Consider a database schema with three relations:
Employee (*eid:integer*, ename:string, age:integer, salary:real)
Works (*eid:integer*, *did:integer*, pct_time:integer)
Department(*did:integer*, dname:string, budget:real, managerid:integer)
Query the view above to find the name of the manager who manages most employees. If thesame employee works in several departments, that employee is counted once in each of the
departments. The manager is included in the count the same as all other employees, i.e., based
on his or her records in the Works table.
Why do I get this error:
ORDER BY SUM (EmpCount) DESC)
*
ERROR at line 6:
ORA-00907: missing right parenthesis
Here is my query:
SELECT distinct(MgrName)
FROM ManagerSummary
WHERE MgrID = (SELECT MgrID
FROM ManagerSummary
GROUP BY MgrID
ORDER BY SUM (EmpCount) DESC
LIMIT 1 );
The view is:
CREATE VIEW ManagerSummary (DeptName, MgrID, MgrName, MgrSalary, EmpCount)
AS SELECT d.dname, d.managerid, e.ename, e.salary,
(SELECT COUNT (w.did)
FROM works w
WHERE w.did = d.did
GROUP BY w.did)
FROM employee e, department d WHERE d.managerid = e.eid;
Thank you
Update: Changing LIMIT 1 for WHERE ROWNUM = 1 doesn't help
Try this
SELECT DISTINCT (MgrName)
FROM ManagerSummary
WHERE MgrID = (SELECT MgrId
FROM ( SELECT MgrId, SUM (empcount) AS maxemp
FROM ManagerSummary
GROUP BY MgrId
ORDER BY SUM (empcount) DESC)
WHERE ROWNUM = 1)
You seem to want the name of the manager that has the most employees.
My guess is that you can do this in Oracle as:
select ms.MgrName
from (select ms.*
from ManagerSummary ms
order by EmpCount desc
) ms
where rownum = 1;
It is hard for me to envision a table called ManagerSummary that would have more than one row per MgrId. That's why I don't think aggregation is necessary.
SELECT mgrname
FROM (SELECT mgrname, numemps
FROM (SELECT mgrname, count(*) numemps
FROM mgrsummary
GROUP BY mgrname)
ORDER BY NUMEMPS desc);
Just noticed - this is based on a view. This is ~not~ going to perform well.
Related
I am trying to use a query to return the count from rows such that the date of the rows matches the maximum date for that column in the table.
Oracle SQL: version 11.2:
The following syntax would seem to be correct (to me), and it compiles and runs. However, instead of returning JUST the count for the maximum, it returns several counts more or less like the "HAIVNG" clause wasn't there.
Select ourDate, Count(1) as OUR_COUNT
from schema1.table1
group by ourDate
HAVING ourDate = max(ourDate) ;
How can this be fixed, please?
You can use:
SELECT MAX(ourDate) AS ourDate,
COUNT(*) KEEP (DENSE_RANK LAST ORDER BY ourDate) AS ourCount
FROM schema1.table1
or:
SELECT ourDate,
COUNT(*) AS our_count
FROM (
SELECT ourDate,
RANK() OVER (ORDER BY ourDate DESC) AS rnk
FROM schema1.table1
)
WHERE rnk = 1
GROUP BY ourDate
Which, for the sample data:
CREATE TABLE table1 (ourDate) AS
SELECT SYSDATE FROM DUAL CONNECT BY LEVEL <= 5 UNION ALL
SELECT SYSDATE - 1 FROM DUAL;
Both output:
OURDATE
OUR_COUNT
2022-06-28 13:35:01
5
db<>fiddle here
I don't know if I understand what you want. Try this:
Select x.ourDate, Count(1) as OUR_COUNT
from schema1.table1 x
where x.ourDate = (select max(y.ourDate) from schema1.table1 y)
group by x.ourDate
One option is to use a subquery which fetches maximum date:
select ourdate, count(*)
from table1
where ourdate = (select max(ourdate)
from table1)
group by ourdate;
Or, a more modern approach (if your database version supports it; 11g doesn't, though):
select ourdate, count(*)
from table1
group by ourdate
order by ourdate desc
fetch first 1 rows only;
You can use this SQL query:
select MAX(ourDate),COUNT(1) as OUR_COUNT
from schema1.table1
where ourDate = (select MAX(ourDate) from schema1.table1)
group by ourDate;
I got one error
ORA-00923: FROM keyword not found where expected
Here is my query
SELECT * FROM
(SELECT *, (SELECT COUNT(*) FROM invoices) AS numberOfRows
FROM invoices ORDER BY Id DESC) WHERE rownum <= 1
I am begginer in Oracle SQL, but as I see here I have FROM keyword and it looks everythink OK.
I try to modify this query something like but still get another error
ORA-00933: SQL command not properly ended
SELECT * FROM
(SELECT COUNT(*) FROM invoices) AS numberOfRows
FROM invoices ORDER BY Id DESC) WHERE rownum <= 1
What is wrong in first select query ? What is missing ? Since I check everything, start from special character ( . , )
Also I try this kind of solution and get error
ORA-00936: missing expression
SELECT * FROM (SELECT , (SELECT COUNT() FROM invoices) AS numberOfRows FROM invoices ORDER BY Id DESC) WHERE rownum <= 1
The railroad diagram in the documentation:
... shows that you can either use * on its own, or <something>.* along with other columns or expressions. So you need to precede your * with the table name or an alias:
SELECT * FROM
(SELECT i.*, (SELECT COUNT(*) FROM invoices) AS numberOfRows
FROM invoices i ORDER BY Id DESC) WHERE rownum <= 1
If you're on a recent version of Oracle you can do this much more simply with:
select i.*, count(*) over () as numberOfRows
from invoices i
order by id desc
fetch first row only
On older version you still need a subquery, but only one level:
select *
from (
select i.*, count(*) over () as numberOfRows
from invoices i
order by id desc
)
where rownum = 1
db<>fiddle
looks like the FROM is missing from this select "SELECT *,"
SELECT * FROM
(SELECT , (SELECT COUNT() FROM invoices) AS numberOfRows
FROM invoices ORDER BY Id DESC) WHERE rownum <= 1
I need to write query to remove duplicate employee ids among 10,000 results
EmpID name
1 x
1 x
2 y
2 y
3 z
4 A
The result should be only:
EmpID name
3 z
4 A
Select * from EMPLOYEE where ?
How can I do this?
You need aggregation :
select e.empid, e.name
from employee e
group by e.empid, e.name
having count(*) = 1;
You can try below -
Select empid,name from EMPLOYEE
group by empid,name
having count(*)=1
I would really ask you to use other than the GROUP BY method as you will be able to fetch other fields of the EMPLOYEE table along with EMPID and NAME.
Using analytical function
SELECT * FROM
(SELECT T.*, COUNT(1) OVER (PARTITION BY EMPID) AS CNT
FROM EMPLOYEE)
WHERE CNT = 1;
Using NOT EXISTS
SELECT * FROM EMPLOYEE T
WHERE NOT EXISTS
(SELECT 1 FROM EMPLOYEE TIN
WHERE TIN.ID = T.ID AND TIN.ROWID <> T.ROWID);
Cheers!!
I have a table department with 3 column(department_name , department_id, department_block_number) so I want to fetch the department_block_number in which maximum number of department is located ? I have two department_block_number 303, 202 and each has 4 and 2 departments respectively? how can i do it?
select q1.department_block_number , max(c)
(select department_block_number , count(department_id)as c from department group by department_block_number)q1,
group by department_block_number ;
select q1.department_block_number , max(c)
(select department_block_number , count(department_id)as c from department group by department_block_number)q1,
group by department_block_number ;
Now i want to show only 303 as it is the block number with maximum departments in it but my query is showing both 303, 202 please help me . If you know some other way so that i can fetch the result so please help
In standard SQL if you are looking for one row, you would do:
select d.department_block_number, count(*)
from department d
group by d.department_block_number
order by count(*) desc
fetch first 1 row only;
Some databases spell fetch first 1 row only as limit 1 or select top (1) or in even more arcane ways.
In older versions of Oracle (fetch is supported in 12c+), you can do:
select department_block_number, cnt
from (select d.department_block_number, count(*) as cnt
from department d
group by d.department_block_number
order by count(*) desc
) d
where rownum = 1;
So I have two tables salary and emp whose definition is shown as below
[
I am tring to create a query that Find the employee who draws the maximum salary and Display the employee details along with the nationality.
I created this query
select empcode,
max(basic) as "Highest Sal"
from salary
join emp on empcode;
Please help with this
Your query uses a simple aggregate max(basic) which would find the highest salary. Except you need to join to the EMP table to display other details. This means you can't use aggregation, because we need to GROUP BY the non-aggregated columns, which would make a nonsense of the query.
Fortunately we can solve the problem with an analytic function. The subquery selects all the relevant information and ranks each employee by salary, with a rank of 1 being the highest paid. We use rank() here because that will handle ties: two employees with the same basic will be in the same rank.
select empcode
, empname
, nationality
, "Highest Sal"
from (
select emp.empcode
, emp.empname
, emp.nationality
, salary.basic as "Highest Sal"
, rank() over (order by salary.basic desc ) as rnk
from salary join emp on emp.empcode = salary.empcode
)
where rnk = 1;
Find the employee who draws the maximum salary
An employee can have multiple salaries in your datamodel. An employee's (total) salary hence is the sum of these. You want to find the maximum salary per employee and show the employee(s) earning that much.
You can use MAX OVER to find the maximum sum:
select e.*, s.total_salary
from emp e
join
(
select
empcode,
sum(basic) as total_salary,
max(sum(basic)) over () as max_total_salary
from salary
) s on s.empcode = e.empcode and s.total_salary = s.max_total_salary
order by e.empcode;
Try this:
SELECT * FROM
(SELECT E.EmpCode, E.EmpName, E.DOB, E.DOJ, E.DeptCode, E.DesgCode, E.PhNo,
E.Qualification, E.Nationality, S.Basic, S.HRA, S.TA, S.UTA, S.OTRate
FROM EMP AS E JOIN SALARY AS S ON (E.EmpCode = S.EmpCode) order by S.Basic desc)
WHERE rownum = 1