XSLT将多个节点组合成单个节点

Question

<RowSet>
 <Row>
  <Location_Long_Desc>Sydney Office</Location_Long_Desc>
  <Location_Code>SYDNEY</Location_Code>
  <Daypart_Long_Desc>Peak Night</Daypart_Long_Desc>
  <Daypart_Code>PEANIG</Daypart_Code>
  <W_20050703_Dlr>30849.3</W_20050703_Dlr>
  <W_20050703_Spots>9</W_20050703_Spots>
  <W_20050710_Dlr>16.35</W_20050710_Dlr>
  <W_20050710_Spots>19</W_20050710_Spots>
 </Row>
</RowSet>

所以,我现在有了这个XML,我需要将W_节点转换为新的单个节点.使用这个XSL

<?xml version="1.0"?>
  <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" indent="yes"/>
  <xsl:template match="node()">
    <xsl:copy>
      <xsl:apply-templates select="node()"/>
    </xsl:copy>
  </xsl:template>
  <xsl:template match="*">
    <xsl:variable name="tmp" select="local-name()"/>
    <xsl:choose>
      <xsl:when test="starts-with($tmp, 'W_') and ends-with($tmp, '_Dlr')">
    <xsl:if test="text() != ''">
          <xsl:element name="Expenditure">
            <xsl:element name="Period">
              <xsl:value-of select="substring($tmp,3,8)"/>
            </xsl:element>
            <xsl:element name="Value">
              <xsl:apply-templates select="node()"/>
            </xsl:element>
            <xsl:element name="Spots">
              <xsl:apply-templates select="//RowSet/Row/W_20050703_Spots/text()"/>
            </xsl:element>
          </xsl:element>
    </xsl:if>
    </xsl:when>
    <xsl:otherwise>
        <xsl:element name="{$tmp}">
          <xsl:apply-templates select="node()"/>
        </xsl:element>
      </xsl:otherwise>
    </xsl:choose>
  </xsl:template>
</xsl:stylesheet>

我几乎可以到达那里,但我有几个问题.

1. 我需要结合W _ ???????? Dlr和W ???????? _点到单个节点但是
2. 我无法弄清楚如何在xpath语句中使用变量,或者我距离应该在哪里.

再一次,我仍然要掌握这一切,所以请温柔;-)

TIA

编辑:02/12/2010 12:00

好,

还有一个问题,取决于数据库级别切换(我忘了所有),Spots节点可能存在也可能不存在.

所以我仍然需要输出虽然它不应该是以下兄弟调用,其中下一个兄弟不是一个有效的_spots节点.

例:

<RowSet>
 <Row>
  <Location_Long_Desc>Sydney Office</Location_Long_Desc>
  <Location_Code>SYDNEY</Location_Code>
  <Daypart_Long_Desc>Peak Night</Daypart_Long_Desc>
  <Daypart_Code>PEANIG</Daypart_Code>
  <W_20050703_Dlr>30849.3</W_20050703_Dlr>
  <W_20050710_Dlr>16.35</W_20050710_Dlr>
 </Row>
</RowSet>

您知道,我通过Oracle软件包调用所有这些

-- get the query context;
v_qryctx := dbms_xmlgen.newcontext(in_sql_query);

dbms_xmlgen.setnullhandling(v_qryctx, 2);
dbms_xmlgen.setrowsettag(v_qryctx, 'RowSet');
dbms_xmlgen.setrowtag(v_qryctx, 'Row');

IF in_export_type = cnst_export_booking
THEN
    dbms_xmlgen.setxslt(v_qryctx, v_booking_export_xsl);

ELSIF in_export_type = cnst_export_expenditure
THEN
    dbms_xmlgen.setxslt(v_qryctx, v_expenditure_export_xsl);

END IF;

Answer 1

这种转变:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

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

 <xsl:template match=
 "*[starts-with(name(),'W_')
  and
   substring-after(substring-after(name(),'_'),'_')='Dlr'
  and
   text()
   ]">
  <Expenditure>
    <Period>
      <xsl:value-of select="substring(name(),3,8)"/>
    </Period>
    <Value>
      <xsl:apply-templates/>
    </Value>
      <xsl:apply-templates mode="extract" select=
      "following-sibling::*[1]
        [starts-with(name(),'W_')
       and
        substring-after(substring-after(name(),'_'),'_')='Spots'
         ]
       "/>
  </Expenditure>
 </xsl:template>

 <xsl:template mode="extract" match=
  "*[starts-with(name(),'W_')
  and
   substring-after(substring-after(name(),'_'),'_')='Spots'
    ]
  ">
   <Spots><xsl:value-of select="."/></Spots>
  </xsl:template>


 <xsl:template match=
  "*[starts-with(name(),'W_')
  and
   substring-after(substring-after(name(),'_'),'_')='Spots'
    ]
  "/>
</xsl:stylesheet>

应用于提供的源XML文档时:

<RowSet>
 <Row>
  <Location_Long_Desc>Sydney Office</Location_Long_Desc>
  <Location_Code>SYDNEY</Location_Code>
  <Daypart_Long_Desc>Peak Night</Daypart_Long_Desc>
  <Daypart_Code>PEANIG</Daypart_Code>
  <W_20050703_Dlr>30849.3</W_20050703_Dlr>
  <W_20050703_Spots>9</W_20050703_Spots>
  <W_20050710_Dlr>16.35</W_20050710_Dlr>
  <W_20050710_Spots>19</W_20050710_Spots>
 </Row>
</RowSet>

产生想要的,正确的结果:

<RowSet>
   <Row>
      <Location_Long_Desc>Sydney Office</Location_Long_Desc>
      <Location_Code>SYDNEY</Location_Code>
      <Daypart_Long_Desc>Peak Night</Daypart_Long_Desc>
      <Daypart_Code>PEANIG</Daypart_Code>
      <Expenditure>
         <Period>20050703</Period>
         <Value>30849.3</Value>
         <Spots>9</Spots>
      </Expenditure>
      <Expenditure>
         <Period>20050710</Period>
         <Value>16.35</Value>
         <Spots>19</Spots>
      </Expenditure>
   </Row>
</RowSet>

当应用于第二个提供的XML文档时,OP在更新中请求:

<RowSet>
 <Row>
  <Location_Long_Desc>Sydney Office</Location_Long_Desc>
  <Location_Code>SYDNEY</Location_Code>
  <Daypart_Long_Desc>Peak Night</Daypart_Long_Desc>
  <Daypart_Code>PEANIG</Daypart_Code>
  <W_20050703_Dlr>30849.3</W_20050703_Dlr>
  <W_20050710_Dlr>16.35</W_20050710_Dlr>
  <W_20050710_Spots>19</W_20050710_Spots>
 </Row>
</RowSet>

再次产生想要的,正确的结果(<Spot>如果直接兄弟不是a,则生成没有元素W_nnnnnnnn_Spots):

<RowSet>
   <Row>
      <Location_Long_Desc>Sydney Office</Location_Long_Desc>
      <Location_Code>SYDNEY</Location_Code>
      <Daypart_Long_Desc>Peak Night</Daypart_Long_Desc>
      <Daypart_Code>PEANIG</Daypart_Code>
      <Expenditure>
         <Period>20050703</Period>
         <Value>30849.3</Value>
      </Expenditure>
      <Expenditure>
         <Period>20050710</Period>
         <Value>16.35</Value>
         <Spots>19</Spots>
      </Expenditure>
   </Row>
</RowSet>

请注意:

1. 使用标识规则 "按原样"复制任何节点.

2. 仅为W_nnnnnnnn_Dlr元素覆盖标识模板.

3. 使用匹配W_nnnnnnnn_Spots元素的空模板覆盖标识模板.

4. 采用标准的XPath函数:name(),starts-with()和substring-after()

5. 该功能ends-with()仅在XPath 2.0(XSLT 2.0)中可用,并且未在此XSLT 1.0解决方案中使用.