搜索和替换一个元素，使用xslt 3，替换短语是相同的

Question

而我有一个xml文件作为输入，比如：

 <?xml version="1.0"?>
    <catalog>
       <book id="bk101">
          <author>Gambardella, Matthew</author>
          <title>XML Developer's Guide</title>
          <genre>Computer</genre>
          <price>44.95</price>
          <publish_date>2000-10-01</publish_date>
          <description>An in-depth look at creating applications 
          with XML.</description>
       </book>
       <book id="bk102">
          <author>Ralls, Kim</author>
          <title>Midnight Rain</title>
          <genre>Fantasy</genre>
          <price>5.95</price>
          <publish_date>2000-12-16</publish_date>
          <description>A former architect battles corporate zombies, 
          an evil sorceress, and her own childhood to become queen 
          of the world.</description>
       </book>
       <book id="bk103">
          <author>Corets, Eva</author>
          <title>Maeve Ascendant</title>
          <genre>Fantasy</genre>
          <price>5.95</price>
          <publish_date>2000-11-17</publish_date>
          <description>After the collapse of a nanotechnology 
          society in England, the young survivors lay the 
          foundation for a new society.</description>
       </book>
    </catalog>

我试图找到在文件中或在xsl本身中包含以下信息的最佳方法：

value to search for: 
An in-depth look at creating applications with XML.
add location:
on the self
value to search for:
A former architect battles corporate zombies, an evil sorceress, and her own childhood to become queen of the world.
add location:
on the self

因此，如果我创建一个逗号分隔的输入文件，它将如下所示：

"An in-depth look at creating applications with XML.","on the self"
"A former architect battles corporate zombies, an evil sorceress, and her own childhood to become queen of the world.","on the self" 

我尝试过使用XSLT2，但不断收到错误，比如变量$search_phrase的值不允许包含多个项目的序列...

所需输出：

<?xml version="1.0"?>
<catalog>
   <book id="bk101">
      <author>Gambardella, Matthew</author>
      <title>XML Developer's Guide</title>
      <genre>Computer</genre>
      <price>44.95</price>
      <publish_date>2000-10-01</publish_date>
      <description>to be checked</description>
      <location>on the self</location>
   </book>
   <book id="bk102">
      <author>Ralls, Kim</author>
      <title>Midnight Rain</title>
      <genre>Fantasy</genre>
      <price>5.95</price>
      <publish_date>2000-12-16</publish_date>
      <description>to be checked</description>
      <location>on the self</location>
   </book>
   <book id="bk103">
      <author>Corets, Eva</author>
      <title>Maeve Ascendant</title>
      <genre>Fantasy</genre>
      <price>5.95</price>
      <publish_date>2000-11-17</publish_date>
      <description>After the collapse of a nanotechnology 
      society in England, the young survivors lay the 
      foundation for a new society.</description>
   </book>
</catalog>

谁能给我举一个xslt-3.0的例子，在那里我可以替换上面的短语，并在任何匹配的地方添加所需的元素？

我需要做的是：

在整个xml文件中，有许多记录可以具有相同的描述。我还需要在描述上进行精确匹配:短语“深入了解如何使用XML创建应用程序，作者是...”不应匹配。在我的例子中，我还描述了其中的区别，例如，“深入了解如何使用XML创建应用程序”。不应该也匹配。因为在我的代码中我使用了小写，这也可能是问题所在，但不确定...只要有匹配，就必须将搜索条件中指定的位置添加到location元素中，该元素目前不存在于xml中的任何记录中。

Answer 1

下面是关于如何将description元素与作为参数传入的字符串序列进行比较的建议(但您也可以从文件中读取它)：

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema"
    exclude-result-prefixes="xs"
    expand-text="yes"
    version="3.0">

  <xsl:param name="new" as="xs:string" select='"on the self"'/>

  <xsl:param name="replace" as="xs:string" select="'to be checked'"/>

  <xsl:param name="search" as="xs:string*"
    select='"An in-depth look at creating applications with XML.",
"A former architect battles corporate zombies, an evil sorceress, and her own childhood to become queen of the world."'/>

  <xsl:mode on-no-match="shallow-copy"/>

  <xsl:output indent="yes"/>
  <xsl:strip-space elements="*"/>

  <xsl:template match="description[. = $search]">
      <xsl:copy>{$replace}</xsl:copy>
      <location>{$new}</location>
  </xsl:template>

</xsl:stylesheet>

在http://xsltfiddle.liberty-development.net/eiQZDbk中运行良好，但仅在编辑样例以使所有描述数据位于一行之后。

如果不是这样，则将模板更改为

  <xsl:template match="description[normalize-space() = $search]">
      <xsl:copy>{$replace}</xsl:copy>
      <location>{$new}</location>
  </xsl:template>

应该有所帮助：http://xsltfiddle.liberty-development.net/eiQZDbk/1

如果您有几个相互关联的术语，那么某种XML格式似乎更适合于结构化数据，因此在

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema"
    exclude-result-prefixes="xs"
    expand-text="yes"
    version="3.0">

  <xsl:param name="data-url" as="xs:string" select="'data.xml'"/>

  <!-- if you want to load from a file use xsl:param name="replacement-doc" select="doc($data-url)" -->
  <xsl:param name="replacement-doc">
    <root>
        <search>
            <term>An in-depth look at creating applications with XML.</term>
            <replacement>to be checked</replacement>
            <new>on the self</new>
        </search>
        <search>
            <term>A former architect battles corporate zombies, an evil sorceress, and her own childhood to become queen of the world.</term>
            <replacement>whatelse</replacement>
            <new>something</new>
        </search>
    </root>
  </xsl:param>

  <xsl:key name="search" match="search" use="term"/>

  <xsl:mode on-no-match="shallow-copy"/>

  <xsl:output indent="yes"/>
  <xsl:strip-space elements="*"/>

  <xsl:template match="description[key('search', normalize-space(), $replacement-doc)]">
      <xsl:variable name="search" select="key('search', normalize-space(), $replacement-doc)"/>
      <xsl:copy>{$search/replacement}</xsl:copy>
      <location>{$search/new}</location>
  </xsl:template>

</xsl:stylesheet>

我已经提出了一些建议，并对模板进行了调整。在线样本在http://xsltfiddle.liberty-development.net/eiQZDbk/2。正如注释中所指出的，您可以采用这种方法从单独的文件加载数据，而不是将其内联保存在XSLT中。