2nd and 5th highest salary

Hi, I'm kind of new to sql. I have a table with the name of the employee and salary.
I would like to find the name of the 2nd higher salary? also I want to get the name of the highest salary 5th
Here's my data

INSERT INTO er4 VALUES ('John', 1000, 123);
INSERT INTO er4 VALUES ('john2', 3000, 234);
INSERT INTO er4 VALUES ('Jean3', 1000, 223);
INSERT INTO er4 VALUES ('john4', 4000, 123);
INSERT INTO er4 VALUES ('john5', 8000, 023);
INSERT INTO er4 VALUES ('john6', 9000, 023);
INSERT INTO er4 VALUES ('john7', 8000, 723);

You can see in the data, there are two employees with the second highest salary. I would like to get one or the other, but I want my query to return a single line

can someone help me write a query to get the name and salary of the 2nd higher and the name and salary of 5th higher

Hello

Part of learning about the functions is to learn what they are good for, and what they are not good for.
As Justin said, MAX is not good for this task.
If you were interested in just the 1st highest salary, you may be able to use MAX like this:

SELECT  *
FROM    er4
WHERE   salary  =      (
               SELECT     MAX (salary)
               FROM     er4
               )
AND     ROWNUM     = 1     -- in case of a tie, pick one row arbitrarily
;

But even if the goal was to find the highest salary, I would use ROW_NUMBER, as Justin did.
Why? The following scenario happens all the time in real life. The client for which you wrote this comes back in 3 months and said: 'this query you wrote for the highest salary works very well! We found it so useful, we would like to do something similar to find the 2nd highest salary, either the 5th highest, or the top 5 or the next 5. "ROW_NUMBER is flexible enough to do all these things. When you have to solve a problem, you'll save time (long-term) If you can predict, or just guessing, what you need to solve later and write something for the problem today that can easily be adapted for potential future similar problems problems.

Tags: Database

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How can I get the second and third highest salary from the emp table

How can I get the second and third highest salary from the emp table in the ecah Department

SQL> ed
Wrote file afiedt.buf

  1  select empno, ename, sal, deptno
  2  from (
  3    select empno, ename, sal, deptno, dense_rank() over (partition by deptno order by sal desc) as rn
  4    from emp
  5    )
  6* where rn in (2,3)
SQL> /

     EMPNO ENAME             SAL     DEPTNO
---------- ---------- ---------- ----------
      7782 CLARK            2450         10
      7934 MILLER           1300         10
      7566 JONES            2975         20
      7876 ADAMS            1100         20
      7499 ALLEN            1600         30
      7844 TURNER           1500         30

6 rows selected.

SQL>

query to retrieve the second highest salary managers

Select * from employee where salary = (select max (salary) in e employees where employee_id = e.manager_id and salary < (select max (salary) in employees))
/

It does not run... can someone suggest a new

987184 wrote:
our teacher was asked to write without y analytical function.dats.

1 list of Manager IDs:

select  manager_id
  from  hr.employees
  where manager_id is not null
/

2 manager info:

select  employee_id,
        first_name,
        last_name,
        salary
  from  hr.employees
  where employee_id in (
                        select  manager_id
                          from  hr.employees
                          where manager_id is not null
                       )
/

3. Manager second highest salary

with managers as (
                  select  employee_id,
                          first_name,
                          last_name,
                          salary
                    from  hr.employees
                    where employee_id in (
                                          select  manager_id
                                            from  hr.employees
                                            where manager_id is not null
                                         )
                 )
select  m1.employee_id,
        m1.first_name,
        m1.last_name,
        max(m1.salary) second_highest_salary
  from  managers m1,
        managers m2
  where m1.salary < m2.salary
  group by m1.employee_id,
           m1.first_name,
           m1.last_name
  having count(distinct m2.salary) = 1
/

And execution:

SQL> select  employee_id,
  2          first_name,
  3          last_name,
  4          salary
  5    from  hr.employees
  6    where employee_id in (
  7                          select  manager_id
  8                            from  hr.employees
  9                            where manager_id is not null
 10                         )
 11    order by salary desc
 12  /

EMPLOYEE_ID FIRST_NAME           LAST_NAME                     SALARY
----------- -------------------- ------------------------- ----------
        100 Steven               King                           24000
        102 Lex                  De Haan                        17000
        101 Neena                Kochhar                        17000
        145 John                 Russell                        14000
        146 Karen                Partners                       13500
        201 Michael              Hartstein                      13000
        108 Nancy                Greenberg                      12008
        205 Shelley              Higgins                        12008
        147 Alberto              Errazuriz                      12000
        114 Den                  Raphaely                       11000
        148 Gerald               Cambrault                      11000

EMPLOYEE_ID FIRST_NAME           LAST_NAME                     SALARY
----------- -------------------- ------------------------- ----------
        149 Eleni                Zlotkey                        10500
        103 Alexander            Hunold                          9000
        121 Adam                 Fripp                           8200
        120 Matthew              Weiss                           8000
        122 Payam                Kaufling                        7900
        123 Shanta               Vollman                         6500
        124 Kevin                Mourgos                         5800

18 rows selected.

SQL> with managers as (
  2                    select  employee_id,
  3                            first_name,
  4                            last_name,
  5                            salary
  6                      from  hr.employees
  7                      where employee_id in (
  8                                            select  manager_id
  9                                              from  hr.employees
 10                                              where manager_id is not null
 11                                           )
 12                   )
 13  select  m1.employee_id,
 14          m1.first_name,
 15          m1.last_name,
 16          max(m1.salary) second_highest_salary
 17    from  managers m1,
 18          managers m2
 19    where m1.salary < m2.salary
 20    group by m1.employee_id,
 21             m1.first_name,
 22             m1.last_name
 23    having count(distinct m2.salary) = 1
 24  /

EMPLOYEE_ID FIRST_NAME           LAST_NAME                 SECOND_HIGHEST_SALARY
----------- -------------------- ------------------------- ---------------------
        101 Neena                Kochhar                                   17000
        102 Lex                  De Haan                                   17000

SQL>

As you can see, most high salary 24000 is won by the King. Second highest salary is 17000 and is won by two managers: Kochhar and De Haan.

SY.

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Hi guys,.

I can't understand this query that we got a Web site. It is used to find the nth highest salary of the employee... Can someone explain to me please each and every part of it...

The Charly is:

Select distinct (a.salary)
EMP a
where 1 = (select count (distinct (salary))
b emp
where a.salary < = b.salary)

The "BOLD" part which is the return value... is my main question about this query...

Help, please

It is called a correlated sub query. The inner query is executed for each iteration of the outer query. For example, if the table emp has 14 rows and then for each salary 14 the inner query will run.
The part

select count(distinct(sal))
from emp b
where a.sal<=b.sal

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Clearly for the King, who gets the highest salary is greater than or equal to 1 (himself). It returns the highest salary.

SQL> select *
  2  from emp a
  3  where 1 >= (select count(distinct(sal))
  4  from emp b
  5  where a.sal<=b.sal)
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     EMPNO ENAME                               JOB              MGR HIREDATE         SAL       COMM     DEPTNO
---------- ----------------------------------- --------- ---------- --------- ---------- ---------- ----------
      7839 KING                                PRESIDENT            17-NOV-81       5000                    10

Salary of the first two.

SQL> ed
Wrote file afiedt.buf

  1  select *
  2  from emp a
  3  where 2 >= (select count(distinct(sal))
  4  from emp b
  5  where a.sal<=b.sal)
  6* order by sal desc
SQL> /

     EMPNO ENAME                               JOB              MGR HIREDATE         SAL       COMM     DEPTNO
---------- ----------------------------------- --------- ---------- --------- ---------- ---------- ----------
      7839 KING                                PRESIDENT            17-NOV-81       5000                    10
      7902 FORD                                ANALYST         7566 03-DEC-81       3000                    20
      7788 SCOTT                               ANALYST         7566 09-DEC-82       3000                    20

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Select
* de (
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I have a requirement to get the employee who gets the 5th highest salary.

I don't want to use query analysis here.

I remember writing a query using ' select * from emp where 5 > (select...) »
Now I forgot logic, please help, thank you.
Well, it depends on how your data are defined.

For example if your data are as mentioned below.
```
emp table -

       ENO     SALARY
---------- ----------
        10       1000
        20       2000
        30       3000
        40       4000
        50       5000
        60       6000
        70       7000
        80       8000
        90       9000
```
In this case, the following query will work-
```
Connected to Oracle Database 11g Release 11.2.0.2.0 

SQL>
SQL> SELECT eno
  1        ,salary
  2    FROM emp a
  3   WHERE 5 = (SELECT COUNT(*)
  4                FROM emp b
  5               WHERE b.salary >= a.salary) -- Please note the sign ">="
  6  /

       ENO     SALARY
---------- ----------
        50       5000

SQL> 
```
So if you are sure that column salary will have unique values, then you can use above query probably. However, this query fails for a few scenarios what column salary contains extensible data (which are possible for most of the implementation for the salary column). Two of these scenarios are
s ' is used mutiple with 5th highest salary i.e. 5000 in the example above
If ' there are several employees with the 3rd highest salary i.e. 3000 in the example above.

examine the data mentioned below. There are several employees with 5th highest salary that is 5000. So expect multiple records output. Above mentioned query will fail for such data. So is the modified query in order to manage these data-
```
emp table -

       ENO     SALARY
---------- ----------
        10       1000
        20       2000
        30       3000
        40       5000        -- salary changed to 5000
        50       5000
        60       6000
        70       7000
        80       8000
        90       9000
```
In this case, the following query will work-
```
Connected to Oracle Database 11g Release 11.2.0.2.0 

SQL>
SQL> SELECT eno
  1        ,salary
  2    FROM emp a
  3   WHERE 5 = (SELECT COUNT(DISTINCT salary )
  4                FROM emp b
  5               WHERE b.salary >= a.salary)
  6  /

       ENO     SALARY
---------- ----------
        40       5000
        50       5000

SQL> 
```
Top query also works for other combinations of data. For example if there are several employees with salary 7000 which is greater than 5 the higher salary. In other words, if there are several employees with the 3rd highest salary, then also there is not any change in the production of the above query. If you want the PERFECT query (without using the analytical functions) then second above application so should be the choice.

I hope this helps.

See you soon,.
Gregory

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