Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

SQL: prevent duplicate generation for repeated aggs #33252

Merged
merged 3 commits into from
Aug 30, 2018
Merged
Show file tree
Hide file tree
Changes from 1 commit
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
Original file line number Diff line number Diff line change
Expand Up @@ -112,6 +112,9 @@ public Aggs addGroups(Collection<GroupByKey> groups) {
}

public Aggs addAgg(LeafAgg agg) {
if (metricAggs.contains(agg)) {
return this;
}
return new Aggs(groups, combine(metricAggs, agg), pipelineAggs);
}

Expand Down
10 changes: 9 additions & 1 deletion x-pack/qa/sql/src/main/resources/agg.sql-spec
Original file line number Diff line number Diff line change
Expand Up @@ -394,4 +394,12 @@ SELECT MIN(salary) min, MAX(salary) max, gender g, languages l, COUNT(*) c FROM
aggMultiWithHavingOnCount
SELECT MIN(salary) min, MAX(salary) max, gender g, COUNT(*) c FROM "test_emp" WHERE languages > 0 GROUP BY g HAVING c > 40 ORDER BY gender;
aggMultiGroupByMultiWithHavingOnCount
SELECT MIN(salary) min, MAX(salary) max, gender g, languages l, COUNT(*) c FROM "test_emp" WHERE languages > 0 GROUP BY g, languages HAVING c > 40 ORDER BY gender, languages;
SELECT MIN(salary) min, MAX(salary) max, gender g, languages l, COUNT(*) c FROM "test_emp" WHERE languages > 0 GROUP BY g, languages HAVING c > 40 ORDER BY gender, languages;

// repetion of same aggs to check whether the generated query contains duplicates or not
aggRepeatFunctionAcrossFields
SELECT MIN(emp_no) AS a, 1 + MIN(emp_no) AS b, ABS(MIN(emp_no)) AS c FROM test_emp;
aggRepeatFunctionBetweenSelectAndHaving
SELECT gender, COUNT(DISTINCT languages) AS c FROM test_emp GROUP BY gender HAVING count(DISTINCT languages) > 0 ORDER BY gender;