如何合并两个 XML 文件 [英] How to merge two XML files

问题描述

我有两个xml文件，需要合并成一个xml.例子如下:

I have two xml files, which need to be merged into one xml. Here is the example:

原始.xml文件:

<employees>
  <employee id="1">
    <name> Name1 </name>
    <email> email1 <email>
  </employee>
  <employee id="2">
    <name> Name2 </name>
    <email> email2 <email>
  </employee>
</employees>

update.xml 文件:

update.xml file:

<employees>
        <employee id="2">
            <name> Name2 </name>
            <email> email_New <email>
        </employee>
        <employee id="3">
            <name> Name3 </name>
            <email> email3 <email>
        </employee>
</employees>

它们应该像这样合并到一个 xml 文件中:

they should be merged to a xml file like this:

<employees>
  <employee id="1">
    <name> Name1 </name>
    <email> email1 <email>
  </employee>  
  <employee id="2">
    <name> Name2 </name>
    <email> email_New <email>
  </employee>
  <employee id="3">
    <name> Name3 </name>
    <email> email3 <email>
  </employee>
</employees>

实际上，我想使用 update.xml 来更新 orginal.xml :

Actually, I'd like to use the update.xml to update the orginal.xml :

1. update.xml 中的新员工应添加到 original.xml

1. new employee in update.xml should be added to original.xml

update.xml 中修改的员工信息应该覆盖相应的员工节点.

modified employee information in update.xml should overwrite the corresponding employee node.

我对 XSLT 有一点了解，但我的知识不足以找出适合合并的 XSLT.

I know a little of XSLT, but my knowledge is not enough to figure out the right XSLT for the merge.

推荐答案

请试一试:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml" indent="yes"/>
  <xsl:param name="fileName" select="'update.xml'" />
  <xsl:param name="updates" select="document($fileName)" />

  <xsl:variable name="updateEmployees" select="$updates/employees/employee" />

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

  <xsl:template match="employees">
    <xsl:copy>
      <xsl:apply-templates select="employee[not(@id = $updateEmployees/@id)]" />
      <xsl:apply-templates select="$updateEmployees" />
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>

当使用您的第一个 XML 作为输入并且 update.xml 存在于同一文件夹中时，这会产生:

When run with your first XML as the input and with the update.xml present in the same folder, this produces:

<employees>
  <employee id="1">
    <name> Name1 </name>
    <email>
      email1 </email>
  </employee>
  <employee id="2">
    <name> Name2 </name>
    <email>
      email_New
    </email>
  </employee>
  <employee id="3">
    <name> Name3 </name>
    <email>
      email3
    </email>
  </employee>
</employees>

fileName 和 updates 被声明为参数，因此如果更新位于名为update.xml"以外的文件中，或者您想要传递更新 XML直接使用，您无需对 XSLT 进行任何修改即可完成此操作.

fileName and updates are declared as parameters, so if the updates are in a file named something other than "update.xml" or you want to pass the update XML in directly, you can do that without making any modifications to the XSLT.

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