This question already has answers here:
MySQL - How to join two tables without duplicates?
(5个答案)
2年前关闭。
如何从没有重复行的三个表中获取记录?(已编辑问题)
下面是我的表结构
这是我的努力
您可以在此处获得进一步的解释:https://www.codeproject.com/Questions/693539/left-join-in-multiple-tables
(5个答案)
2年前关闭。
如何从没有重复行的三个表中获取记录?(已编辑问题)
下面是我的表结构
--
-- table structure for table `users`
--
CREATE TABLE `users`(
`user_id` int(11)NOT NULL,
`name` varchar(50) NOT NULL,
`email` varchar(50) NOT NULL,
`password` varchar(300) NOT NULL,
`time_joined` time NOT NULL,
`date_joined` date NOT NULL
)ENGINE=MyISAM DEFAULT CHARSET=latin1;
--
-- table structure for table `activity`
--
CREATE TABLE `activity`(
`activity_id` int(11)NOT NULL,
`user_id` int(11) NOT NULL,
`time_loged` time NOT NULL,
`time_out` time NOT NULL
)ENGINE=MyISAM DEFAULT CHARSET=latin1;
--
-- table structure for table `timeout`
--
CREATE TABLE `timeout`(
`timeout_id` int(11)NOT NULL,
`user_id` int(11)NOT NULL,
`time_out` time NOT NULL
)ENGINE=MyISAM DEFAULT CHARSET=latin1;
这是我的努力
$id=$_SESSION['user'];
$query = $conn->query("SELECT * FROM users left join timeout on users.user_id=timeout.user_id left join activity on users.user_id=activity.user_id WHERE users.user_id='$id'");
while($row = $query->fetch(PDO::FETCH_ASSOC))
{
最佳答案
在'activity'和'timeout'上做第二个左联接,因此查询应如下所示:
SELECT * FROM users
LEFT JOIN timeout ON users.user_id = timeout.user_id
LEFT JOIN activity ON timeout.user_id = activity.user_id
WHERE users.user_id = '$id'
您可以在此处获得进一步的解释:https://www.codeproject.com/Questions/693539/left-join-in-multiple-tables
08-04 20:42