Question

如何将不同的模板应用于具有相同名称的不同子节点？

我得到了一个我无法改变的xml。

为了过度简化一个例子（真正的xml比这复杂得多），我想将不同的模板应用于员工和经理，但他们的子节点都是＆＃34;员工＆＃34;。下面的xslt不起作用，因为有2个Employee模板。

仅供参考，我是xslts的新手，所以如果事实证明这很容易解决，我会道歉。

提前致谢！任何帮助将不胜感激。

XML

<Company>
    <CompanyId>1</CompanyId>
    <CompanyName>Company B</CompanyName>
    <Employees>
        <Employee>
            <EmployeeId>1</EmployeeId>
            <FirstName>Jane</FirstName>
            <LastName>Doe</LastName>
        </Employee>
        <Employee>
            <EmployeeId>2</EmployeeId>
            <FirstName>James</FirstName>
            <LastName>Smith</LastName>
        </Employee>
    </Employees>
    <Managers>
        <Employee>
            <EmployeeId>3</EmployeeId>
            <FirstName>Michael</FirstName>
            <LastName>Johnson</LastName>
            <DepartmentName>Sales</DepartmentName>
            <NumberOfStaff>20</NumberOfStaff>
        </Employee>
        <Employee>
            <EmployeeId>2</EmployeeId>
            <FirstName>James</FirstName>
            <LastName>Smith</LastName>
            <DepartmentName>IT</DepartmentName>
            <NumberOfStaff>50</NumberOfStaff>
        </Employee>
    </Managers>
</Company>

XSLT

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
                xmlns:ms="urn:schemas-microsoft-com:xslt"
                xmlns:dt="urn:schemas-microsoft-com:datatypes">

  <xsl:output method="xml" indent="yes"/>

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

  <xsl:template match="Company">
    COMPANY INFO<br />
    Company Id: <xsl:value-of select="CompanyId" /><br />
    Company Name: <xsl:value-of select="CompanyName" /><br /> 

    <br />
    EMPLOYEES<br />
    <xsl:apply-templates select="/Company/Employees/Employee"/>

    <br />
    MANAGERS<br />  
    <xsl:apply-templates select="/Company/Managers/Employee"/>

    <br />
  </xsl:template>

  <xsl:template match="Employee">
    <br />
    Employee Id: <xsl:value-of select="EmployeeId" /><br />
    Employee Name:  <xsl:value-of select="LastName" />,<xsl:value-of select="FirstName" /><br /> 
  </xsl:template>

  <xsl:template match="Employee">
    <br />
    Manager Id: <xsl:value-of select="EmployeeId" /><br />
    Manager Name:  <xsl:value-of select="LastName" />,<xsl:value-of select="FirstName" /><br /> 
    Department Name: <xsl:value-of select="DepartmentName" /><br /> 
    Number of Staff: <xsl:value-of select="NumberOfStaff" /><br /> 
  </xsl:template>

</xsl:stylesheet>

期望的输出

 COMPANY INFO
 Company Id: 1
 Company Name: Company B

 EMPLOYEES

 Employee Id: 1
 Employee Name: Doe,Jane

 Employee Id: 2
 Employee Name: Smith,James

 MANAGERS

 Manager Id: 3
 Manager Name: Johnson,Michael
 Department Name: Sales
 Number of Staff: 20

 Manager Id: 2
 Manager Name: Smith,James
 Department Name: IT
 Number of Staff: 50

Answer 1

您可以在模板

的匹配属性中指定父级

对于经理......

<xsl:template match="Managers/Employee">

对于员工......

<xsl:template match="Employees/Employee">

请注意，虽然不一定在这种情况下，另一种选择是使我们成为＆＃34;模式＆＃34;

<xsl:apply-templates select="/Company/Employees/Employee" mode="Employees" />

<xsl:apply-templates select="/Company/Managers/Employee" mode="Managers" />

然后你可以像这样编写模板匹配：

<xsl:template match="Employee" mode="Employees">

<xsl:template match="Employee" mode="Managers">

XSLT应用模板同名不同的节点

1 个答案: