表1：employee_detail：
id name
1 ABC
2 CCC
3 FFF
4 ggg
5 jjj
表2：效果评估
id date_of_join isAppraisalcomplete emp_id
1 1-07-2010 Yes 1
2 09-6-2010 Yes 2
3 10-7-2012 Yes 3
4 23-8-2015 No 4
5 07-11-2018 No 5
表3：Financial_details
id salary hike emp_id p_a_id year
1 11000 12 1 1 2016
2 11000 9 1 1 2017
3 11000 11 1 1 2016
4 11000 10 1 1 2017
2 33000 15 2 2 2016
3 36000 10 2 2 2017
4 31000 15 2 2 2016
5 44001 10 2 2 2017
..........
..........
..........
期望输出：
Emp_id Date_of_join isAppraisalcomplete Salary Hike year
1 1-07-2010 Yes 11000 12 2016
1 1-07-2010 Yes 11000 9 2017
2 09-06-2010 Yes 33000 15 2016
2 09-06-2010 Yes 36000 10 2017
..........
..............
查询我用于：
select * from financial_details bsd inner join performance_appraisal fi on fi.emp_id = bsd.emp_id limit 2;
我的结果：
Emp_id Date_of_join isAppraisalcomplete Salary Hike year
1 1-07-2010 Yes 11000 12 2016
1 1-07-2010 Yes 11000 9 2017
在增加限制时，它显示emp_id中的所有记录，例如emp_id 1不仅显示前两个记录，还显示所有记录。
如何使用MySql从表中获取前两个记录以及加入另一个表。
在使用where条件时，按desc限制2进行订购，我得到了一条记录（一位雇员）的准确结果。但是实际上我正在尝试从financial_details表和performance_appraisal表联接中获取前两个唯一数据的所有记录（所有员工）。请帮忙。
编辑：
查询：
CREATE TABLE employee_detail
(
id int,
name varchar(255)
);
CREATE TABLE performance_appraisal
(
id int,
date_of_join varchar(255),
isAppraisalcomplete varchar(255),
emp_id int
);
CREATE TABLE financial_details
(
id int,
salary varchar(255),
hike varchar(255),
emp_id int,
p_a_id int,
t_year varchar(255)
);
insert into employee_detail (id, name) values (1,"abc");
insert into employee_detail (id, name) values (2,"def");
insert into employee_detail (id, name) values (3,"ghi");
insert into performance_appraisal (id, date_of_join, isAppraisalcomplete, emp_id) values (1, "1-07-2010", "Yes", 1);
insert into performance_appraisal (id, date_of_join, isAppraisalcomplete, emp_id) values (2, "09-6-2010", "Yes", 2);
insert into performance_appraisal (id, date_of_join, isAppraisalcomplete, emp_id) values (3, "10-7-2012", "Yes", 3);
insert into performance_appraisal (id, date_of_join, isAppraisalcomplete, emp_id) values (4, "23-8-2015", "No", 4);
insert into performance_appraisal (id, date_of_join, isAppraisalcomplete, emp_id) values (5, "07-11-2018", "No", 5);
insert into financial_details (id, salary, hike, emp_id,p_a_id, t_year) values (1, "11000", "12", 1,1,"2016");
insert into financial_details (id, salary, hike, emp_id,p_a_id, t_year) values (2, "12000", "9", 1,1,"2017");
insert into financial_details (id, salary, hike, emp_id,p_a_id, t_year) values (3, "10500", "11", 1,1,"2016");
insert into financial_details (id, salary, hike, emp_id,p_a_id, t_year) values (4, "11400", "10", 1,1,"2017");
insert into financial_details (id, salary, hike, emp_id,p_a_id, t_year) values (5, "36000", "15", 2,2,"2016");
insert into financial_details (id, salary, hike, emp_id,p_a_id, t_year) values (6, "36000", "15", 2,2,"2017");
insert into financial_details (id, salary, hike, emp_id,p_a_id, t_year) values (7, "31000", "15", 2,2,"2016");
insert into financial_details (id, salary, hike, emp_id,p_a_id, t_year) values (8, "44000", "15", 2,2,"2017");
答案 0 :(得分：2)
这里有一些需要考虑的问题，尽管按照书面规定，此解决方案仅适用于8.0之前的版本...
SELECT emp_id
, id
FROM
( SELECT emp_id
, id
, CASE WHEN @prev = emp_id THEN @i:=@i+1 ELSE @i:=1 END i
, @prev:=emp_id prev
FROM financial_details
, (SELECT @prev:=null,@i:=0) vars
ORDER
BY emp_id
, id
) x
WHERE i <= 2;
+--------+------+
| emp_id | id |
+--------+------+
| 1 | 1 |
| 1 | 2 |
| 2 | 5 |
| 2 | 6 |
+--------+------+
答案 1 :(得分：0)
您可以在MySQL 8+中使用窗口函数：
req.hostname答案 2 :(得分：0)
感谢Strawberry，我们需要在输出中的where子句中添加答案。
select * from financial_details bsd inner join performance_appraisal fi on fi.emp_id = bsd.emp_id where bsd.id in (SELECT id
FROM ( SELECT emp_id, id, CASE WHEN @prev = emp_id THEN @i:=@i+1 ELSE @i:=1 END i, @prev:=emp_id prev
FROM financial_details, (SELECT @prev:=null,@i:=0) vars ORDER BY emp_id, id
) x
WHERE i <= 2);