xsl:for-each inside xsl:for-each in table rows and columns - xslt

Question

I've just started learning XML/XSL and I've hit a roadblock in one of my assignments. Tried Googling and searching over here but I can't seem to find a question that has a solution that is basic. so what I am trying is to display rows of bucket-type and room-types associated with it. can somebody please help
<list-inventory list-count="2">
<list list-type="Standard" list-order = "1" count-Types = "3">
<types type="BEN2D"></room>
<types type="BESH2D"></room>
<types type="HNK"></room>
</list>
<list list-type="Deluxe" list-order = "2" count-Types = "3">
<types type="SNK"></room>
<types type="TESTKD"></room>
<types type="TESTKD"></room>
</list>
<list-inventory>
I want table as below
Standard | Deluxe
BEN2D |SNK
BESH2D |TESTKD
HNK |TESTKD
I tried below xsl code but i see all list-type in single column and only 1st is being printing for all list-type:
<xsl:for-each select="/contents/list-inventory/list">
<tr>
<td class="alt-th" style="border:1px solid black">
<xsl:value-of select="#list-type"/>
</td>
</tr>
<tr>
<td style="border:1px solid black">
<xsl:for-each select="/contents/list-inventory/list/types">
<span><xsl:value-of select="#type"/></span>
<xsl:if test="position()!=last()">
<br/>
</xsl:if>
</xsl:for-each>
</td>
</tr>
</xsl:for-each>
Can someone help me with xsl:for-each inside a xsl:for-each

Answer 1

It's too bad you did not post your expected result as code. I would assume that you want a separate row for each pair of values. As I stated in the comments, this is far from being trivial.
However, you could make it simpler if you are willing to settle for a single data row, where each cell contains all the values of the corresponding list (your attempt seems to suggest that this is what you actually tried to accomplish).
So, given a well-formed XML input:
XML
<list-inventory list-count="2">
<list list-type="Standard" list-order = "1" count-Types = "3">
<types type="BEN2D"/>
<types type="BESH2D"/>
<types type="HNK"/>
</list>
<list list-type="Deluxe" list-order = "2" count-Types = "3">
<types type="SNK"/>
<types type="TESTKD"/>
<types type="TESTKD"/>
</list>
</list-inventory>
you could do simply:
XSLT 1.0
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:template match="/list-inventory">
<table border="1">
<!-- header -->
<tr>
<xsl:for-each select="list">
<th>
<xsl:value-of select="#list-type"/>
</th>
</xsl:for-each>
</tr>
<!-- body -->
<tr>
<xsl:for-each select="list">
<td>
<xsl:for-each select="types">
<xsl:value-of select="#type"/>
<br/>
</xsl:for-each>
</td>
</xsl:for-each>
</tr>
</table>
</xsl:template>
</xsl:stylesheet>
to get:
Result
<?xml version="1.0" encoding="UTF-8"?>
<table border="1">
<tr>
<th>Standard</th>
<th>Deluxe</th>
</tr>
<tr>
<td>BEN2D<br/>BESH2D<br/>HNK<br/>
</td>
<td>SNK<br/>TESTKD<br/>TESTKD<br/>
</td>
</tr>
</table>
which would render as:

Answer 2

I'm not sure how general you want your solution to be (i.e what inputs does it have to handle other than the example shown), but I would do something like:
<xsl:template match="list-inventory">
<xsl:variable name="list2" select="list[2]/types"/>
<xsl:for-each select="list[1]/types">
<xsl:variable name="position" select="position()"/>
<tr>
<td><xsl:value-of select="#type"/></td>
<td><xsl:value-of select="$list2[$position]/#type"/></td>
</tr>
</xsl:for-each>
</xsl:template>
Both the variables here are needed to avoid problems with context: the effect of an XPath expression depends on the context at the time it is evaluated, so you can evaluate a variable to capture information at the time you're in the right context.