XSLT: Get Element Name of "Tokenized Item" - xslt

Question

Is there a way to gather element name of a tokenized value? I have been trying to do it but it is giving me an error "[Saxon-PE 9.6.0.7] XPTY0004: Required item type of first argument of name() is node(); supplied value has item type xs:string"
Here are my sample set of data:
<SET>
<REAL_TAGNAME> 1 2 3 4 </REAL_TAGNAME>
</SET>
If I have use this code:
<xsl:for-each select="SET/REAL_TAGNAME">
<xsl:for-each select="tokenize(normalize-space(.),'\s+')">
<Hardcode_Tag>
<xsl:value-of select="."/>
</Hardcode_Tag>
</xsl:for-each>
</xsl:for-each>
then I will successfully have the following:
<Hardcode_Tag>1</Hardcode_Tag>
<Hardcode_Tag>2</Hardcode_Tag>
<Hardcode_Tag>3</Hardcode_Tag>
<Hardcode_Tag>4</Hardcode_Tag>
But I want to move away from hard-coding and would like to use its original tag name to have something like:
<REAL_TAGNAME>1</REAL_TAGNAME>
<REAL_TAGNAME>2</REAL_TAGNAME>
<REAL_TAGNAME>3</REAL_TAGNAME>
<REAL_TAGNAME>4</REAL_TAGNAME>
While I try below with the xsl:element, it keeps giving me an error mentioned above:
<xsl:for-each select="SET/REAL_TAGNAME">
<xsl:for-each select="tokenize(normalize-space(.),'\s+')">
<xsl:element name="{name(.)}">
<xsl:value-of select="."/>
</xsl:element>
</xsl:for-each>
</xsl:for-each>
does anyone have any idea on how I can fix this? Thanks in advance for your help!

Answer 1

Within the xsl:for-each, the expression . refers to the current item, i.e. the current token extracted from the string value of the element. If you want to remember the name of the element that was the current element before you entered the for-each, just set a variable before the inner xsl:for-each:
<xsl:for-each select="SET/REAL_TAGNAME">
<xsl:variable name="element-name" select="name(.)"/>
<xsl:for-each select="tokenize(normalize-space(.),'\s+')">
<xsl:element name="{$element-name}">
<xsl:value-of select="."/>
</xsl:element>
</xsl:for-each>
</xsl:for-each>
In the code you show, of course, the value of $element-name will invariably be 'REAL_TAGNAME'. I'm assuming that's not true in the general case.