6

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式（例如，它让我很痛苦，它MAX只是用来获取一个不平凡的价值）。有没有人有一个很酷的想法？

4

3 回答 3

8 
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

于 2012-10-11T17:05:04.817 回答
4 
SELECT  country
,       profession
,       sum(case when postalcode is not null then 1 end) as num_have
,       sum(case when postalcode is null then 1 end) as num_not_have
FROM    citizens
GROUP BY 
        country
,       profession
于 2012-10-11T16:58:10.867 回答
4 
SELECT Country, Profession, 
    count(Country) as num_have, count(*) - count(PostalCode) as num_not_have
FROM citizens
GROUP BY Country, Profession
于 2012-10-11T16:59:03.860 回答