Hi All I have been looking for some code that will list A-Z content on my site have found the following which displays everything I need but want it to point to a specific Content Item and list child items not everything... But struggling to see how or where I can do that with the following XSLT

<xsl:param name="currentPage" />

<xsl:template match="/">
    <xsl:variable name="letters" select="umbraco.library:Split('A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z', ',')" />

    <xsl:apply-templates select="$letters" mode="list" />
</xsl:template>


<xsl:template match="value" mode="index">
    <li>
        <a href="#{text()}">
            <xsl:value-of select="text()"/>
        </a>
    </li>
</xsl:template>

<xsl:template match="values" mode="list">
    <div class="wrapper">
        <xsl:apply-templates select="value" mode="list" />
    </div>
</xsl:template>

<xsl:template match="value" mode="list">
    <xsl:variable name="letter" select="text()" />
    <xsl:for-each select="$currentPage/@*[1]">
        <xsl:if test="key('document-starts-with', $letter)">
            <div id="{$letter}">
                <h2>
                    <xsl:value-of select="$letter"/>
                </h2>
                <ul>
                    <xsl:apply-templates select="key('document-starts-with', $letter)" mode="list" />
                </ul>
            </div>
        </xsl:if>
    </xsl:for-each>
</xsl:template>

<xsl:template match="*[@isDoc]" mode="list">
    <li>
        <a href="{umbraco.library:NiceUrl(@id)}">
            <xsl:value-of select="@nodeName"/>
        </a>
    </li>
</xsl:template>