Editing Relational algebra (section)

==== Full outer join ====

The '''outer join''' (⟗) or '''full outer join''' in effect combines the results of the left and right outer joins.

The full outer join is written as ''R'' ⟗ ''S'' where ''R'' and ''S'' are [[relation (database)|relation]]s.{{efn|In [[Unicode]], the Full Outer join symbol is ⟗ (U+27D7).}} The result of the full outer join is the set of all combinations of tuples in ''R'' and ''S'' that are equal on their common attribute names, in addition to tuples in ''S'' that have no matching tuples in ''R'' and tuples in ''R'' that have no matching tuples in ''S'' in their common attribute names.{{citation needed|date=April 2022}}

For an example consider the tables ''Employee'' and ''Dept'' and their full outer join:

{{col-begin|width=auto; margin:0.5em auto}}
{{col-break}}
{| class="wikitable"
 |+           ''Employee''
 |-
 ! Name   !! EmpId !! DeptName
 |-
 | Harry  || 3415  || Finance
 |-
 | Sally  || 2241  || Sales
 |-
 | George || 3401  || Finance
 |-
 | Harriet || 2202 || Sales
 |-
 | Tim || 1123 || Executive
|}
{{col-break|gap=2em}}
{| class="wikitable"
 |+           ''Dept''
 |-
 ! DeptName   !! Manager
 |-
 | Sales  || Harriet
 |-
 | Production || Charles
|}
{{col-break|gap=2em}}
{| class="wikitable"
 |+  ''Employee'' ⟗  ''Dept''
 |-
 ! Name   !! EmpId !! DeptName !! Manager
 |-
 | Harry  || 3415  || Finance || ω
 |-
 | Sally  || 2241   || Sales || Harriet
 |-
 | George || 3401  || Finance || ω
 |-
 | Harriet || 2202 || Sales || Harriet
 |-
 | Tim || 1123 || Executive || ω
 |-
 | ω || ω || Production  || Charles
|}
{{col-end}}

In the resulting relation, tuples in ''R'' which have no common values in common attribute names with tuples in ''S'' take a ''null'' value, ''ω''. Tuples in ''S'' which have no common values in common attribute names with tuples in ''R'' also take a ''null'' value, ''ω''.

The full outer join can be simulated using the left and right outer joins (and hence the natural join and set union) as follows:

:''R'' ⟗ ''S'' = (''R'' ⟕ ''S'') &cup; (''R'' ⟖ ''S'')