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我有两张桌子projects和project_hours，一对多（一个项目，很多小时）。

这是我的两张表：

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，并得到一个总小时数，所以我有一个总小时数的最终项目列表。因此，如果我有 1 个项目，project_hours 中有 2 条记录，我想取回 1 行，而不是 2 行。

这是我尝试过的。我得到的是 2 行，每行的时间跨度小于 1 小时，因此 current_hours 显示为0. 我能做些什么来总结这两行？获得 1.50 或类似的东西？

select *, datediff(start_time, end_time) * 60 as current_hours from projects 
left join project_hours using(project_id)
where client_id = 2
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1 回答 1

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开始了！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
于 2013-04-25T17:11:29.417 回答