How to find the row with 7th highest salary from employee table in MySQL?
I have tried it this way but unable to get the exact query.
SELECT MAX(salary) FROM employee
WHERE salary NOT IN
(SELECT MAX(salary) FROM employee)
What a brief post!!! Try this though,
select *
from(
select distinct salary
from employee
order by salary desc limit 7
) t
order by salary
limit 1
maybe you can use this
SELECT * FROM employe ORDER BY salary DESC LIMIT 7
hope this will help you
I got the answer.
SELECT *
FROM one one1
WHERE ( 7 ) = ( SELECT COUNT( one2.salary )
FROM one one2
WHERE one2.salary >= one1.salary
)
SELECT *
FROM employees emp
WHERE 7 =
(SELECT COUNT(DISTINCT salary)
FROM employees
WHERE emp.salary<=salary );
Related
I have a salary table in which I am trying to return determine the lowest salary earned and by which industry for each year however despite getting the correct lowest salary earned I am receiving the wrong industry name.
I am aware that it is due to the fact that I have utilized GROUP BY without placing a constraint(?) on it hence it is returning me the wrong value but I am not sure how I can solve it.
SALARY TABLE
salaryID
salaryAmount
salaryYear
industryName (ForeignKey)
Can someone please guide me on the right path?
**(Problem Code)**
SELECT MIN(S.salary), S.industryName, S.salaryYear
FROM salary
GROUP BY S.salaryYear;
**(Attempted solution)**
SELECT S.salary
FROM salary
INNER JOIN
SELECT (min(S1.amount)), S1.year, S1.industryName, S1.salaryId
FROM salary S1
GROUP BY S1.year
ON S.salaryId = S1.salaryId);
Use a proper GROUP BY. Any non-aggregated columns must be included in GROUP BY.
SELECT MIN(amount), year
FROM salary
GROUP BY year
If you want to include industryName,
SELECT amount, year, industryName, salaryId
FROM (
SELECT amount, year, industryName, salaryId
, ROW_NUMBER() OVER(PARTITION BY year ORDER BY amount) AS rn
FROM salary
) a
WHERE rn = 1
Pre-MySQL 8 version
SELECT *
FROM salary s
INNER JOIN (
SELECT MIN(amount) AS minAmount, year
FROM salary
GROUP BY year
) m ON m.minAmount = s.amount AND m.year = s.year
I think you need a self-join :
SELECT s1.industryName, s2.min_salary, s2.salaryYear
FROM salary s1
JOIN
(
SELECT MIN(salary) as min_salary, salaryYear
FROM salary
GROUP BY salaryYear
) s2
ON s1.salary = s2.min_salary
AND s1.salaryYear = s2.salaryYear;
The Demo of this query with your sample data
I want to find name of that persons who worked most in a month. but the query doesn't returning max value from sum of value
I'm new in mysql
SELECT
x.name,
sec_to_time(MAX(x.sum_time)) maximum
FROM (
SELECT
name,
SUM(TIME_TO_SEC(ending_time) - TIME_TO_SEC(starting_time)) sum_time
FROM working_hours wh, employees
WHERE wh.employees_id = employees.id
AND project_id IS NOT NULL
GROUP BY employees_id
) x
GROUP BY x.name;
this is my query. i want to show just name of that persons who worked most in a month. but it returns all persons who worked in a month
Try making these changes to your query:
change name to MAX(name)
qualify employees_id with wh.employees_id
SELECT
x.name,
sec_to_time(MAX(x.sum_time)) maximum
FROM (
SELECT
MAX(name) AS name,
SUM(TIME_TO_SEC(ending_time) - TIME_TO_SEC(starting_time)) sum_time
FROM working_hours wh, employees
WHERE wh.employees_id = employees.id
AND project_id IS NOT NULL
GROUP BY wh.employees_id
) x
group by x.name;
Simply use Order by LIMIT -
SELECT X1.name, X1.maximum
FROM (SELECT name, SUM(TIME_TO_SEC(ending_time) - TIME_TO_SEC(starting_time)) maximum
FROM working_hours wh, employees
WHERE wh.employees_id=employees.id
GROUP BY name) X1
JOIN (SELECT SUM(TIME_TO_SEC(ending_time) - TIME_TO_SEC(starting_time)) sum_time
FROM working_hours wh, employees
WHERE wh.employees_id=employees.id
AND project_id is not null
GROUP BY employees_id
ORDER BY sum_time DESC
LIMIT 1) X2 ON X2.sum_time = X1.maximum;
I use MySQL
I wanted to show some records like Total Salary per designation , avg salary etc. All are working except : I want to show person name with heighest salary in group. When I use the query as below , I give error : Invalid use of group function . I ran it many times using different tweaks , ordering of query , still giving error. What may be the issue here ?
SELECT designation , SUM(salary) as salary_as_per_designation ,
count(*) as num_employees ,
avg(salary) as avg_salary ,
MIN(salary) as min_salary_of_emp_per_designation,
MAX(salary) as max_salary_of_emp_per_designation ,
(SELECT name FROM articles where salary = max(salary) GROUP BY designation LIMIT 1) as emp_name_min_sal
FROM articles GROUP BY designation;
The weired thing is when i use on 2nd last line above like .... where salary = 10000... 100 as Some random value query works but gives wrong result , I mean wrong name for the user with heighest salary in group.
My table structure is as follows :
CREATE TABLE IF NOT EXISTS `articles` (
`id` int(12) NOT NULL,
`name` varchar(256) NOT NULL,
`salary` double NOT NULL,
`designation` enum('tech_support','developer','tester','designer') NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=9 ;
Thanks.
SELECT ilv.*,
(SELECT GROUP_CONCAT(name)
FROM articles
where salary = ilv.max_salary_of_emp_per_designation
) as emp_name_min_sal
FROM
(
SELECT designation , SUM(salary) as salary_as_per_designation ,
count(*) as num_employees ,
avg(salary) as avg_salary ,
MIN(salary) as min_salary_of_emp_per_designation,
MAX(salary) as max_salary_of_emp_per_designation
FROM articles
GROUP BY designation
) AS ilv
Does this work? Note sub-select conditions, you're looking for salaries within the designation grouping?
SELECT designation,
SUM(salary) as salary_as_per_designation,
count(*) as num_employees,
avg(salary) as avg_salary,
MIN(salary) as min_salary_of_emp_per_designation,
MAX(salary) as max_salary_of_emp_per_designation,
(SELECT max(name)
FROM articles
where designation = a1.designation
and salary = (select max(salary) from articles
where designation = a1.designation)) as emp_name_min_sal
FROM articles a1
GROUP BY designation;
Use this query instead,
SELECT MAX(salary) into #max_sal FROM articles;
SELECT designation , SUM(salary) as salary_as_per_designation ,
count(*) as num_employees ,
avg(salary) as avg_salary ,
MIN(salary) as min_salary_of_emp_per_designation,
MAX(salary) as max_salary_of_emp_per_designation ,
(SELECT name FROM articles where salary = #max_sal GROUP BY designation LIMIT 1) as emp_name_min_sal
FROM articles GROUP BY designation;
Hope this will work, thank you.
I have a table with 3 columns:
Name department salary
How can I determine using one query to find 3rd highest salary in each department?
One way is to LIMIT a correlated subquery, but it's not especially efficient:
SELECT department, (
SELECT salary
FROM my_table t2
WHERE t2.department = t1.department
ORDER BY salary DESC
LIMIT 2, 1
)
FROM my_table t1
GROUP BY department
In addition to eggyal's excellent answer, here's a query that will give you the names, too, of those that have salary equal to the third (in each department):
SELECT
t.name, t.department, t.salary AS third_salary
FROM
( SELECT DISTINCT department
FROM tableX
) AS d
JOIN
tableX AS t
ON t.department = d.department
AND t.salary =
( SELECT tt.salary -- notice that this
FROM tableX AS tt -- subquery is
WHERE tt.department = d.department -- exactly the same as
ORDER BY tt.salary DESC -- the one in
LIMIT 1 OFFSET 2 -- #eggyal's answer
) ;
This RANK question is similar to this one:
MySQL, Get users rank
I you can thy this:
SELECT s.*,
(
SELECT COUNT(*)
FROM salaries si
WHERE si.salary >= s.salary AND si.department = s.department
) AS rank
FROM salaries s
WHERE s.rank = 3
Try this:
SELECT name, department, salary
FROM (SELECT name, department, salary, IF(#dept=(#dept:=department), #auto:=#auto+1, #auto:=1) indx
FROM employee e, (SELECT #dept:=0, #auto:=1) A
ORDER BY department, salary DESC ) AS A
WHERE indx = 3;
the database schema looks like
employee(employee_name,street,city)
works(employee_name,company_name,salary)
company(company_name,city)
manages(employee_name,manager_name)
the query needed to do is:
find the company that has the most employees.
I could find out the maximum count by the query:
SELECT max( cnt ) max_cnt
FROM (
SELECT count( employee_name ) cnt, company_name
FROM works
GROUP BY company_name
)w1;
But now I can't find out the name of the company. If anyone has some idea please share.
To get the entire row containing the maximum value you can use ORDER BY ... DESC LIMIT 1 instead of MAX:
SELECT company_name, cnt
FROM (
SELECT company_name, count(employee_name) AS cnt
FROM works
GROUP BY company_name
) w1
ORDER BY cnt DESC
LIMIT 1
How about something like:
SELECT count( employee_name ) cnt, company_name
FROM works
GROUP BY company_name
ORDER BY cnt DESC
LIMIT 1;
Edit:
Corrected above for MySQL
SELECT company_name,count(*) as cnt
FROM works
GROUP BY company_name
ORDER BY cnt DESC
select company_name
from works
group by company_name
having count(distinct employee_name)>=all(select count(distinct employee_name)
from works
group by company_name )
Here's the working query
Select * from(SELECT count(EmpName)cnt, CName FROM works GROUP BY CName Order By cnt desc) where ROWNUM = 1;
This looks like a course question.
If more than one companies have the same largest number of employees the query with LIMIT doesn't work. "ORDER BY" didn't filter out useless info. Thus we have the following solution
SELECT company_name FROM
(SELECT company_name, count(employee_name) cnt
FROM works
GROUP BY company_name)
JOIN
(SELECT max(cnt) max_cnt
FROM (
SELECT count(employee_name) cnt
FROM works
GROUP BY company_name
)) ON cnt = max_cnt
select company_name from works_for
group by company_name
having count(employee_name) = (select max(count(employee_name))from works_for
group by company_name);
In Oracle
select company_name, count(*) as count
from works
group by company_name
having count(*) >= all(select count(*) from works group by company_name)