sql - 计算 NULL 和非 NULL 行的巧妙方法？

Question

我一直在用子选择和一些聚合函数计算 NULL 和非 NULL 列

CREATE TEMPORARY TABLE citizens(name text, country text,profession text,postalcode text);

INSERT INTO citizens VALUES
 ('Fred', 'USA', 'Professor', NULL),
 ('Amy', 'USA', 'Professor', NULL),
 ('Ted', 'USA', 'Professor', 90210),
 ('Barb', 'USA', 'Lawyer', 10248),
 ('Wally', 'USA', 'Lawyer', NULL),
 ('Fred', 'Canada', 'Professor', 'S0H'),
 ('Charles', 'Canada', 'Professor', 'S4L'),
 ('Nancy', 'Canada', 'Lawyer', NULL),
 ('Linda', 'Canada', 'Professor', NULL),
 ('Steph', 'France', 'Lawyer', 75008 ),
 ('Arnold', 'France', 'Lawyer', 75008 ),
 ('Penny', 'France', 'Lawyer', 75008 ),
 ('Harry', 'France', 'Lawyer', NULL);

SELECT country,
    profession,
    MAX(have_postalcode::int*num) AS num_have,
    MAX((1-have_postalcode::int)*num) AS num_not_have
FROM
    (
    SELECT country, profession,
        COUNT(*) AS num,
        (postalcode IS NOT NULL) AS have_postalcode
    FROM citizens
    GROUP BY country, profession, have_postalcode
    ) AS d
GROUP BY country, profession

结果

USA     Professor   1   2
Canada  Lawyer      0   1
USA     Lawyer      1   1
France  Lawyer      3   1
Canada  Professor   2   1

但似乎应该有一种更巧妙的方法(例如，MAX 仅用于获取一个非平凡的值，这让我很痛苦)。有人有什么好主意吗？

Answer 1

SELECT country, profession,
        COUNT(postalcode) AS num_have
      , (COUNT(*) - COUNT(postalcode)) AS num_not_have
FROM citizens
GROUP BY country, profession;

http://sqlfiddle.com/#!1/17a9d/15