[CALCITE-2914] Improve how LatticeSuggester deduces foreign keys

• Type: Bug
• Status: Closed
• Priority: Major
• Resolution: Fixed
• Fix Version/s: 1.20.0
• Component/s: core
• Labels: pull-request-available
• Assignee: Julian Hyde
• Reporter: Julian Hyde
• Created: 12/Mar/19 00:48
• Updated: 27/Oct/19 19:38
• Resolved: 22/Apr/19 21:05

Description

Improve how LatticeSuggester deduces foreign keys.

Currently LatticeSuggester uses SqlStatisticProvider and deduces that a join between two tables is "many to one" if the left table has more rows than the right table. Clearly this is wrong, because it does not take into account the keys of the join.

I think we will change the method

double tableCardinality(List<String> qualifiedTableName)

to

double tableCardinality(RelOptTable table)

and add a method

boolean isForeignKey(RelOptTable fromTable, List<String> fromColumns,
   RelOptTable toTable, List<String> toColumns)

For example, isForeignKey(empTable, ["DEPTNO"], deptTable, ["DEPTNO"]) would return true.

One valid implementation might look at the catalog for foreign key constraints; another might execute a query to see whether every row in emp is matched by exaclty one row in dept. Here is a query that might do it:

select emp.deptno
from (select distinct deptno from emp) as emp
left join dept using (deptno)
group by emp.deptno
having count(dept.deptno) != 1

If the query returns any rows it is not a foreign key.

Issue Links

• relates to: CALCITE-3449 Sync the table name logic from TableScan into the TableModify
• links to: GitHub Pull Request #1141