如何用一个空格替换空格序列但不在 XSLT 中修剪?

问题描述：

函数 normalize-space 删除前导和尾随空格，并用单个空格替换空格字符序列.如何在 XSLT 1.0 中仅用单个空格替换空白字符序列?例如，"..xy..\n\t..z."(空格替换为一个点以提高可读性)应该变成 ".xyz".

The function normalize-space removes leading and trailing whitespace and replaces sequences of whitespace characters by a single space. How can I only replaces sequences of whitespace characters by a single space in XSLT 1.0? For instance "..x.y...\n\t..z." (spaces replaced by a dot for readability) should become ".x.y.z.".

答

如果没有 Becker 的方法，您可以使用一些不鼓励字符作为标记:

Without Becker's method, you could use some discouraged character as mark:

translate(normalize-space(concat('&#x7F;',.,'&#x7F;')),'&#x7F;','')

注意:三个函数调用...

或者使用任何字符但重复某些表达式:

Or with any character but repeating some expression:

substring(
   normalize-space(concat('.',.,'.')),
   2,
   string-length(normalize-space(concat('.',.,'.'))) - 2
)

在 XSLT 中，您可以轻松地声明一个变量:

In XSLT you can easily declare a variable:

<xsl:variable name="vNormalize" select="normalize-space(concat('.',.,'.'))"/>
<xsl:value-of select="susbtring($vNormalize,2,string-length($vNormalize)-2)"/>

如何用一个空格替换空格序列但不在 XSLT 中修剪?

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