使用XSLT转换将XML转换为XML

Question

XML输入文件：

<employees>
    <employee empId ="5001">
        <firstName>Ram</firstName>
        <middleName>Prakash</middleName>
        <lastName>Sharma</lastName>
        <department>IT</department>
        <address>    
            <address1>H7 H Block</address1>
            <address2>Sector 63</address2>    
            <city>Noida</city>    
            <state>UP</state>    
            <pincode>201301</pincode>    
        </address>    
        <dateOfJoining>01/01/2017</dateOfJoining >    
        <grade>R</grade>    
    </employee>    
</employees>

XML输入文件转换为XML输出文件。将XML输入文件的firstName、middleName和lastName标签合并为XML Ouput文件的name标签，将XML输入文件的address1、address2、city、state和pincode标签合并为XML Ouput文件的address标签。

<employees>    
    <employee empId ="5001">    
        <name>Ram Prakash Sharma</name>    
        <department>IT</department>    
        <address>H7 H Block Sector 63 Noida UP 201301</address>    
        <dateOfJoining>01/01/2017</dateOfJoining >    
        <grade>R</grade>    
    </employee>    
</employees>

我几乎转换了代码，但我在这里与empId作斗争。我已经在XSLT文件中手动输入了empId。谁能帮帮我吗？

XSLT文件：

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <!-- TODO: Auto-generated template -->
    <xsl:template match="employees">
        <employees>
            <employee empId="5001">
                <xsl:apply-templates />
            </employee>
        </employees>
    </xsl:template>    
    <xsl:template match="employee">    
        <name>
            <xsl:value-of
                select="concat(firstName, ' ', middleName, ' ', lastName)" />
        </name>    
        <department>
            <xsl:value-of select="department" />
        </department>    
        <address>
            <xsl:value-of select="address/address1" />
            <xsl:text> </xsl:text>
            <xsl:value-of select="address/address2" />
            <xsl:text> </xsl:text>
            <xsl:value-of select="address/city" />
            <xsl:text> </xsl:text>
            <xsl:value-of select="address/state" />
            <xsl:text> </xsl:text>
            <xsl:value-of select="address/pincode" />
        </address>    
        <dateOfJoining>
            <xsl:value-of select="dateOfJoining" />
        </dateOfJoining>    
        <grade>
            <xsl:value-of select="grade" />
        </grade>    
    </xsl:template>
</xsl:stylesheet>

Answer 1

只需将employee元素移到employee模板，并使用{@empid}属性值模板设置empid属性：

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:template match="employees">
    <employees>
      <xsl:apply-templates />
    </employees>
  </xsl:template>
  <xsl:template match="employee">
    <employee id="{@empId}">
      <name>
        <xsl:value-of select="concat(firstName, ' ', middleName, ' ', lastName)" />
      </name>
      <department>
        <xsl:value-of select="department" />
      </department>
      <address>
            <xsl:value-of select="address/address1" />
            <xsl:text> </xsl:text>
            <xsl:value-of select="address/address2" />
            <xsl:text> </xsl:text>
            <xsl:value-of select="address/city" />
            <xsl:text> </xsl:text>
            <xsl:value-of select="address/state" />
            <xsl:text> </xsl:text>
            <xsl:value-of select="address/pincode" />
        </address>
      <dateOfJoining>
        <xsl:value-of select="dateOfJoining" />
      </dateOfJoining>
      <grade>
        <xsl:value-of select="grade" />
      </grade>
    </employee>
  </xsl:template>
</xsl:stylesheet>