每天为每个用户获取最后一次注销（MAX（），GROUP BY）[postgresql]

Question

我有很多用户的多次注销。我想获取每个用户每天的最后一次注销。

SELECT t_logouts.log_id,
t_users.last_name || ' ' || t_users.first_name,
t_logouts.log_date,
MAX(t_logouts.log_time),
(DATE_PART('hour', '18:00:00'::time - t_logouts.log_time ) * 60 +
 DATE_PART('minute', '18:00:00'::time - t_logouts.log_time )) / 60
FROM t_logouts
LEFT JOIN t_users
ON t_users.user_id = t_logouts.user_id
GROUP BY DATE(t_logouts.log_date), t_users.user_id
ORDER BY t_logouts.log_date  DESC, t_logouts.log_time  ASC ;

抛出错误：

LINE 1: SELECT t_logouts.log_id,
               ^
SQL state: 42803
Character: 8

据我了解，多重聚合存在一些问题。我试过使用本手册： https://www.postgresqltutorial.com/postgresql-group-by/ https://www.postgresqltutorial.com/postgresql-max-function/

我正在尝试使用的表。

CREATE TABLE t_users (
    user_id SERIAL PRIMARY KEY,
    nickname name UNIQUE not null check (length(nickname) < 64),
    first_name VARCHAR NOT NULL check (length(first_name) < 64),
    last_name VARCHAR NOT NULL check (length(last_name) < 64),
    email VARCHAR DEFAULT NULL check (email ~* '^.+@.+\..+$' )
);
CREATE TABLE t_logins (
    log_id SERIAL PRIMARY KEY,
    user_id INTEGER NOT NULL,
        FOREIGN KEY (user_id)
        REFERENCES t_users (user_id)
        ON UPDATE CASCADE ON DELETE RESTRICT,
    log_date DATE NOT NULL,
    log_time TIME NOT NULL
);
CREATE TABLE t_logouts (
    log_id SERIAL PRIMARY KEY,
    user_id INTEGER NOT NULL,
        FOREIGN KEY (user_id)
        REFERENCES t_users (user_id)
        ON UPDATE CASCADE ON DELETE RESTRICT,
    log_date DATE NOT NULL,
    log_time TIME NOT NULL
);

Answer 1

这通常使用 Postgres 中的 distinct on () 完成：

SELECT distinct on (t_users.user_id) 
         t_logouts.log_id,
         t_users.last_name || ' ' || t_users.first_name,
         t_logouts.log_date,
         t_logouts.log_time,
         extract(epoch from '18:00:00'::time - t_logouts.log_time) / 3600
FROM t_logouts
  LEFT JOIN t_users ON t_users.user_id = t_logouts.user_id
ORDER BY t_users.user_id, t_logouts.log_date DESC;

每天为每个用户获取最后一次注销（MAX（），GROUP BY）[postgresql]

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