【发布时间】:2014-09-16 08:41:13
【问题描述】:
假设一个主“作业”表和两个对应的“日志”表(一个用于服务器事件,另一个用于用户事件,每个表中存储的数据完全不同)。
从两个“日志”表(如果有的话)中返回选择的“作业”记录和最新的相应日志记录(具有多个字段)的最佳方式是什么。
确实从MySQL Order before Group by获得了一些灵感
以下 SQL 将创建一些示例表/数据...
CREATE TABLE job (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` tinytext NOT NULL,
PRIMARY KEY (id)
);
CREATE TABLE job_log_server (
`id` int(11) NOT NULL AUTO_INCREMENT,
`job_id` int(11) NOT NULL,
`event` tinytext NOT NULL,
`ip` tinytext NOT NULL,
`created` datetime NOT NULL,
PRIMARY KEY (id),
KEY job_id (job_id)
);
CREATE TABLE job_log_user (
`id` int(11) NOT NULL AUTO_INCREMENT,
`job_id` int(11) NOT NULL,
`event` tinytext NOT NULL,
`user_id` int(11) NOT NULL,
`created` datetime NOT NULL,
PRIMARY KEY (id),
KEY job_id (job_id)
);
INSERT INTO job VALUES (1, 'Job A');
INSERT INTO job VALUES (2, 'Job B');
INSERT INTO job VALUES (3, 'Job C');
INSERT INTO job VALUES (4, 'Job D');
INSERT INTO job_log_server VALUES (1, 2, 'Job B Event 1', '127.0.0.1', '2000-01-01 00:00:01');
INSERT INTO job_log_server VALUES (2, 2, 'Job B Event 2', '127.0.0.1', '2000-01-01 00:00:02');
INSERT INTO job_log_server VALUES (3, 2, 'Job B Event 3*', '127.0.0.1', '2000-01-01 00:00:03');
INSERT INTO job_log_server VALUES (4, 3, 'Job C Event 1*', '127.0.0.1', '2000-01-01 00:00:04');
INSERT INTO job_log_user VALUES (1, 1, 'Job A Event 1', 5, '2000-01-01 00:00:01');
INSERT INTO job_log_user VALUES (2, 1, 'Job A Event 2*', 5, '2000-01-01 00:00:02');
INSERT INTO job_log_user VALUES (3, 2, 'Job B Event 1*', 5, '2000-01-01 00:00:03');
INSERT INTO job_log_user VALUES (4, 4, 'Job D Event 1', 5, '2000-01-01 00:00:04');
INSERT INTO job_log_user VALUES (5, 4, 'Job D Event 2', 5, '2000-01-01 00:00:05');
INSERT INTO job_log_user VALUES (6, 4, 'Job D Event 3*', 5, '2000-01-01 00:00:06');
一个选项(仅从每个表返回 1 个字段)是使用嵌套子查询...但 ORDER BY 必须在对 GROUP BY (x2) 的单独查询中完成:
SELECT
*
FROM
(
SELECT
s2.*,
jlu.event AS user_event
FROM
(
SELECT
*
FROM
(
SELECT
j.id,
j.name,
jls.event AS server_event
FROM
job AS j
LEFT JOIN
job_log_server AS jls ON jls.job_id = j.id
ORDER BY
jls.created DESC
) AS s1
GROUP BY
s1.id
) AS s2
LEFT JOIN
job_log_user AS jlu ON jlu.job_id = s2.id
ORDER BY
jlu.created DESC
) AS s3
GROUP BY
s3.id;
这实际上似乎表现得很好......只是不太容易理解。
或者您可以尝试在两个单独的子查询中返回和排序日志记录:
SELECT
j.id,
j.name,
jls2.event AS server_event,
jlu2.event AS user_event
FROM
job AS j
LEFT JOIN
(
SELECT
jls.job_id,
jls.event
FROM
job_log_server AS jls
ORDER BY
jls.created DESC
) AS jls2 ON jls2.job_id = j.id
LEFT JOIN
(
SELECT
jlu.job_id,
jlu.event
FROM
job_log_user AS jlu
ORDER BY
jlu.created DESC
) AS jlu2 ON jlu2.job_id = j.id
GROUP BY
j.id;
但这似乎需要更长的时间才能运行......可能是因为它添加到临时表中的记录数量,然后大部分被忽略(为了保持这个简短,我没有添加任何作业表的条件,否则只会返回活动作业)。
不确定我是否遗漏了任何明显的东西。
【问题讨论】:
标签: mysql sql group-by subquery sql-order-by