Question

我有两张表projects和project_hours，一对多（一个项目，多个小时）。

以下是我的两张表：

CREATE TABLE `projects` (
    `project_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
    `client_id` int(10) unsigned NOT NULL,
    `project_name` char(50) NOT NULL,
    `project_date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
    PRIMARY KEY (`project_id`),
    KEY `project_owner` (`client_id`),
    CONSTRAINT `projects_ibfk_1` FOREIGN KEY (`client_id`) REFERENCES `clients` (`client_id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=utf8;

CREATE TABLE `project_hours` (
    `hours_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
    `project_id` int(10) unsigned NOT NULL,
    `start_time` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
    `end_time` datetime NOT NULL,
    PRIMARY KEY (`hours_id`),
    KEY `project_id` (`project_id`),
    CONSTRAINT `project_hours_ibfk_1` FOREIGN KEY (`project_id`) REFERENCES `projects` (`project_id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

我想要做的是选择所有项目，并获得小时数的总和，因此我有一个总小时的项目最终列表。因此，如果我在project_hours中有1个项目，2个记录，我想要返回1行，而不是2行。

这是我尝试过的。我得到的是2行，每个时间跨度不到1小时，因此current_hours显示为0。我该怎么做才能将两行相加？得到1.50或其他东西？

select *, datediff(start_time, end_time) * 60 as current_hours from projects 
left join project_hours using(project_id)
where client_id = 2

Answer 1

我们走吧！这似乎符合我的要求：

select *, sum(time_to_sec(timediff(end_time, start_time))) / 60 / 60 as current_hours from projects 
left join project_hours using(project_id)
where client_id = 2
group by project_id

MySQL多个日期之间的小时数

1 个答案: