How to find all not matching elements in XPath expression of XSL template?-CodePudding

This is my XML:

<f>
  <a>10</a>
  <a>2</a>
  <a>4</a>
  <a>13</a>
  <b>55</b>
  <b>4</b>
</f>

I'm trying to match elements <b/> which are not equal to <a/>:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
  <xsl:template match="b[. != //a]">
    <xsl:text>found!</xsl:text>
  </xsl:template>
</xsl:stylesheet>

However, it doesn't work. I believe, my expression . != //a is wrong, since //a matches only the first occurrence of <a/>. What is the right expression?

It's a test sample, please don't suggest other possible ways of solving this task. I only need a fix to the XPath expression.

CodePudding user response：

Most XPath users using e.g. . != $sequence want rather not(. = $sequence) e.g. not(. = //a) i.e. the current . item is not equal to any a element.