'])"/>

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

<xsl:template match="attrGroupMany[@name = 'temperatureInformation']">
    <xsl:copy>
        <xsl:apply-templates select="@*"/>
        <xsl:apply-templates select="row[generate-id() = generate-id(key('group', concat(generate-id(ancestor::pos), '|', attr[@name = 'temperatureCode'], '|', attrQualMany[@name = 'temperature']))[1])]"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="attrGroupMany[@name = 'temperatureStats']">
    <xsl:copy>
        <xsl:apply-templates select="@* | key('group', concat(generate-id(ancestor::pos), '|',../attr[@name = 'temperatureCode'], '|', ../attrQualMany[@name = 'temperature']))/attrGroupMany[@name = 'temperatureStats']/row"/>
    </xsl:copy>
</xsl:template>

</xsl:stylesheet>

Sample Message 2 which doesnt work with above XSLT is 与以上XSLT不兼容的示例消息2是 

<document>
<body>
    <party>
        <gtin>1000909090</gtin>
        <pos>
            <attrGroupMany name="temperatureInformation">
                <row>
                    <gtin>1000909090</gtin>
                    <attr name="temperatureCode">STORAGE</attr>
                    <attrQualMany name="temperature">
                        <value qual="FAH">10</value>
                        <value qual="CC">20</value>
                    </attrQualMany>
                </row>
                <row>
                    <attr name="temperatureCode">STORAGE</attr>
                    <attrQualMany name="temperature">
                        <value qual="FAH">10</value>
                        <value qual="CC">20</value>
                    </attrQualMany>
                    <attrGroupMany name="temperatureStats">    <!--  optional group -->
                        <row>
                            <attr name="StatsCode">CODE3</attr>
                        </row>
                        <row>
                            <attr name="StatsCode">CODE4</attr>
                        </row>
                    </attrGroupMany>
                </row>
                <row>
                    <attr name="temperatureCode">HANDLING</attr>
                    <attrQualMany name="temperature">
                        <value qual="FAH">10</value>                    
                    </attrQualMany>
                    <attrGroupMany name="temperatureStats">    <!--  optional group -->
                        <row>
                            <attr name="StatsCode">CODE5</attr>
                        </row>
                        <row>
                            <attr name="StatsCode">CODE6</attr>
                        </row>
                    </attrGroupMany>
                </row>
                <row>
                    <attr name="temperatureCode">HANDLING</attr>
                    <attrGroupMany name="temperatureStats">      <!--  optional group -->
                        <row>
                            <attr name="StatsCode">CODE7</attr>
                        </row>
                        <row>
                            <attr name="StatsCode">CODE8</attr>
                        </row>
                    </attrGroupMany>
                </row>
            </attrGroupMany>
        </pos>
    </party>    
</body>
</document>

Can someone please let me know how to handle optional node in match n merge 有人可以让我知道如何处理比赛n合并中的可选节点

The expected output for sample message 2 is 示例消息2的预期输出为 

<?xml version="1.0" encoding="UTF-8"?>
<document>
<body>
  <party>
     <gtin>1000909090</gtin>
     <pos>
        <attrGroupMany name="temperatureInformation">
           <row>
              <gtin>1000909090</gtin>
              <attr name="temperatureCode">STORAGE</attr>
              <attrQualMany name="temperature">
                 <value qual="FAH">10</value>
                 <value qual="CC">20</value>
              </attrQualMany>
              <attrGroupMany name="temperatureStats">                   
                 <row>
                    <attr name="StatsCode">CODE3</attr>
                 </row>
                 <row>
                    <attr name="StatsCode">CODE4</attr>
                 </row>
              </attrGroupMany>
           </row>
           <row>
              <attr name="temperatureCode">HANDLING</attr>
              <attrQualMany name="temperature">
                 <value qual="FAH">10</value>
              </attrQualMany>
              <attrGroupMany name="temperatureStats">
                 <row>
                    <attr name="StatsCode">CODE5</attr>
                 </row>
                 <row>
                    <attr name="StatsCode">CODE6</attr>
                 </row>
              </attrGroupMany>
           </row>
           <row>
              <attr name="temperatureCode">HANDLING</attr>
              <attrGroupMany name="temperatureStats">
                 <row>
                    <attr name="StatsCode">CODE7</attr>
                 </row>
                 <row>
                    <attr name="StatsCode">CODE8</attr>
                 </row>
              </attrGroupMany>
           </row>
        </attrGroupMany>
     </pos>
  </party>
</body>
</document>

Can someone please let me know how to handle optional node in match n merge. 有人可以让我知道如何在比赛n合并中处理可选节点。 Thanks 谢谢

2 个解决方案

解决方案1
 1 已采纳  2016-08-17 12:50:59

You haven't explained the logic behind what you are trying to do, but from looking at the XSLT from this question, and previous questions too, you are grouping the attrGroupMany[@name = 'temperatureInformation']/row elements but a concatenation of the ancestor pos together with "temperatureCode" and "temperature". 您尚未解释要执行的操作背后的逻辑，但是通过查看此问题以及以前的问题的XSLT，您可以将attrGroupMany[@name = 'temperatureInformation']/row元素分组，但是将祖先pos以及“ temperatureCode”和“ temperature”。

Then for each such distinct row it looks like you want to add in all the <attrGroupMany name="temperatureStats"> elements. 然后，对于每个这样的不同row ，您似乎都想添加所有<attrGroupMany name="temperatureStats">元素。 Having a template matching this element is not going to work if you say it is optional. 如果您说模板是可选的，那么与该元素匹配的模板将无法工作。 Instead, have a template matching the parent row and use that to select all the child elements from all elements in the keys. 而是要有一个与父row匹配的模板，并使用该模板从键中的所有元素中选择所有子元素。 

<xsl:template match="attrGroupMany[@name = 'temperatureInformation']/row">
    <xsl:copy>
        <xsl:apply-templates select="@*|node()[not(self::attrGroupMany[@name = 'temperatureStats'])]"/> 
        <attrGroupMany name="temperatureStats">
            <xsl:apply-templates select="key('group', concat(generate-id(ancestor::pos), '|', attr[@name = 'temperatureCode'], '|', attrQualMany[@name = 'temperature']))/attrGroupMany[@name = 'temperatureStats']/row"/>
        </attrGroupMany>
    </xsl:copy>
</xsl:template>

I am assuming all the StatsCode are distinct here. 我假设所有StatsCode在这里都是不同的。 If there could be duplicated, and you want to remove such duplicates, you need to say so in your question. 如果可能有重复项，并且您想删除此类重复项，则需要在问题中这样说。

Try this XSLT: 试试这个XSLT： 

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">

<xsl:output method="xml" indent="yes" />

<xsl:key name="group" match="party/pos/attrGroupMany[@name = 'temperatureInformation']/row"
    use="concat(generate-id(ancestor::pos), '|', attr[@name = 'temperatureCode'], '|', attrQualMany[@name = 'temperature'])"/>

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

<xsl:template match="attrGroupMany[@name = 'temperatureInformation']">
    <xsl:copy>
        <xsl:apply-templates select="@*"/>
        <xsl:apply-templates select="row[generate-id() = generate-id(key('group', concat(generate-id(ancestor::pos), '|', attr[@name = 'temperatureCode'], '|', attrQualMany[@name = 'temperature']))[1])]"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="attrGroupMany[@name = 'temperatureInformation']/row">
    <xsl:copy>
        <xsl:apply-templates select="@*|node()[not(self::attrGroupMany[@name = 'temperatureStats'])]"/> 
        <attrGroupMany name="temperatureStats">
            <xsl:apply-templates select="key('group', concat(generate-id(ancestor::pos), '|', attr[@name = 'temperatureCode'], '|', attrQualMany[@name = 'temperature']))/attrGroupMany[@name = 'temperatureStats']/row"/>
        </attrGroupMany>
    </xsl:copy>
</xsl:template>

</xsl:stylesheet>

解决方案2
 0  2016-08-19 06:10:35

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"    version="1.0">

<xsl:output method="xml" indent="yes" />

<xsl:key name="grouptemperatureInformation" match="party/pos/attrGroupMany[@name = 'temperatureInformation']/row"
use="concat(generate-id(ancestor::pos), '|', attr[@name = 'temperatureCode'], '|', attrQualMany[@name = 'temperature'])"/>

<xsl:key name="grouptemperatureStats" 
         match="party/pos/attrGroupMany[@name = 'temperatureInformation']/row/attrGroupMany[@name = 'temperatureStats']/row"
         use="concat(generate-id(ancestor::pos), '|', ../../../attr[@name = 'temperatureCode'], '|', ../../../attrQualMany[@name = 'temperature'], '|', attr[@name = 'StatsCode'])"/>


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

<xsl:template match="attrGroupMany[@name = 'temperatureInformation']">
    <xsl:copy>
        <xsl:apply-templates select="@*"/>
        <xsl:apply-templates select="row[generate-id() = generate-id(key('grouptemperatureInformation', concat(generate-id(ancestor::pos), '|', attr[@name = 'temperatureCode'], '|', attrQualMany[@name = 'temperature']))[1])]"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="attrGroupMany[@name = 'temperatureInformation']/row">
    <xsl:variable name="group" select="key('grouptemperatureInformation', concat(generate-id(ancestor::pos), '|', attr[@name = 'temperatureCode'], '|', attrQualMany[@name = 'temperature']))/attrGroupMany[@name='temperatureStats']/row" />
    <xsl:copy>
        <xsl:apply-templates select="@*|node()[not(self::attrGroupMany[@name = 'temperatureStats'])]"/> 
        <attrGroupMany name="temperatureStats">             
            <xsl:apply-templates select="@* | $group[generate-id() = generate-id(key('grouptemperatureStats', concat(generate-id(ancestor::pos), '|', ../../../attr[@name = 'temperatureCode'], '|', ../../../attrQualMany[@name = 'temperature'], '|', attr[@name = 'StatsCode']))[1])]"/> 
        </attrGroupMany>
    </xsl:copy>
 </xsl:template>
 </xsl:stylesheet>

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