xml XSLT - 在输出中用转义文本替换撇号

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时间:2020-09-06 12:37:11  来源:igfitidea点击:

XSLT - Replace apostrophe with escaped text in output

xmlxsltsitemap

提问by Neil Fenwick

I'm writing an XSLT template that need to output a valid xml file for an xml Sitemap.

我正在编写一个 XSLT 模板,它需要为 xml 站点地图输出一个有效的 xml 文件。

<url>
<loc>
    <xsl:value-of select="umbraco.library:NiceUrl($node/@id)"/>
</loc>
<lastmod>
    <xsl:value-of select="concat($node/@updateDate,'+00:00')"/>
</lastmod>
</url>

Unfortunately, Url that is output contains an apostrophe - /what's-new.aspx

不幸的是,输出的 Url 包含一个撇号 - /what's-new.aspx

I need to escape the ' to &apos;for google Sitemap. Unfortunately every attempt I've tried treats the string '&apos;' as if it was ''' which is invalid - frustrating. XSLT can drive me mad sometimes.

我需要转义 ' 到&apos;谷歌站点地图。不幸的是,我尝试过的每次尝试都将字符串 ' &apos;' 视为无效的 ''' - 令人沮丧。XSLT 有时会让我发疯。

Any ideas for a technique? (Assume I can find my way around XSLT 1.0 templates and functions)

对技术有什么想法吗?(假设我可以找到解决 XSLT 1.0 模板和函数的方法)

回答by Welbog

So you have 'in your input, but you need the string &nbsp;in your output?

所以你有'你的输入,但你需要&nbsp;输出中的字符串?

In your XSL file, replace &apos;with &amp;apos;, using this find/replace implementation(unless you are using XSLT 2.0):

在您的 XSL 文件中,替换&apos;&amp;apos;,使用此查找/替换实现(除非您使用的是 XSLT 2.0):

<xsl:template name="string-replace-all">
  <xsl:param name="text"/>
  <xsl:param name="replace"/>
  <xsl:param name="by"/>
  <xsl:choose>
    <xsl:when test="contains($text,$replace)">
      <xsl:value-of select="substring-before($text,$replace)"/>
      <xsl:value-of select="$by"/>
      <xsl:call-template name="string-replace-all">
        <xsl:with-param name="text" select="substring-after($text,$replace)"/>
        <xsl:with-param name="replace" select="$replace"/>
        <xsl:with-param name="by" select="$by"/>
      </xsl:call-template>
    </xsl:when>
    <xsl:otherwise>
      <xsl:value-of select="$text"/>
    </xsl:otherwise>
  </xsl:choose>
</xsl:template>

Call it this way:

这样称呼它:

<loc>
  <xsl:call-template name="string-replace-all">
    <xsl:with-param name="text" select="umbraco.library:NiceUrl($node/@id)"/>
    <xsl:with-param name="replace" select="&apos;"/>
    <xsl:with-param name="by" select="&amp;apos;"/>
  </xsl:call-template>
</loc>

The problem is &apos;is interpreted by XSL as '. &amp;apos;will be interpreted as &apos;.

问题被&apos;XSL 解释为'. &amp;apos;将被解释为&apos;.

回答by Myster

The simple way to remove unwanted characters from your URL is to change the rules umbraco uses when it generates the NiceUrl.

从 URL 中删除不需要的字符的简单方法是更改​​ umbraco 在生成 NiceUrl 时使用的规则。

Edit the config/umbracoSettings.config

编辑 config/umbracoSettings.config

add a rule to remove all apostrophes from NiceUrls like so:

添加规则以从 NiceUrls 中删除所有撇号,如下所示:

<urlReplacing>
    ...
    <char org="'"></char>     <!-- replace ' with nothing -->
    ...
</urlReplacing>

Note: The contents of the "org" attribute is replaced with the contents of the element, here's another example:

注意:“org”属性的内容替换为元素的内容,这里是另一个例子:

<char org="+">plus</char> <!-- replace + with the word plus -->

回答by Gaurav Balyan

This will work, you just need to change TWO params as given below

这将起作用,您只需要更改下面给出的两个参数

<xsl:with-param name="replace">&apos;</xsl:with-param>
<xsl:with-param name="by" >AnyString</xsl:with-param>

回答by Stephen Denne

Have you tried setting disable-output-escapingto yes for your xsl:value-of element:

您是否尝试将xsl:value-of 元素的disable-output-escaping设置为 yes:

<xsl:value-of disable-output-escaping="yes" select="umbraco.library:NiceUrl($node/@id)"/>

Actually - this is probably the opposite of what you want.

实际上 - 这可能与您想要的相反。

How about wrapping the xsl:value-of in an xsl:text element?

将 xsl:value-of 包装在 xsl:text 元素中怎么样?

<xsl:text><xsl:value-of select="umbraco.library:NiceUrl($node/@id)"/></xsl:text>

Perhaps you should try to translate 'to &amp;apos;

也许你应该试着翻译'&amp;apos;