Majority function in SQL

问题

I am writing a SQL command to find majority of value(a value is returned as majority if its occurence in the table is more then 50%) in a table. For e.g. a Emp(name, age, dept) with values in table: (32, 33, 45, 45 ,45 21, 45)

So if user wants to get majority value of field age, he will run the command:

SELECT majority(age) FROM Emp

Now at the backend the corresponding SQL command generated will be like:

SELECT age FROM Emp GROUP BY age HAVING COUNT(*) > ((SELECT COUNT(*) FROM Emp) / 2);

This query returns 45 as result.

But for a user query like:

SELECT majority(age), dept FROM Emp GROUP BY dept

Then I am not sure how to create similar query with group by feature, or should I create some other query? Thanks in advance

回答1:

(Assuming you are using Oracle - because in MySQL you can't write your own aggregates anyway)

I think you can do this with window functions:

This is the equivalent to the first example:

select distinct age
from (
  select age, 
         dept,
         count(*) over () as total_count, 
         count(*) over (partition by age) as age_count
  from emp
) 
where age_count >= total_count / 2;

By simply adding the department to the partition (group) definition this should give you what you want:

select distinct age
from (
  select age, 
         dept,
         count(*) over (partition by dept) as dept_count, 
         count(*) over (partition by dept, age) as age_count
  from emp
) 
where age_count >= dept_count / 2;

回答2:

What you are asking to perform is termed creating a user-defined aggregate function. This is a not a trivial task, but here is a tutuorial for doing so in Oracle.

User-defined aggregate functions in oracle 9i

来源：https://stackoverflow.com/questions/22533290/majority-function-in-sql

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sql

plsql

group-by

oracle-sqldeveloper