XSLT select all text except one child node text

Question

I have xml like this:

<article>
   <title> Test title - <literal> Compulsory - </literal> <fn> ABC </fn> 
   <comments> a comment</comments>
   </title>
</article>

I want to get all child node + self text in a variable e.g.

$full_title = "Test title - Compulsory - ABC"

Except comments node text.

Following is my unsuccessful try where i miss title node text.

<xsl:template name="test">
    <xsl:variable name="full_title" select="article/title/*[not(self::comments)][1]" />
    <xsl:variable name="width" select="45" /> 
                <xsl:choose>
                    <xsl:when test="string-length($full_title) &gt;    $width">
                        <xsl:value-of select="concat(substring($full_title,1,$width),'..')"/>
                    </xsl:when>
                    <xsl:otherwise>
                        <xsl:value-of select="$full_title"/>    
                    </xsl:otherwise>
            </xsl:choose>
</xsl:template>

Answer 1

Change * to node(). That will select both elements and text nodes that are children of the <title> element. Then take out the [1] since you want all children of <title>:

<xsl:variable name="full_title"
    select="string-join(article/title/node()[not(self::comments)], '')" />

A more reliable way to do it, so that you won't get tripped up if you have multiple levels under <title> and <comments> elements occur as grandchildren, would be this:

<xsl:variable name="full_title"
    select="string-join(article/title//text()[not(ancestor::comments)], '')" />

Update:

Since you want the variable to hold a string value, and since you're passing it to functions like concat() and string-length() which cannot take a sequence of multiple nodes as a first argument, using string-join(..., '') around the sequence converts it to a string by concatenating the string values of each node.