在 XSLT 中与可选节点匹配和合并 [英] match and merge in XSLT with optional nodes

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问题描述

我有一个包含多个父节点的示例 XML 消息.要求是如果两个父节点相同，则合并子节点.这在所有节点都存在时工作正常，但在可选节点不存在时不起作用

I have a sample XML message which contains multiple parent node. The requirement is if the two parent node are same, merge the child node. This works fine when all the nodes are present but doesn't work when the optional node is absent

示例消息:1 存在可选节点

Sample Message:1 With Optional Nodes present

<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>

下面的示例 XSLT 可以很好地从 'attrGroupMany name="temperatureInformation" '

The below sample XSLT works fine to remove the duplicate from 'attrGroupMany name="temperatureInformation" '

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>

不适用于以上 XSLT 的示例消息 2 是

Sample Message 2 which doesnt work with above XSLT is

<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>

谁能告诉我如何处理匹配 n 合并中的可选节点

Can someone please let me know how to handle optional node in match n merge

示例消息 2 的预期输出是

The expected output for sample message 2 is 

<?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>

有人可以让我知道如何处理匹配 n 合并中的可选节点.谢谢

Can someone please let me know how to handle optional node in match n merge. Thanks

推荐答案

您还没有解释您尝试执行的操作背后的逻辑，但是从这个问题以及之前的问题中查看 XSLT，您正在分组attrGroupMany[@name = 'temperatureInformation']/row 元素，但是祖先 pos 与temperatureCode"和温度"的串联.

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".

然后对于每个这样不同的 row，您似乎想要添加所有 元素.如果您说它是可选的，那么拥有与此元素匹配的模板将不起作用.相反，有一个与父 row 匹配的模板，并使用它从键中的所有元素中选择所有子元素.

Then for each such distinct row it looks like you want to add in all the <attrGroupMany name="temperatureStats"> elements. Having a template matching this element is not going to work if you say it is optional. Instead, have a template matching the parent row and use that to select all the child elements from all elements in the keys.

<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>

我假设所有 StatsCode 在这里都是不同的.如果可能有重复，并且您想删除此类重复，则需要在问题中说明.

I am assuming all the StatsCode are distinct here. If there could be duplicated, and you want to remove such duplicates, you need to say so in your question.

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>

这篇关于在 XSLT 中与可选节点匹配和合并的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！

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