I made up this weird example trying to illustrate what I want to do (it's kind of stupid, but bear with me):
Consider the following table:
EMPLOYEES
married, certified and religious are just boolean fields (in case of Oracle, they are of type NUMBER(1,0)).
I need to come up with SQL that displays for each hire_year, count of married, certified and religious employees within the following salary categories:
- A
SALARY > 2000
- B
SALARY BETWEEN 1000 AND 2000
- C
SALARY < 1000
Based on the above dataset, here is what I expect to get:
So far, I've only come up with the following SQL:
SELECT
COUNT(CASE WHEN married = 1 THEN 1 END) as MARRIED,
COUNT(CASE WHEN certified = 1 THEN 1 END) as certified,
COUNT(CASE WHEN religious = 1 THEN 1 END) as religious,
hire_year
FROM employees
GROUP BY hire_year;
The result of executing this SQL is:
Which is almost what I need, but I also need to divide these counters further down into the groups based on a salary range.
I guess that some analytic function, that divides groups into the buckets based on some SQL expression would help, but I can't figure out which one. I tried with NTILE, but it expects a positive constant as a parameter, rather than an SQL expression (such as SALARY BETWEEN X and Y
).
Nope, no need for analytic functions; they're difficult to have in the same query as an aggregate function anyway.
You're looking for the case
statement again, you just have to put it in the GROUP BY.
select hire_year
, sum(married) as married
, sum(certified) as certified
, sum(religious) as religious
, case when salary > 2000 then 'A'
when salary >= 1000 then 'B'
else 'C' end as salary_class
from employees
group by hire_year
, case when salary > 2000 then 'A'
when salary >= 1000 then 'B'
else 'C' end
Note that I've changed your count(case when...)
to sum()
. This is because you're using a boolean 1/0 so this'll work in the same manner but it's a lot cleaner.
For the same reason I've ignored your between
in your salary calculation; there's no particular need for it as if the salary is greater than 2000 the first CASE has already been fulfilled.
来源:https://stackoverflow.com/questions/12331800/apply-count-function-on-a-subgroup-of-groups