Re: Transforming XML to XML (diff. format) using XSL

                    <xsl:sort select="bug_id"/>

That would select the bug_id element  child from the current field
element.
However you don't have such an element, what you do have is

   <field name="bug_id">1</field>

which would be field[@name='bug_id'] from the parent row element.

you have this xsl:sort in the xsl:for-each that is iterating over
fields, but i suspect that you want it on the outer xsl:for-each
that is iterating over the rows, to sort teh rows into bug id order.

Actually I don't think you want the inner for-each at all.

something like

 <xsl:for-each select="mysqldump/database/table/row"> 
 <xsl:sort select="field[@name='bug_id']"/>
 <xsl:if test= "field[@name='status_id']='7']">
                         <xsl:if test="field[@name='resolution']='1']">
                           <ol>
                           <li><xsl:value-of select="field[@name='title']"/></li>
                           </ol>
                       </xsl:if>
                     </xsl:if>
...

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