替换各种XML文件中的字符串

Question

给出以下xml文件,知道结构和内容可以更改:

<something>
  <parent>
    <child>Bird is the word 1.</child>
    <child>Curd is the word 2.</child>
    <child>Nerd is the word 3.</child>
  </parent>
  <parent>
    <child>Bird is the word 4.</child>
    <child>Word is the word 5.</child>
    <child>Bird is the word 6.</child>
  </parent>
</something>

我想用一种方法来使用xquery(甚至xslt)来替换所提供的字符串的所有实例.例如,将"Bird"替换为"Dog".因此结果将是:

<something>
  <parent>
    <child>Dog is the word 1.</child>
    <child>Curd is the word 2.</child>
    <child>Nerd is the word 3.</child>
  </parent>
  <parent>
    <child>Dog is the word 4.</child>
    <child>Word is the word 5.</child>
    <child>Dog is the word 6.</child>
  </parent>
</something>

我不知道这是否可能.我所做的每一次尝试都消除了标签.我甚至尝试了这个例子(http://geekswithblogs.net/Erik/archive/2008/04/01/120915.aspx),但是文本不是整个文档.

请帮忙!

UPDATE

我尝试使用xslt 2.0建议,因为它似乎最合适.在尝试根据我的情况对其进行修改时,我一直在干涸.

我想传入一个xml参数来定义替换.所以,像这样修改xslt:

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output indent="yes"/>
  <xsl:strip-space elements="*"/>
  <xsl:param name="list">
<words>
  <word>
        <search>Bird</search>
    <replace>Dog</replace>
  </word>
      <word>
        <search>word</search>
    <replace>man</replace>
  </word>
</words>
  </xsl:param>


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

<xsl:template match="text()">
  <xsl:param name="chosen" select="." />
<xsl:for-each select="$list//word">
  <xsl:variable name="search"><xsl:value-of select="search" /></xsl:variable>
  <xsl:analyze-string select="$chosen" regex="{$search}">
    <xsl:matching-substring><xsl:value-of select="replace" /></xsl:matching-substring>
    <xsl:non-matching-substring><xsl:value-of select="$chosen"/></xsl:non-matching-substring>
  </xsl:analyze-string>
</xsl:for-each>
  </xsl:template>
</xsl:stylesheet>

结果是:

<something>
  <parent>
    <child>Bird is the word 1.Bird is the word 1.</child>
    <child>Curd is the word 2.Curd is the word 2.</child>
    <child>Nerd is the word 3.Nerd is the word 3.</child>
  </parent>
  <parent>
    <child>Bird is the word 4.Bird is the word 4.</child>
    <child>Word is the word 5.Word is the word 5.</child>
    <child>Bird is the word 6.Bird is the word 6.</child>
  </parent>
</something>

不用说,但是,我不希望它重复,也不正确.

请帮忙!

Answer 1

如果XQuery和XSLT都是一个选项,那么你可能正在使用XSLT 2.0处理器.如果是这样,这应该工作:

XSLT 2.0

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output indent="yes"/>
    <xsl:strip-space elements="*"/>

    <xsl:param name="search" select="'Bird'"/>
    <xsl:param name="replace" select="'Dog'"/>

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

    <xsl:template match="text()">
        <xsl:analyze-string select="." regex="{$search}">
            <xsl:matching-substring><xsl:value-of select="$replace"/></xsl:matching-substring>
            <xsl:non-matching-substring><xsl:value-of select="."/></xsl:non-matching-substring>
        </xsl:analyze-string>
    </xsl:template>

</xsl:stylesheet>

使用问题中的XML输入,此XSLT生成以下输出:

<something>
   <parent>
      <child>Dog is the word 1.</child>
      <child>Curd is the word 2.</child>
      <child>Nerd is the word 3.</child>
   </parent>
   <parent>
      <child>Dog is the word 4.</child>
      <child>Word is the word 5.</child>
      <child>Dog is the word 6.</child>
   </parent>
</something>

注意:在创建输出时,不会更改元素/属性/注释/处理指令.

编辑

你得到重复的原因是因为你xsl:for-each正在循环这两个word元素.如果你有3,它将输出文本3次.

你只需要以不同的方式构建正则表达式:

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output indent="yes"/>
    <xsl:strip-space elements="*"/>
    <xsl:param name="list">
        <words>
            <word>
                <search>Bird</search>
                <replace>Dog</replace>
            </word>
            <word>
                <search>word</search>
                <replace>man</replace>
            </word>
        </words>
    </xsl:param>

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

    <xsl:template match="text()">
        <xsl:variable name="search" select="concat('(',string-join($list/words/word/search,'|'),')')"/>
        <xsl:analyze-string select="." regex="{$search}">
            <xsl:matching-substring>
                <xsl:value-of select="$list/words/word[search=current()]/replace"/>
            </xsl:matching-substring>
            <xsl:non-matching-substring>
                <xsl:value-of select="."/>
            </xsl:non-matching-substring>
        </xsl:analyze-string>
    </xsl:template>
</xsl:stylesheet>

这将产生:

<something>
   <parent>
      <child>Dog is the man 1.</child>
      <child>Curd is the man 2.</child>
      <child>Nerd is the man 3.</child>
   </parent>
   <parent>
      <child>Dog is the man 4.</child>
      <child>Word is the man 5.</child>
      <child>Dog is the man 6.</child>
   </parent>
</something>