Question

我有这个xml文件....我必须检查SE / SSE的状态...如果它是活动的那么它将进入详细信息标签....它将读取状态...如果它是活动的，然后它将读取详细信息标记，除非它将丢弃该详细信息节点。就像那样，如果SSE状态是＆＃34; InACTIVE＆＃34;，则无需在该节点内读取。

<Employees>
<Employee>
    <SE>
        <Name>bikash</Name>
        <dept>DY</dept>
        <status>ACTIVE</status>
     <Details dataStr="list">
        <status>ACTIVE</status>
        <address>India</address>
        <streetNo>19</streetNo>
     </Details>
     <Details dataStr="list">
        <status>InACTIVE</status>
        <address>CHINA</address>
        <streetNo>20</streetNo>
     </Details>
     <area>BLORE</area>
     <SEIdCount>1</SEIdCount>
  </SE>
  <SSE>
        <status>InACTIVE</status>
     <emplNo>23</emplNo>
     <Details dataStr="list">
        <status>InActive</status>
        <absent>y</absent>
     </Details>
     <Details dataStr="list">
        <status>Active</status>
        <name>anu</NAME>
     </Details>
    <area>CHN</area>
    <SEIdCount>2</SEIdCount>
  </SSE>
 </Employee> 
</Employees>

我的预期回应是这个

<Employees>
<Employee>
    <SE>
        <Name>bikash</Name>
        <dept>DY</dept>
        <status>ACTIVE</status>
     <Details dataStr="list">
        <status>ACTIVE</status>
        <address>India</address>
        <streetNo>19</streetNo>
     </Details>
     <area>BLORE</area>
     <SEIdCount>1</SEIdCount>
  </SE>
 </Employee> 
</Employees>

Answer 1

尝试：

XSLT 1.0

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

<xsl:template match="/Employees">
    <Employees>
        <xsl:for-each select="Employee/SE">
            <Employee>
                <SE>
                    <xsl:copy-of select="Name | dept | status | Details[status='Active'] | area | SEIdCount"/>
                </SE>
            </Employee> 
        </xsl:for-each> 
    </Employees>
</xsl:template>

</xsl:stylesheet>

应用于格式良好的XML输入：

<Employees>
  <Employee>
    <SE>
      <Name>bikash</Name>
      <dept>DY</dept>
      <status>Active</status>
      <Details dataStr="list">
        <status>Active</status>
        <address>India</address>
        <streetNo>19</streetNo>
      </Details>
      <Details dataStr="list">
        <status>InACTIVE</status>
        <address>CHINA</address>
        <streetNo>20</streetNo>
      </Details>
      <area>BLORE</area>
      <SEIdCount>1</SEIdCount>
    </SE>
    <SSE>
      <status>InACTIVE</status>
      <emplNo>23</emplNo>
      <Details dataStr="list">
        <status>InActive</status>
        <absent>y</absent>
      </Details>
      <Details dataStr="list">
        <status>Active</status>
        <name>anu</name>
      </Details>
      <area>CHN</area>
      <SEIdCount>2</SEIdCount>
    </SSE>
  </Employee>
</Employees>

这将返回：

<?xml version="1.0" encoding="UTF-8"?>
<Employees>
  <Employee>
    <SE>
      <Name>bikash</Name>
      <dept>DY</dept>
      <status>Active</status>
      <Details dataStr="list">
        <status>Active</status>
        <address>India</address>
        <streetNo>19</streetNo>
      </Details>
      <area>BLORE</area>
      <SEIdCount>1</SEIdCount>
    </SE>
  </Employee>
</Employees>

注意：XML区分大小写：

select="Name" 不选择<name>，＆＃34;有效＆＃34;是不与＆＃34; ACTIVE＆＃34;。

相同

Answer 2

使用XSLT 1.0 如果您没有SE节点中的数字子节点已修复

，您也可以尝试这一个

  <xsl:template match="/*/Employee">
    <Employees>
      <Employee>
        <SE>
          <xsl:for-each select="*">
            <xsl:choose>
              <xsl:when test="status='ACTIVE'">
                <xsl:for-each select="./*">
                  <xsl:choose>
                    <xsl:when test="local-name(.)='Details' and ./status='ACTIVE'">
                      <xsl:copy-of select="."/>
                    </xsl:when>
                    <xsl:when test="local-name(.)='Details' and ./status='InACTIVE'"/>
                    <xsl:otherwise>
                      <xsl:copy-of select="."/>
                    </xsl:otherwise>
                  </xsl:choose>
                </xsl:for-each>
              </xsl:when>
            </xsl:choose>
          </xsl:for-each>
        </SE>
      </Employee>
    </Employees>
  </xsl:template>

如何复制父节点以及具有特定属性的父节点的所有子节点应该同时存在于父节点和子节点中

2 个答案:

注意：XML区分大小写：