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	<title>
	Comments on: SQL Puzzle: SQL Advance Query &#8211; Report on Manager &#038; Employee Nth level hierarchy	</title>
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	<link>https://www.dbrnd.com/2019/05/sql-puzzle-sql-advance-query-report-on-manager-employee-nth-level-hierarchy/</link>
	<description>BIGData &#124; NoSQL &#124; MSSQL &#124; MySQL &#124; PostgreSQL</description>
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		<title>
		By: koustav		</title>
		<link>https://www.dbrnd.com/2019/05/sql-puzzle-sql-advance-query-report-on-manager-employee-nth-level-hierarchy/#comment-7861</link>

		<dc:creator><![CDATA[koustav]]></dc:creator>
		<pubDate>Wed, 30 Oct 2019 10:17:48 +0000</pubDate>
		<guid isPermaLink="false">https://www.dbrnd.com/?p=5865#comment-7861</guid>

					<description><![CDATA[select * from 
(select empname as EmpName,empid as Hierarchy ,empname as FullHierarchyName  from Employees where empname not in(select emp1.empname  from Employees emp1,Employees emp2
where emp1.managerid=emp2.empid)
union all

select emp1.empname,emp1.empid,emp3.empname &#124;&#124;&#039;..&#039;&#124;&#124;emp2.empname&#124;&#124;&#039;..&#039;&#124;&#124;emp1.empname FullHierarchyName  from Employees emp1,Employees emp2,Employees emp3
where emp1.managerid=emp2.empid
and emp2.managerid=emp3.empid 
union 
select emp1.empname, emp1.empid,emp2.empname&#124;&#124;&#039;..&#039;&#124;&#124;emp1.empname FullHierarchyName  from Employees emp1,Employees emp2
where emp1.managerid=emp2.empid and emp1.empname  not in(select emp1.empname  from Employees emp1,Employees emp2,Employees emp3
where emp1.managerid=emp2.empid
and emp2.managerid=emp3.empid )

)T1
order by 3]]></description>
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(select empname as EmpName,empid as Hierarchy ,empname as FullHierarchyName  from Employees where empname not in(select emp1.empname  from Employees emp1,Employees emp2<br />
where emp1.managerid=emp2.empid)<br />
union all</p>
<p>select emp1.empname,emp1.empid,emp3.empname ||&#8217;..&#8217;||emp2.empname||&#8217;..&#8217;||emp1.empname FullHierarchyName  from Employees emp1,Employees emp2,Employees emp3<br />
where emp1.managerid=emp2.empid<br />
and emp2.managerid=emp3.empid<br />
union<br />
select emp1.empname, emp1.empid,emp2.empname||&#8217;..&#8217;||emp1.empname FullHierarchyName  from Employees emp1,Employees emp2<br />
where emp1.managerid=emp2.empid and emp1.empname  not in(select emp1.empname  from Employees emp1,Employees emp2,Employees emp3<br />
where emp1.managerid=emp2.empid<br />
and emp2.managerid=emp3.empid )</p>
<p>)T1<br />
order by 3</p>
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