在 2 个表中，获取 DB2 中行数最多的表

Question

我有一个 DB2 数据库，其中包含表和动物之间的关系。有两张桌子，猴子和鹦鹉。 猴子长这样：

CREATE TYPE MonkeyType UNDER AnimalType AS
    (
    chimp CHAR(1)
    )
    MODE DB2SQL
    OVERRIDING METHOD food() RETURNS VARCHAR(10),
    OVERRIDING METHOD intelligence() RETURNS INT
    @

鹦鹉长这样：

CREATE TYPE ParrotType UNDER AnimalType AS
    (
    numcolors INT
    )
    MODE DB2SQL
    OVERRIDING METHOD food() RETURNS VARCHAR(10),
    OVERRIDING METHOD intelligence() RETURNS INT
    @

动物长这样：

CREATE TYPE AnimalType AS
    (
    name VARCHAR(10),
    birthyear INT
    )
    INSTANTIABLE
    MODE DB2SQL
    METHOD food() RETURNS VARCHAR(10),
    METHOD intelligence() RETURNS INT
    @

表的 DDL：

CREATE TABLE Animal OF AnimalType
        (REF IS Oid user GENERATED)@

CREATE TABLE Monkey OF MonkeyType UNDER Animal
    INHERIT SELECT PRIVILEGES@

CREATE TABLE Parrot OF ParrotType UNDER Animal
    INHERIT SELECT PRIVILEGES@

每种动物都有自己最喜欢的食物，猴子有香蕉，鹦鹉有种子。

我需要一个查询来找出鹦鹉和猴子最常吃的食物。这与询问哪张桌子的行数最多（鹦鹉或猴子）并返回最大桌子最喜欢的食物是一样的。

我试过查询：

SELECT * 
FROM (SELECT COUNT(*) AS monkey_count FROM Monkey), 
    (SELECT COUNT(*) AS parrot_count FROM Parrot)

这给了我：

MONKEY_COUNT PARROT_COUNT
------------ ------------
           5            2

但我不知道如何使用此信息来创建完整的查询。

Answer 1

select temp.food 
from (    (select distinct Oid->food() AS food, (select count(*) from Monkey) as amount from Monkey) 
              union 
          (select distinct Oid->food() AS food, (select count(*) from Parrot) as amount from Parrot)) AS temp 
where temp.amount >= all 
                ((select count(*) from monkey) union (select count(*) from Parrot))