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我有兩個表projects
和project_hours
,與一對多(一個項目,許多小時)。MySQL之間的多個日期之間的小時數總和
這裏是我的兩個表:
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;
我想這樣做的是選擇的所有項目,並且得到的時間的總和,所以我有項目的總小時最終名單。所以如果我有1個項目,2個記錄在project_hours中,我想返回1行,而不是2行。
這是我試過的。我得到的是2行,每個時間跨度小於1小時,因此current_hours顯示爲0
。我能做些什麼來將這兩行相加?得到1.50什麼的?
select *, datediff(start_time, end_time) * 60 as current_hours from projects
left join project_hours using(project_id)
where client_id = 2
datediff返回一個不同的天。你的查詢將永遠不會工作,因爲'some_number_of_days * 60'是** NOT **小時。 – 2013-04-25 17:05:56
哈哈,是的,我剛剛意識到這一點。 – 2013-04-25 17:07:23
我想我找到了答案 – 2013-04-25 17:10:04