How to remove string wrapped in double square brackets [[abc]]

XML:

   <root>
    <a>This is first para [[the first para1]]</a>
    <a>This is second para [[the second para2]]</a>
   </root>

XSLT 2.0 I tried:

   <xsl:template match='@* | node()'>
        <xsl:copy>
          <xsl:apply-templates select='@* | node()'/>
        </xsl:copy>        
  </xsl:template>
  <xsl:template match="*[contains(text(),'[[')]"></xsl:template>

Expected result:

  <root>
   <a> This is first para</a>
   <a> This is second para</a>
  </root>

Thank you in advance for help.

>Solution :

This could do it for you in XSLT 2.0. Please note that in your sample input, the second text entry is missing the finishing ‘]]’.

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
  version="2.0"
  exclude-result-prefixes="#all">
  
  <xsl:template match='@* | node()'>
    <xsl:copy>
      <xsl:apply-templates select='@* | node()'/>
    </xsl:copy>        
  </xsl:template>
  
  <xsl:template match="text()">
    <xsl:sequence select="replace(., ' [\[][\[].*\]\]', '')" />
  </xsl:template>
  
</xsl:stylesheet>

In the event that the missing ‘]]’ in the second entry is intentional, you could make the ending ‘]]’ optional in the match pattern:

  <xsl:template match="text()">
    <xsl:sequence select="replace(., '[\[][\[].*(\]\])?', '')" />
  </xsl:template>

This latter provides this output with your given input:

<root>
  <a>This is first para </a>
  <a>This is second para </a>
</root>