簡體   English   中英

xslt 1.0中的xml節點分組

[英]Grouping of xml nodes in xslt 1.0

輸入:

輸入XML

<Root>
    <Number>1</Number>
    <Reference>100</Reference>

    <Number>2</Number>
    <Reference>101</Reference>

    <Number>3</Number>
    <Reference>100</Reference>

    <Number>4</Number>
    <Reference>102</Reference>

    <Number>5</Number>
    <Reference>100</Reference>
</Root>

預期產量:

<Root>
    <Number>1</Number>
    <Reference>100</Reference>
    <RefNumber>1</RefNumber>

    <Number>2</Number>
    <Reference>101</Reference>
    <RefNumber>1</RefNumber>

    <Number>3</Number>
    <Reference>100</Reference>
    <RefNumber>2</RefNumber>

    <Number>4</Number>
    <Reference>102</Reference>
    <RefNumber>1</RefNumber>

    <Number>5</Number>
    <Reference>100</Reference>
    <RefNumber>3</RefNumber>
</Root>

如何在xslt 1.0的輸出中基於根/引用進行分組並向RefNumber添加順序號?

提前致謝

一種方法是使用xsl:number 只要與Reference元素匹配,就復制該元素,並添加RefNumber元素,其中RefNumber元素的數量應具有相同的值:

<xsl:template match="Reference">
    <xsl:copy-of select="." />
    <xsl:variable name="Ref" select="." />
    <RefNumber><xsl:number count="Reference[. = $Ref]" /></RefNumber>
</xsl:template>

這是完整的XSLT

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" indent="yes"/>

    <xsl:template match="Reference">
        <xsl:copy-of select="." />
        <xsl:variable name="Ref" select="." />
        <RefNumber><xsl:number count="Reference[. = $Ref]" /></RefNumber>
    </xsl:template>

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

當應用於示例XML時,將輸出以下內容:

<Root>
    <Number>1</Number>
    <Reference>100</Reference>
    <RefNumber>1</RefNumber>

    <Number>2</Number>
    <Reference>101</Reference>
    <RefNumber>1</RefNumber>

    <Number>3</Number>
    <Reference>100</Reference>
    <RefNumber>2</RefNumber>

    <Number>4</Number>
    <Reference>102</Reference>
    <RefNumber>1</RefNumber>

    <Number>5</Number>
    <Reference>100</Reference>
    <RefNumber>3</RefNumber>
</Root>

請注意使用身份轉換模板來復制其他現有節點。

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM