XSLT 1.0:将一个值分组为多个值 - 使用密钥?

时间:2013-11-27 16:41:20

标签: xslt grouping

我有以下简化的XML结构:

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<DESADV>
<M_DESADV>
    <G_SG10>
        <G_SG13>
            <S_GIN>
                <id>GIN</id>
                <D_7405_6>BJ</D_7405_6>
                <C_C208_2>
                    <D_7402_8>373500699550026556</D_7402_8>
                </C_C208_2>
            </S_GIN>
        </G_SG13>
    </G_SG10>
    <G_SG10>
        <G_SG17>
            <S_PIA>
                <id>PIA</id>
                <D_4347>1</D_4347>
                <C_C212_2>
                    <D_7140_2>110005</D_7140_2>
                    <D_7143_4>SA</D_7143_4>
                </C_C212_2>
            </S_PIA>
        </G_SG17>
    </G_SG10>
    <G_SG10>
        <G_SG17>
            <S_PIA>
                <id>PIA</id>
                <D_4347>1</D_4347>
                <C_C212_2>
                    <D_7140_2>123456</D_7140_2>
                    <D_7143_4>SA</D_7143_4>
                </C_C212_2>
            </S_PIA>
        </G_SG17>
    </G_SG10>
    <G_SG10>
        <G_SG13>
            <S_GIN>
                <id>GIN</id>
                <D_7405_6>BJ</D_7405_6>
                <C_C208_2>
                    <D_7402_8>3735009999999999</D_7402_8>
                </C_C208_2>
            </S_GIN>
        </G_SG13>
    </G_SG10>
    <G_SG10>
        <G_SG17>
            <S_PIA>
                <id>PIA</id>
                <D_4347>1</D_4347>
                <C_C212_2>
                    <D_7140_2>632154</D_7140_2>
                    <D_7143_4>SA</D_7143_4>
                </C_C212_2>
            </S_PIA>
        </G_SG17>
    </G_SG10>
    <G_SG10>
        <G_SG17>
            <S_PIA>
                <id>PIA</id>
                <D_4347>1</D_4347>
                <C_C212_2>
                    <D_7140_2>887796</D_7140_2>
                    <D_7143_4>SA</D_7143_4>
                </C_C212_2>
            </S_PIA>
        </G_SG17>
    </G_SG10>
</M_DESADV>
</DESADV>

我需要进行映射,以便获得以下XML输出:

<?xml version="1.0" encoding="UTF-8"?>
<list>
<number>373500699550026556 110005</number>
<number>373500699550026556 123456</number>
<number>3735009999999999 632154</number>
<number>3735009999999999 887796</number>
</list>

我无法获得该输出。我有以下XSLT,但我会工作:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">

<xsl:template match="text()"/>

<xsl:template match="/">
    <list>
    <xsl:apply-templates/>
    </list>
</xsl:template>

<xsl:template match="M_DESADV">

    <xsl:for-each select="G_SG10/G_SG17/S_PIA/C_C212_2[D_7143_4='SA']/D_7140_2">
        <number>
            <xsl:value-of select="concat(
                parent::C_C212_2/parent::S_PIA/parent::G_SG17/parent::G_SG10/parent::M_DESADV
                /G_SG10/G_SG13/S_GIN[D_7405_6='BJ']/C_C208_2/D_7402_8,' ',
                .)
                "/>
        </number>
    </xsl:for-each>
</xsl:template>



</xsl:stylesheet>

但是我的输出看起来像这不是我所期望的:

<?xml version="1.0" encoding="UTF-8"?>
<list>
<number>373500699550026556 110005</number>
<number>373500699550026556 123456</number>
<number>373500699550026556 632154</number>
<number>373500699550026556 887796</number>
</list>

这可以通过Muenchian Grouping解决吗?我尝试设置一个键,但它要么选择 “D_7402_8”或仅“D_7140_2”。

有人能告诉我如何解决这个问题吗? 感谢你并致以真诚的问候, 彼得

1 个答案:

答案 0 :(得分:1)

我认为你不需要钥匙,你只需找到每个G_SG10 - 那个有一个G_SG17,它最近的兄弟G_SG10 - 该-具有-α - G_SG13

<xsl:for-each select="G_SG10/G_SG17/S_PIA/C_C212_2[D_7143_4='SA']/D_7140_2">
    <number>
        <xsl:value-of select="concat(
            ancestor::G_SG10/preceding-sibling::G_SG10[G_SG13][1]
            /G_SG13/S_GIN[D_7405_6='BJ']/C_C208_2/D_7402_8,' ',
            .)
            "/>
    </number>
</xsl:for-each>

如果真实XML中的条件比“有G_SG13”更复杂,那么您只需要在preceding-sibling::步骤的第一个谓词中对相关条件进行编码,例如

preceding-sibling::G_SG10[G_SG13/S_GIN/D_7405_6='BJ'][1]