Question

可能重复：
How to make this XSL transformation consider only for siblings with the same id?

我有这个输入XML：

<map>
    <region>
        <gridA id="1">
            <blockA id="01" method="build">                 
                <building1 id="x" method="build">
                    <otherchild>a</otherchild>
                </building1>
                <building1 id="x" method="build"> <!-- this one will be removed -->
                    <otherchild>a</otherchild>
                </building1>
            </blockA>    

            <blockA id="01">                 
                <building1 id="x" method="modify"> 
                    <otherchild>a</otherchild>
                </building1>
                <building1 id="x" method="build"> <!-- this one will be kept (prev node have same id but diff method so it's not considered as successive -->
                    <otherchild>a</otherchild>
                </building1>
            </blockA>    

            <blockA id="02">
                <building3 id="y" method="modify">
                    <otherchild>b</otherchild>
                </building3>
                <building2 id="x" method="demolish"/>
            </blockA>      

            <blockA id="01">                
                <building1 id="y" method="build"> <!-- this one will be kept (diff id) -->
                    <otherchild>a</otherchild>
                </building1>
                <building1 id="x" method="build"> <!-- this one will be removed -->
                    <otherchild>a</otherchild>
                </building1>
            </blockA>

            <blockA id="02">                
                <building3 id="y" method="modify"> <!-- this one will be removed -->
                    <otherchild>b</otherchild>
                </building3>
                <building2 id="x" method="demolish"/> <!-- this one will be removed -->
            </blockA>          
        </gridA>   

        <gridA id="2">
            <blockA id="01" method="build">                 
                <building1 id="x" method="build">
                    <otherchild>a</otherchild>
                </building1>
                <building1 id="x" method="build"> <!-- this one will be removed -->
                    <otherchild>a</otherchild>
                </building1>
                <building1 id="x" method="build"> <!-- this one will be kept (diff children) -->
                    <otherchild>b</otherchild>
                </building1>
            </blockA>                              
            <blockA id="01">                
                <building1 id="x" method="build"> <!-- this one will be removed -->
                    <otherchild>b</otherchild>
                </building1>
            </blockA> 
        </gridA>
        <gridB id="1">
            ...and so on..
        </gridB>
    </region>    
</map>

预期产出：

<map>
    <region>
        <gridA id="1">
            <blockA id="01" method="build">                 
                <building1 id="x" method="build">
                    <otherchild>a</otherchild>
                </building1>
            </blockA>    

            <blockA id="01">                 
                <building1 id="x" method="modify"> 
                    <otherchild>a</otherchild>
                </building1>
                <building1 id="x" method="build"> <!-- this one will be kept (prev node have same id but diff method so it's not considered as successive -->
                    <otherchild>a</otherchild>
                </building1>
            </blockA>    

            <blockA id="02">
                <building3 id="y" method="modify">
                    <otherchild>b</otherchild>
                </building3>
                <building2 id="x" method="demolish"/>
            </blockA>      

            <blockA id="01">                
                <building1 id="y" method="build"> <!-- this one will be kept (diff id) -->
                    <otherchild>a</otherchild>
                </building1>
            </blockA>

            <blockA id="02"/>                         
        </gridA>   

        <gridA id="2">
            <blockA id="01" method="build">                 
                <building1 id="x" method="build">
                    <otherchild>a</otherchild>
                </building1>

                <building1 id="x" method="build"> <!-- this one will be kept (diff children) -->
                    <otherchild>b</otherchild>
                </building1>
            </blockA>                              
            <blockA id="01"/>  
        </gridA>
        <gridB id="1">
            ...and so on..
        </gridB>
    </region>    
</map>

到目前为止的XSLT：

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output indent="yes"/> <xsl:strip-space elements="*"/>

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

    <xsl:template match="region/*/*/*
         [deep-equal(.,preceding::*[name()=current()/name()]
                       [@id = current()/@id]
                       [../../@id = current()/../../@id][1])]" />
</xsl:stylesheet>

现在XSLT的问题在于它无法区分兄弟姐妹中发生的重复（即具有相同id的blockA）。

我需要删除一个被视为重复的节点。

具有相同name和id的两个节点如果一个接一个地出现并且具有相同的method，将被视为重复和children。

例如：

<elem id="1" method="a" />
<elem id="1" method="a" /> <!-- this is repetitive for id=1-->
<elem id="1" method="b" /> 
<elem id="1" method="a" /> <!-- this is the new boundary for removal id=1-->
<elem id="2" method="a" /> 
<elem id="1" method="a" /> <!-- this is repetitive for id=1 -->
<elem id="2" method="a" /> <!-- this is repetitive for id=2 -->

将简化为：

<elem id="1" method="a" />
<elem id="1" method="b" />
<elem id="1" method="a" /> <!-- this is the new boundary for removal id=1-->
<elem id="2" method="a" />

- 每次same id的连续节点都有different method时，重置boundary的下一次删除id。

我们需要考虑一个父母或兄弟姐妹（两个或多个具有相同元素名称和id的父节点）的重复项，即（例如：blockX）
如果被比较的两个节点没有共享相同的gridX级别，那么它们不应被视为要删除的重复项
所有后续重复项都将被删除，直到它变得不重复：请参阅示例的最后一部分，注释虽然它已elem id=1 method=a，但由于前一个节点为{{1}而未被删除这使它成为一个非重复的节点。
因此密钥首先查看具有相同id的连续节点，然后检查方法和子节点是否相同。

对不起，我不知道怎么用更容易理解的句子解释这个问题。

我希望解释清楚，不要混淆。

请告诉我如何使用XSLT实现此类转换。非常感谢你的帮助。

约翰

如何使用XSLT转换此XML？

0 个答案: