First, the query will return the nth
lowest salary value. To return the nth
highest salary value you must change t.sal <= sal
to t.sal >= sal
.
Next, this query works by first finding the distinct list of salary values as one derived table and then determines the number of employees that have a salary less than each one in this list. t.sal <= sal
is taking the derived table (which most databases would require have an alias) and comparing each value against the outer emp
table. It should be noted that this will return multiple rows in the case of a tie.
To manually trace the output, we need some inputs:
Alice | 200
Bob | 100
Charlie | 200
Danielle | 150
Select Distinct sal
From emp
Gives us
200
100
150
Now we analyze each row in the outer table
Alice - There are 3 distinct salary values less than or equal to 200
Bob - 1 rows <= 100
Charlie - 3 rows <= 200
Danielle - 2 row <= 150
Thus, for each salary value we get the following counts (and reordered by count):
Bob 1
Danielle 2
Charlie 3
Alice 3
The most important aspect that I think you are overlooking is that the outer emp
table is correlated to the inner count calculation (which is why it is called a correlated subquery). I.e., for each row in the outer emp
table, a new count is calculated for that row's salary via t.sal <= sal
. Again, most database systems would require the inner most query to have an alias like so (note the As Z
alias):
Select sal
From emp As t
Where &n = (
Select Count(Z.sal)
From (
Select Distinct sal
From emp
) As Z
Where t.sal <= Z.sal
)
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