Xslt 根据上下文选择子字符串

我有这段XML，希望在

章节之后立即获得编号

<para>Insolvency Rules, r 12.12, which gives the court a broad discretion, unfettered by the English equivalent of the heads of Order 11, r 1(1) (which are now to be found, in England, in CPR, Chapter 6, disapplied in the insolvency context by Insolvency Rules, r 12.12(1)). </para>

但是我只想要
6

请告诉我该怎么做

我不想用这种说法

<xsl:value-of select="substring-before(substring-after(current(),'Chapter'), ,',')"/>



因为每种情况下的章节号都会有所不同，介于1和15之间。
尝试以下方法：

    <xsl:variable name="vS" select="concat(substring-after(current(),'Chapter '),'Z')"/>
    <xsl:value-of select=
           "substring-before(translate($vS,translate($vS,'0123456789',''),'Z'),'Z')"/>



这是基于：感谢
@迪米特里·诺瓦切夫

更新：如果“章节”后的空间数量未知，您可以使用以下内容：

<xsl:variable name="vS" select="concat(substring-after(current(),'Chapter'),'Z')"/>

<xsl:value-of select=
        " translate(
        substring-before(translate($vS,translate($vS,' 0123456789',''),'Z'),'Z')
     , ' ','')"/>



    <xsl:variable name="vS" select="concat(substring-after(current(),'Chapter '),'Z')"/>
    <xsl:value-of select=
           "substring-before(translate($vS,translate($vS,'0123456789',''),'Z'),'Z')"/>

<xsl:variable name="vS" select="concat(substring-after(current(),'Chapter'),'Z')"/>

<xsl:value-of select=
        " translate(
        substring-before(translate($vS,translate($vS,' 0123456789',''),'Z'),'Z')
     , ' ','')"/>