Postgres SQL 查询按性别分组。如何显示行不存在的值？

Question

我有一个查询计算每天发送的邮件总数，并按性别分组。

如果某天没有消息发送给男性，则不会有男性行。

在这种情况下，如何为男性插入计数值 0？

查询：

SELECT d.date dated, count(se.id), gender
FROM    (
        select to_char(date_trunc('day', (current_date - offs)), 'YYYY-MM-DD') AS date
        FROM generate_series(0, 365, 1) AS offs
        ) d
    LEFT OUTER JOIN (
        SELECT id, customer_id, client_report.insert_time, profile.gender, profile.house_income, profile.address AS postcode, profile.age AS age_group, profile.is_employed, profile.is_married, profile.no_children, profile.no_cars, profile.shopping_frequency
        FROM common.client_report
        JOIN common.profile
        ON client_report.profile_id = profile.uuid 
        WHERE sms_status = 'SUCCESS' 
        ) se
ON (d.date=to_char(date_trunc('day', se.insert_time), 'YYYY-MM-DD'))
WHERE customer_id = 25::int AND d.date::date BETWEEN '2017-07-03'::date AND '2017-08-01'::date
GROUP BY d.date, gender
ORDER BY d.date ASC

我希望结果一致，比如：

    day, count, gender
    2017-07-01, 10, Female
    2017-07-01, 5, Male
    2017-07-02, 0, Female
    2017-07-02, 8, Male

Answer 1

您应该 cross join 日期列表和性别列表。

类似于：

SELECT d.date dated, count(se.id), d.gender
FROM    (
        select to_char(date_trunc('day', (current_date - offs)), 'YYYY-MM-DD') AS date, gender
        FROM generate_series(0, 365, 1) AS offs
        CROSS JOIN (VALUES('Male'), ('Female')) AS genders(gender)
        ) d
    LEFT OUTER JOIN (
        SELECT id, customer_id, client_report.insert_time, profile.gender, profile.house_income, profile.address AS postcode, profile.age AS age_group, profile.is_employed, profile.is_married, profile.no_children, profile.no_cars, profile.shopping_frequency
        FROM common.client_report
        JOIN common.profile
        ON client_report.profile_id = profile.uuid 
        WHERE sms_status = 'SUCCESS' 
        ) se
ON (d.date=to_char(date_trunc('day', se.insert_time), 'YYYY-MM-DD')) AND d.gender = se.gender
WHERE customer_id = 25::int AND d.date::date BETWEEN '2017-07-03'::date AND '2017-08-01'::date
GROUP BY d.date, d.gender
ORDER BY d.date ASC

Answer 2

为了简单起见，我们假设您使用 genders:

创建了一个 table

CREATE TABLE genders
(
    gender text primary key
) ;

INSERT INTO genders (gender)
VALUES
   ('MALE'),
   ('FEMALE') ;

如果您想要所有 dates/genders 的答案，您需要 cross join 所有可能的日期和性别，然后 LEFT JOIN 使用实际回答您问题的数据:

SELECT
    calendar.cal_date AS date, coalesce(gender_count, 0) AS gender_count, genders.gender
FROM
    -- Virtual table with all dates/genders
    genders
    CROSS JOIN
    (SELECT 
        generate_series(/*now()*/ date '2017-07-10' - interval '7 days' /* or '1 year' */, 
                        /*now()*/ date '2017-07-10', 
                        interval '1 day')::date AS cal_date
    ) AS calendar

    -- The table with the actual data
    LEFT JOIN
    (
    SELECT
        date, gender, count(gender) AS gender_count
    FROM
        client_report
        JOIN profile ON profile.uuid = client_report.profile_id
    WHERE
        -- conditions to actually select data
        customer_id = 25
        AND sms_status = 'SUCCESS'
    GROUP BY
        date, gender
    ) AS counts 
        ON counts.date = calendar.cal_date AND counts.gender = genders.gender
ORDER BY
    calendar.cal_date, genders.gender ;

您会得到如下答案：

date       | gender_count | gender
:--------- | -----------: | :-----
2017-07-03 |            1 | FEMALE
2017-07-03 |            2 | MALE  
2017-07-04 |            2 | FEMALE
2017-07-04 |            0 | MALE  
2017-07-05 |            0 | FEMALE
2017-07-05 |            0 | MALE  
2017-07-06 |            0 | FEMALE
2017-07-06 |            2 | MALE  
2017-07-07 |            2 | FEMALE
2017-07-07 |            0 | MALE  
2017-07-08 |            0 | FEMALE
2017-07-08 |            1 | MALE  
2017-07-09 |            0 | FEMALE
2017-07-09 |            0 | MALE  
2017-07-10 |            0 | FEMALE
2017-07-10 |            0 | MALE

请注意，我没有用一整年，而是只用了 8 天。

您可以在 dbfiddle here

查看带有示例数据的完整示例

注意：我忽略了insert_date，只关注日期，因为我不知道每个都有哪个确切的含义。我认为你的观点是如何使用交叉连接，其余的与场景无关。

Postgres SQL 查询按性别分组。如何显示行不存在的值？

Postgres SQL query grouped by gender. How to show values where a row does not exist?

postgresql

calendar

group-by

not-exists

cross-join