Question

我有一张雇员和工资表，用这种方式定义：

"name" (type: VARCHAR)
"salary" (type: INTEGER)

我可以使用什么查询来获得此表中第二高的薪水？

Answer 1

这是一个解释关系的人。

Name    Salary
Jim       6
Foo       5
Bar       5
Steve     4

SELECT name, salary
FROM employees
WHERE salary = (SELECT MAX(salary) FROM employees WHERE salary < (SELECT MAX(salary) FROM employees))

Result --> Bar 5, Foo 5

修改我拿了Manoj的第二篇文章，调整它，使它更具人性化。对我来说 n-1 并不直观;但是，使用我想要的值，2 =第2，3 =第3等等。

/* looking for 2nd highest salary -- notice the '=2' */ SELECT name,salary FROM employees WHERE salary = (SELECT DISTINCT(salary) FROM employees as e1 WHERE (SELECT COUNT(DISTINCT(salary))=2 FROM employees as e2 WHERE e1.salary <= e2.salary)) ORDER BY name Result --> Bar 5, Foo 5

Answer 2

第二高薪的直接答案

SELECT name, salary
FROM employees ORDER BY `employees`.`salary` DESC LIMIT 1 , 1

另一个有趣的解决方案

SELECT salary 
FROM emp 
WHERE salary = (SELECT DISTINCT(salary) 
                FROM emp as e1 
                WHERE (n) = (SELECT COUNT(DISTINCT(salary)) 
                             FROM emp as e2 
                             WHERE e1.salary <= e2.salary))

Answer 3

似乎我很晚才回答这个问题。这个衬垫如何获得相同的输出？

SELECT DISTINCT salary FROM employees ORDER BY salary DESC LIMIT 1,1 ;

示例小提琴：https://www.db-fiddle.com/f/v4gZUMFbuYorB27AH9yBKy/0

Answer 4

create table svalue (
name varchar(5),
value int
) engine = myisam;

insert into svalue value ('aaa',30),('bbb',10),('ccc',30),('ddd',20);

select * from svalue where value = (
select value 
from svalue
group by value
order by  value desc limit 1,1)

Answer 5

FOR SECOND END LAST：

SELECT name, salary
    FROM employee 
    ORDER BY salary DESC
    LIMIT 1 , 1

至于第三次：

SELECT name, salary
    FROM employee 
    ORDER BY salary DESC
    LIMIT 2 , 1

Answer 6

显示具有第二大标记值的记录：

SELECT username, mark
FROM tbl_one
WHERE mark = (
    SELECT DISTINCT mark
    FROM tbl_one
    ORDER by mark desc
    LIMIT 1,1
);

Answer 7

简单的解决方案

SELECT * FROM TBLNAME ORDER BY COLNAME ASC LIMIT (n - x), 1

注意：n =列

中的记录总数

  x = value 2nd, 3rd, 4th highest etc

e.g

//to find employee with 7th highest salary

n = 100
x = 7

SELECT * FROM tbl_employee ORDER BY salary ASC LIMIT 93, 1

希望这会有所帮助

Answer 8

找到另一个有趣的解决方案

SELECT salary 
FROM emp 
WHERE salary = (SELECT DISTINCT(salary) 
                FROM emp as e1 
                WHERE (n) = (SELECT COUNT(DISTINCT(salary)) 
                             FROM emp as e2 
                             WHERE e1.salary <= e2.salary))

对不起。忘了写。 n是你想要的第n个薪水。

Answer 9

获得第二高的值：

SELECT `salary` FROM `employees` ORDER BY `salary` DESC LIMIT 1, 1;

Answer 10

要获得第二高薪，只需使用以下查询

SELECT salary FROM employees
ORDER BY salary DESC LIMIT 1,1;

Answer 11

您可以使用以下提到的查询

SELECT emp.name, emp.salary 
FROM employees emp 
WHERE 2 = (SELECT COUNT(DISTINCT salary) 
           FROM employees 
           WHERE emp.salary<=salary
          );

您可以将2更改为所需的最高记录。

Answer 12

SELECT name, salary 
FROM employees 
where
 salary = (SELECT (salary) FROM employees GROUP BY salary DESC LIMIT 1,1)

Answer 13

要获得* N *最高值，最好使用此解决方案：

SELECT * FROM `employees`  WHERE salary =
         (SELECT DISTINCT(salary) FROM `employees` 
         ORDER BY salary DESC LIMIT {N-1},1);

或者您可以尝试：

SELECT * FROM `employees` e1 WHERE 
        (N-1) = (SELECT COUNT(DISTINCT(salary)) 
        FROM `employees` e2 
        WHERE e1.salary < e2.salary );

N = 2，第二高 N = 3表示第三高等等。

Answer 14

获得第二高薪

 select max(salary) from salary where salary not in (select top 1 salary from salary order by salary desc)

获得第三高薪

 select max(salary) from salary where salary not in (select top 2 salary from salary order by salary desc)

依旧......

Answer 15

简单的解决方案如下面的查询：

select max(salary) as salary from employees where salary<(select max(salary) from employees);

Answer 16

SELECT MAX(salary) salary
FROM tbl
WHERE salary <
  (SELECT MAX(salary)
   FROM tbl);

Answer 17

SELECT DISTINCT Salary
FROM emp
ORDER BY salary DESC
LIMIT 1 , 1

此查询也会给出重复记录的第二高薪。

Answer 18

尝试这个获得第n个最高工资

我在发布之前试过这个＆amp;它工作正常

例如。找到第10个最高薪水替换限制9,1 ;

mysql> select name,salary from emp group by salary desc limit n-1,1;

Answer 19

试试这个：

SELECT DISTINCT(`salary`)
    FROM `employee`

    ORDER BY `salary` DEC
    LIMIT 1,1

Answer 20

with alias as
(
select name,salary,row_number() over(order by salary desc ) as rn from employees
)
select name,salary from alias where rn=n--n being the nth highest salary

Answer 21

SELECT username, salary
FROM tblname
GROUP by salary
ORDER by salary desc
LIMIT 0,1 ;

Answer 22

SELECT name, salary
FROM EMPLOYEES
WHERE salary = ( 
SELECT DISTINCT salary
FROM EMPLOYEES
ORDER BY salary DESC 
LIMIT 1 , 1 )

Answer 23

SELECT name, salary
FROM employees
order by salary desc limit 1,1

这个查询应该可以完成你的工作。首先，我们以递减的方式对表格进行排序，以便薪水最高的人位于顶部，第二个位于第二位置。现在limit a,b表示跳过起始a元素，然后打印下一个b元素。所以在这种情况下你应该使用limit 1,1。

希望这有帮助。

Answer 24

SELECT name,salary FROM employee
WHERE salary = (SELECT DISTINCT(salary) FROM employee ORDER BY salary DESC LIMIT 1,1) ORDER BY name

Answer 25

SELECT SALARY 
FROM (SELECT * 
      FROM EMPLOYEE 
      ORDER BY SALARY 
      DESC LIMIT ***2***) AS TOP_SALARY 
ORDER BY SALARY ASC 
LIMIT 1

Answer 26

SELECT MIN(id) as id FROM students where id>(SELECT MIN(id) FROM students);

Answer 27

试试这个：

package nestedcontrollerexample;

import javafx.beans.binding.Bindings;
import javafx.scene.control.ListCell;
import javafx.scene.control.ListView;
import javafx.util.Callback;

public class PersonListCellFactory implements Callback<ListView<Person>, ListCell<Person>> {
    @Override
    public ListCell<Person> call(ListView<Person> param) {
        return new ListCell<Person>() {
            @Override
            public void updateItem(Person person, boolean empty) {
                super.updateItem(person, empty);
                textProperty().unbind();
                if (person != null) {
                    textProperty().bind(Bindings.format("%s %s", person.firstNameProperty(), person.lastNameProperty()));
                }
            }
        };
    }
}

获取MySQL表中的第二高值

27 个答案: