Question

我正在尝试加入两个表格。第一个表包含员工信息，如员工编号名称等。第二个表包含时钟信息，如员工编号，状态（时钟输入或输出），上次更新日期时间，时间输入，时间输出，小时数天。

我有这样的东西，但数据库中的真实信息

employees_tbl(
   employee_id INT NOT NULL AUTO_INCREMENT,
   employee_first VARCHAR(30) NOT NULL,
   employee_last VARCHAR(30) NOT NULL,
   PRIMARY KEY ( employee_id )
);
insert into employee (employee_id, employee_first, employee_last) values (1, First, Name);
insert into employee (employee_id, employee_first, employee_last) values (2, Second, Name);
insert into employee (employee_id, employee_first, employee_last) values (3, Third, Name);
insert into employee (employee_id, employee_first, employee_last) values (4, Fourth, Name);
insert into employee (employee_id, employee_first, employee_last) values (5, Fifth, Name);

employee number, status(clocked in or out), datetime row was last updated, time clocked in, time clocked out, hours
transactions_tbl(
   employee_id INT NOT NULL
   status VARCHAR(3) NOT NULL,
   datetime DATETIME NOT NULL,
   clockin TIME,
   clockout TIME,
   hours TIME,
   PRIMARY KEY ( employee_id )
);

INSERT INTO transactions (employee_id, status, datetime, clockin, clockout, hours) VALUES (1, "OUT", "2015-08-10 05:00:00", "2015-08-10 04:00:00", "2015-08-10 05:00:00", "01:00:00");
INSERT INTO transactions (employee_id, status, datetime, clockin, clockout, hours) VALUES (2, "IN", "2015-08-11 05:00:00", "2015-08-11 04:00:00", , );
INSERT INTO transactions (employee_id, status, datetime, clockin, clockout, hours) VALUES (3, "IN", "2015-08-11 05:00:00", "2015-08-10 04:00:00", "2015-08-10 05:00:00", "01:00:00");

我正在寻找的输出是每个员工都应该拥有他们的名字，如果他们当天有进入或进出的信息，那么应该显示信息。如果他们没有进入或离开当天，它应该只显示他们最后一次进出时间。它应该始终显示员工的最后记录，这意味着最近的活动。如果他们从未进入或退出，它仍应显示其名称。

我使用的查询是：

'SELECT employee_id, employee_first FROM employees e left join (SELECT * FROM transactions ORDER BY datetime DESC) as t1 on e.employee_id = t1.employee_id GROUP BY t1.employee_id'

然而，它只给了我前3名员工，而没有给员工4及以后。

我在mysql和php中编码。我会在这里将结果显示为表格，但无法弄清楚如何正确地制作表格格式。一切都只是陷入了混乱，我甚至无法解决问题。但希望你明白这一点。

Answer 1

这是您的查询：

SELECT e.employee_id, e.employee_first
FROM employees e left join
     (SELECT * FROM transactions ORDER BY datetime DESC) as t1
     on e.employee_id = t1.employee_id
GROUP BY t1.employee_id;

它有几个主要错误。第一个是在t1.employee_id中使用GROUP BY。这是来自第二个表，而不是第一个，所以它可能是NULL。其次，子查询中有ORDER BY，显然希望能做某事。我还假设您确实需要来自t1的列。

根据您描述的内容，我认为您想要：

SELECT e.employee_id, e.employee_first, t.*
FROM employees e left join
     (SELECT employee_id, MAX(datetime) as maxdt
      FROM transactions 
      GROUP BY employee_id
     ) tmax
     on e.employee_id = tmax.employee_id left join
     transactions t
     on tmax.employee_id = t.employee_id and tmax.maxdt = t.datetime;

Answer 2

您正在从Transactions表开始执行左连接。它只有3名员工。您的代码工作正常。如果您想要获得所有员工，请在事务表上进行正确的联接。

Answer 3

为什么不

SELECT employee_id, employee_first
FROM employees e
LEFT JOIN transactions t1 on e.employee_id = t1.employee_id
GROUP BY t1.employee_id

哦，顺便说一句，你之所以得到3个结果而不是5个结果，是因为你在事务表的GROUP BY employee_id上做了t1。如果您将其删除或将其更改为e.employee_id，则应返回所需的值。

Mysql左连接似乎不起作用

3 个答案: