SQL query that summarizes each group - sql

Question

I need help with this. I need to summarize each group of tests. The result should look like this:
name_of_the_group | all_test_cases | passed_test_cases | total_value
numerical stability 4 4 80
memory usage 3 2 20
corner cases 0 0 0
performance 2 0 0
columns:
name_of_the_group (all groups need to be displayed even if there are no corresponding tests)
all_test_cases (number of tests in the group)
passed_test_cases (number of test cases with the status OK)
total_value (total value of passed tests in this group)
Here is the fiddle example: http://sqlfiddle.com/#!17/e96109

Answer 1

Below SQL query provide a solution to all test cases
When there is a name entry in table test_groups but missing in test_cases.
Order by test_value desc first and then by name lexicographically increasing
order.
SELECT CASE
WHEN g.name = c.group_name THEN c.group_name
ELSE g.name
END AS name,
count(c.group_name) AS all_test_cases,
count(CASE
WHEN c.status = 'OK' THEN 1
END) AS all_test_cases,
sum(CASE
WHEN c.status = 'OK' THEN g.test_value
ELSE 0
END) AS total_value
FROM test_groups g
LEFT JOIN test_cases c ON g.name = c.group_name
GROUP BY CASE
WHEN g.name = c.group_name THEN c.group_name
ELSE g.name
END
ORDER BY total_value DESC,
CASE
WHEN g.name = c.group_name THEN c.group_name
ELSE g.name
END

Answer 2

simple conditional aggregation:
select
t.name,
count(*) as total,
sum(case when c.status = 'OK' then 1 else 0 end) as passed,
sum(case when c.status = 'OK' then test_value end) as score
from
test_groups t
left join test_cases c
on t.name = c.group_name
group by t.name
Fiddle

Answer 3

with all_test_cases as (select name, count(id) as cnt from (
select a.name, a.test_value, b.id, b.status from test_groups a
left join test_cases b on a.name = b.group_name) total_test
group by name),
passed_test_cases as (
select name, count(name) as cnt from (
select a.name, a.test_value, b.id, b.status from test_groups a
join test_cases b on a.name = b.group_name
and b.status = 'OK') OK_test
group by name)
select tg.name, atc.cnt as all_test_cases , coalesce(ptc.cnt, 0) as passed_test_cases, coalesce(ptc.cnt * tg.test_value, 0) as total_value
from test_groups tg
left join all_test_cases atc on tg.name = atc.name
left join passed_test_cases ptc on tg.name = ptc.name
order by total_value desc, name

Answer 4

Select t.name,
sum(case when c.status = 'OK' or c.status = 'ERROR' then 1 else 0 end) as all_test_cases,
sum(case when c.status = 'OK' then 1 else 0 end) as passed_test_cases,
sum(case when c.status = 'OK' then 1 else 0 end) * t.test_value as total_value
from
test_groups t
left join test_cases c
on t.name = c.group_name
group by t.name,t.test_value
order by ,t.total_value,t.name

Answer 5

WITH A AS (SELECT group_name AS name, COUNT(status) AS all_test_cases,
SUM(CASE
WHEN status = 'OK' THEN 1
ELSE 0
END) AS passed_test_cases
FROM test_cases
GROUP BY group_name)
SELECT g.name, IFNULL(c.all_test_cases,0),
IFNULL(c.passed_test_cases,0),
IFNULL(g.test_value*c.passed_test_cases,0) AS total_value
FROM test_groups AS g
LEFT JOIN A AS c
ON g.name = c.name
GROUP BY g.name
ORDER BY total_value DESC;

Answer 6

Select Name Name_Of_The_Group ,Count(Status) All_Test_Cases,
Sum(Case When Status = 'OK' Then 1 Else 0 End) passed_test_cases ,
Sum(Case When Status = 'OK' Then test_value End) total_value
From Test_Cases Full Outer Join Test_Groups On Name = Group_Name
Group By Group_Name,Name
order by Name_Of_The_Group;

Answer 7

Try the following query:
select t.name,
count(c.*)as total,
sum(case when c.status='OK'then 1 else 0 end) as passed,
Case when sum(case when c.status='OK'then test_value end)Is NULL then 0
Else sum(case when c.status='OK'then test_value end)End as total_value
from test_groups t
left join test_cases c on t.name=c.group_name
group by t.name
order by total_value desc,t.name;