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)
Related
I am using PHP MyAdmin. I want the result of this query to be Willard Wildcats and 7, but I can't figure out how to do it by using the max function rather than the order by and limit functions. I need to use the max function
Select Team_Name, Count(*)
From Number_3
Group By Team_Name
The result of this query should be
Willard Wildcats and 7
Use LIMIT clause :
SELECT Team_Name, Count(*) AS CNT
FROM Number_3
GROUP BY Team_Name
ORDER BY CNT DESC
LIMIT 1;
If you are limited with LIMIT clause then use subquery :
SELECT Team_Name, Count(*) AS CNT
FROM Number_3
GROUP BY Team_Name
HAVING Count(*) = (SELECT MAX(CNT)
FROM (SELECT Team_Name, Count(*) AS CNT
FROM Number_3
GROUP BY Team_Name
) t
);
If you are working latest version of MySQL then it will be more easier with ranking function :
SELECT t.*
FROM (SELECT Team_Name, Count(*) AS CNT,
DENSE_RANK() OVER (ORDER BY Count(*) DESC) AS Seq
FROM Number_3
GROUP BY Team_Name
) t
WHERE Seq = 1;
You need to write where condition and select a column using max function
Select Team_Name, Count(*) From Number_3 where Team_Name=(select max(Count(*)) from Number_3)
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 );
I have a table with
orderNumber(pk) , customerNumber , comment
I have to count the maximum order placed by a user and show its user ID and MAX count . I have following Query
It shows the count Right but it takes the first CustomerNumber in the table
SELECT maxCount.customerNumber , MAX(`counted`) FROM
(
SELECT customerNumber, COUNT(*) AS `counted`
FROM `orders`
GROUP BY `customerNumber`
)as maxCount
Thanks & regards
Just use ORDER BY with your inner query:
SELECT customerNumber, COUNT(*) AS `counted`
FROM `orders`
GROUP BY `customerNumber`
ORDER BY COUNT(*) DESC
LIMIT 1
If you want to return all customer numbers in the event of a tie, you can use a HAVING clause with a subquery which identifies the maximum count:
SELECT customerNumber, COUNT(*) AS counted
FROM orders
GROUP BY customerNumber
HAVING COUNT(*) = (SELECT MAX(t.counted) FROM (SELECT COUNT(*) AS counted
FROM orders
GROUP BY customerNumber) t)
Demo here:
SQLFiddle
I have a list of login logs from our website, however I am needing to see which user ID has had the most IP's logged into it. Our table is as follows:
userid, ip, date (unix)
I need it to output which userid's have had the most IP's logged into them.
I've tried something such as:
SELECT
userID
FROM loginLogs
GROUP BY userID
HAVING COUNT( DISTINCT ip ) > 1
But that just shows a list of user ID's.
Select userID, count(distinct ip)
from loginLogs
Group by 1
Order by 2 desc
Maybe like this?
SELECT `userID`, count(`ip`) cnt FROM `loginLogs` GROUP BY `userID` HAVING cnt > 1
You can just order by distinct values, descending;
SELECT userID, COUNT(DISTINCT ip) `distinct IP#s`
FROM loginLogs
GROUP BY userID
ORDER BY `distinct IP#s` DESC;
An SQLfiddle to test with.
SELECT userID, COUNT(*) AS count FROM loginLogs
GROUP BY userId ORDER BY count DESC
This will give you all of your users from most logged in to the least. Use LIMIT 1 if you want to limit the results.
You have to order those results order by COUNT( DISTINCT ip ) desc and take the first Limit 0, 1
SELECT `userID`
FROM `loginLogs`
GROUP BY `userID`
ORDER BY COUNT( DISTINCT `ip` ) desc
LIMIT 0, 1
You could wrap what you have in a subquery to get the list of userIDs and distinct IPs, as well.
SELECT DISTINCT ll.`userID`, ll.`ip`
FROM ( SELECT `userID`, COUNT( 1 ) AS Cnt
FROM `loginLogs`
GROUP BY `userID`
HAVING COUNT( DISTINCT `ip` ) > 1 ) id
LEFT JOIN `loginLogs` ll
ON id.`userID` = ll.`userID`
ORDER BY id.`Cnt`;
If you just want to see the user with the most ips and you also want to see the list of ips, you can use GROUP_CONCAT():
SELECT `userID`, group_concat(DISTINCT `ip`)
FROM `loginLogs`
GROUP BY `userID`
ORDER BY COUNT( DISTINCT `ip` ) DESC
LIMIT 1
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;