我有以下输入XML， 由于某种原因，我的xslt无法对其进行排序。 请指教。 感谢您的见解。 这是完整的XML .................................................. ....
我想根据D_628字段对G_SHL段的内容进行排序。
`<?xml version="1.0" encoding="UTF-8"?>
<LIST>
<S_ISA>
<D_I01>00</D_I01>
<D_I02/>
<D_I03>00</D_I03>
<D_I04/>
<D_I05>01</D_I05>
<D_I06></D_I06>
<D_I05_2>ZZ</D_I05_2>
<D_I07></D_I07>
<D_I08>160427</D_I08>
<D_I09>1106</D_I09>
<D_I10>U</D_I10>
<D_I11>00401</D_I11>
<D_I12>000000001</D_I12>
<D_I13>0</D_I13>
<D_I14/>
<D_I15>></D_I15>
<S_GS>
<D_479>SH</D_479>
<D_142></D_142>
<D_124></D_124>
<D_373>20160427</D_373>
<D_337>1106</D_337>
<D_28>1</D_28>
<D_455>X</D_455>
<D_480>004010</D_480>
<S_ST>
<D_143>856</D_143>
<D_329>0001</D_329>
<S_BSN>
<D_353>00</D_353>
<D_396>0081664420</D_396>
<D_373>20160426</D_373>
<D_337>1347</D_337>
<D_1005>0001</D_1005>
</S_BSN>
<G_SHL>
<S_HL>
<D_628>0000001</D_628>
<D_734>0000000</D_734>
<D_735>S</D_735>
<S_PRF>
<D_324>SITT10-1447195769627</D_324>
<D_328>SITT10-1447195769627</D_328>
</S_PRF>
<S_MEA>
<D_737>PD</D_737>
<D_738>G</D_738>
<D_739>0.081</D_739>
<C_C001>
<D_355>KG</D_355>
</C_C001>
</S_MEA>
<S_TD1>
<D_103>PKG</D_103>
<D_80>00002</D_80>
<D_187>G</D_187>
<D_81>00000.18</D_81>
<D_355>LB</D_355>
</S_TD1>
<S_TD5>
<D_133>O</D_133>
<D_66>2</D_66>
<D_368>CC</D_368>
</S_TD5>
<S_REF>
<D_128>WH</D_128>
<D_127>WH:</D_127>
</S_REF>
<S_REF>
<D_128>RN</D_128>
<D_127>RN:</D_127>
</S_REF>
<S_DTM>
<D_374>011</D_374>
<D_373>20160414</D_373>
</S_DTM>
<G_SN1>
<S_N1>
<D_98>ST</D_98>
<D_93>C3333</D_93>
<S_N3>
<D_166>TEST</D_166>
</S_N3>
<S_N4>
<D_19>TEST</D_19>
<D_116>12345</D_116>
<D_26>US</D_26>
</S_N4>
</S_N1>
</G_SN1>
</S_HL>
</G_SHL>
<G_SHL>
<S_HL>
<D_628>0000002</D_628>
<D_734>0000001</D_734>
<D_735>O</D_735>
<S_MAN>
<D_88>CP</D_88>
<D_87></D_87>
</S_MAN>
</S_HL>
</G_SHL>
<G_SHL>
<S_HL>
<D_628>0000003</D_628>
<D_734>0000002</D_734>
<D_735>P</D_735>
</S_HL>
</G_SHL>
<G_SHL>
<S_HL>
<D_628>0000005</D_628>
<D_734>0000002</D_734>
<D_735>P</D_735>
</S_HL>
</G_SHL>
<G_SHL>
<S_HL>
<D_628>0000004</D_628>
<D_734>0000003</D_734>
<D_735>I</D_735>
<S_LIN>
<D_350>000010</D_350>
<D_235>SK</D_235>
<D_234></D_234>
<D_235_2>BP</D_235_2>
<D_234_2></D_234_2>
</S_LIN>
<S_SN1>
<D_350></D_350>
<D_382>2.000</D_382>
<D_355>EA</D_355>
<D_668>AC</D_668>
</S_SN1>
</S_HL>
</G_SHL>
<G_SHL>
<S_HL>
<D_628>0000006</D_628>
<D_734>0000005</D_734>
<D_735>I</D_735>
<S_LIN>
<D_350>000020</D_350>
<D_235></D_235>
<D_234></D_234>
<D_235_2>BP</D_235_2>
<D_234_2></D_234_2>
</S_LIN>
<S_SN1>
<D_382>1.000</D_382>
<D_355>EA</D_355>
<D_668>AC</D_668>
</S_SN1>
</S_HL>
</G_SHL>
<S_CTT>
<D_354>1</D_354>
</S_CTT>
<S_SE>
<D_96>[]</D_96>
<D_329>0001</D_329>
</S_SE>
</S_ST>
<S_GE>
<D_97>0</D_97>
<D_28>1</D_28>
</S_GE>
</S_GS>
<S_IEA>
<D_I16>01</D_I16>
<D_I12>000000001</D_I12>
</S_IEA>
</S_ISA>
</LIST>`
答案 0 :(得分：0)
尝试
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="/LIST/S_ISA/S_GS/S_ST">
<xsl:copy>
<xsl:apply-templates select="@*"/>
<xsl:apply-templates select="G_SHL">
<xsl:sort select="S_HL/D_628" data-type="number" order="descending"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
我无法验证路径/LIST/S_ISA/S_GS/S_ST，但假设您在真实XML中拥有这些祖先元素。
如果您希望将兄弟元素保留在其原始输入位置，那么XSLT 2.0中的一种方法是首先使用for-each-group group-adjacent识别要排序的相邻元素：
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="xs"
version="2.0">
<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="/LIST/S_ISA/S_GS/S_ST">
<xsl:copy>
<xsl:apply-templates select="@*"/>
<xsl:for-each-group select="*" group-adjacent="boolean(self::G_SHL)">
<xsl:choose>
<xsl:when test="current-grouping-key()">
<xsl:apply-templates select="current-group()">
<xsl:sort select="S_HL/D_628" data-type="number" order="descending"/>
</xsl:apply-templates>
</xsl:when>
<xsl:otherwise>
<xsl:apply-templates select="current-group()"></xsl:apply-templates>
</xsl:otherwise>
</xsl:choose>
</xsl:for-each-group>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>