I have a table employees with columns:
company_id,
id,
opt_state (ceased_membership, ignition, opted_out, opted_in),
opt_out_on.
I want to query all companies where all employees opt-state is in ('ceased_membership', 'ignition', 'opted_out') and the date opt_out_on when last employee left.
I have tried this but it didn't work
select company_id from employees where id=all(select id from
employees
where opt_state in ('ceased_membership', 'ignition','opted_out')
Then I wrote this query below, which worked very well and gave me the resolution I was looking for. However, I'd like to ask here if this can be done differently, more elegantly.
SELECT
e.company_id
, max_opt_out
FROM (
SELECT DISTINCT
company_id
, count(id)
OVER (
PARTITION BY company_id ) opt_out
FROM employees
WHERE opt_state IN ('ceased_membership', 'ignition', 'opted_out')) e
LEFT JOIN (
SELECT
company_id
, count(id) opt_in
, max(opt_out_on) max_opt_out
FROM employees
GROUP BY company_id) S
ON e.company_id = s.company_id
WHERE e.opt_out = s.opt_in;
This seems like a good time to use the HAVING clause
SELECT company_id, max(opt_out_on)
FROM employees e
GROUP BY company_id
HAVING bool_and( opt_state in ('ceased_membership', 'ignition','opted_out'));
HAVING in a bit like a WHERE but the condition apples to whole GROUPS
bool_and is an agregate function that is only true when all the records in the group are result in true.
I'd say that you want to query a maximum out_out_on for each company that only have employees in a set of states, which means that do not have any employee not in a set of states.
So, translated to SQL:
select company_id, max(opt_out_on)
from employees e
where not exists(
select 1 from employees
where company_id=e.company_id
and opt_state not in ('ceased_membership', 'ignition','opted_out')
)
group by company_id;
Related
The domain is:
company (id, name, adress)
employee (id, name, adress, company_id, expertise_id)
dependantrelative (id, name, employee_id)
expertise (id, name, class)
I want to know how to get the number of dependantrelatives of each employee who are unique experts in their respective companies.
The Query below does not return the correct answer. Can you help me?
SELECT DISTINCT dependantrelative.employee_id
, COUNT(*) AS qty_dependantrelatives
FROM dependantrelative
INNER JOIN employee
ON employee.id = dependantrelative.employee_id
GROUP BY dependantrelative.employee_id
I just tried out the Query below and it works, but I want to know if there is a faster and simple way of getting the answer.
SELECT employee.id
,COUNT(dependantrelative.employee_id) AS qty_dependantrelatives
FROM (
SELECT employee.company_id
, employee.expertise_id AS expert
, COUNT(employee.expertise_id)
FROM employee
GROUP BY employee.company_id
, employee.expertise_id
HAVING COUNT(employee.expertise_id)<2
) AS uniexpert
LEFT JOIN employee
ON employee.expertise_id = uniexpert.expert
LEFT JOIN salesorderdetail
ON dependantrelative.employee_id = employee.id
GROUP BY employee.id
ORDER BY employee.id
I need to write a sql code that probably is very simple but I am very new to it.
I need to find all the records from one table that have matching id (but no more than one) from the other table. eg. one table contains records of the employees and the second one with employees' telephone numbers. i need to find all employees with only one telephone no
Sample data would be nice. In absence of:
SELECT
employees.employee_id
FROM
employees
LEFT JOIN
(SELECT distinct on(employee_id) employee_id FROM emp_phone) AS phone
ON
employees.employee_id = phone.employee_id
WHERE
phone.employee_id IS NOT NULL;
You need a join of the 2 tables, group by employee and the condition in the having clause:
SELECT e.employee_id, e.name
FROM employees e INNER JOIN numbers n
ON e.employee_id = n.employee_id
GROUP BY e.employee_id, e.name
HAVING COUNT(*) = 1;
If there can be more than a few numbers per employee in the table with the employees' telephone numbers (calling it tel), then it's cheaper to avoid GROUP BY and HAVING which has to process all rows. Find employees with "unique" numbers using a self-anti-join with NOT EXISTS.
While you don't need more than the employee_id and their unique phone number, you don't even have to involve the employee table at all:
SELECT *
FROM tel t
WHERE NOT EXISTS (
SELECT FROM tel
WHERE employee_id = t.employee_id
AND tel_number <> t.tel_number -- or use PK column
);
If you need additional columns from the employee table:
SELECT * -- or any columns you need
FROM (
SELECT employee_id AS id, tel_number -- or any columns you need
FROM tel t
WHERE NOT EXISTS (
SELECT FROM tel
WHERE employee_id = t.employee_id
AND tel_number <> t.tel_number -- or use PK column
)
) t
JOIN employee e USING (id);
The column alias in the subquery (employee_id AS id) is just for convenience. Then the outer join condition can be USING (id), and the ID column is only included once in the result, even with SELECT * ...
Simpler with a smart naming convention that uses employee_id for the employee ID everywhere. But it's a widespread anti-pattern to use employee.id instead.
Related:
JOIN table if condition is satisfied, else perform no join
I'v provided following query to find common records in 2 data sets but it's difficult for me to make sure about correctness of my query because of that I have a lot of data records in my DB.
Is it OK to implement Intersect between "Customers" & "Employees" tables using UNION ALL and apply GROUP BY on the result like below?
SELECT D.Country, D.Region, D.City
FROM (SELECT DISTINCT Country, Region, City
FROM Customers
UNION ALL
SELECT DISTINCT Country, Region, City
FROM Employees) AS D
GROUP BY D.Country, D.Region, D.City
HAVING COUNT(*) = 2;
So can we say that any record which exists in the result of this query also exists in the Intersect set between "Customers & Employees" tables AND any record that exists in Intersect set between "Customers & Employees" tables will be in the result of this query too?
So is it right to say any record in result of this query is in
"Intersect" set between "Customers & Employees" "AND" any record that
exist in "Intersect" set between "Customers & Employees" is in result
of this query too?
YES.
... Yes, but it won't be as efficient because you are filtering out duplicates three times instead of once. In your query you're
Using DISTINCT to pull unique records from employees
Using DISTINCT to pull unique records from customers
Combining both queries using UNION ALL
Using GROUP BY in your outer query to to filter the records you retrieved in steps 1,2 and 3.
Using INTERSECT will return identical results but more efficiently. To see for yourself you can create the sample data below and run both queries:
use tempdb
go
if object_id('dbo.customers') is not null drop table dbo.customers;
if object_id('dbo.employees') is not null drop table dbo.employees;
create table dbo.customers
(
customerId int identity,
country varchar(50),
region varchar(50),
city varchar(100)
);
create table dbo.employees
(
employeeId int identity,
country varchar(50),
region varchar(50),
city varchar(100)
);
insert dbo.customers(country, region, city)
values ('us', 'N/E', 'New York'), ('us', 'N/W', 'Seattle'),('us', 'Midwest', 'Chicago');
insert dbo.employees
values ('us', 'S/E', 'Miami'), ('us', 'N/W', 'Portland'),('us', 'Midwest', 'Chicago');
Run these queries:
SELECT D.Country, D.Region, D.City
FROM
(
SELECT DISTINCT Country, Region, City
FROM Customers
UNION ALL
SELECT DISTINCT Country, Region, City
FROM Employees
) AS D
GROUP BY D.Country, D.Region, D.City
HAVING COUNT(*) = 2;
SELECT Country, Region, City
FROM dbo.customers
INTERSECT
SELECT Country, Region, City
FROM dbo.employees;
Results:
Country Region City
----------- ---------- ----------
us Midwest Chicago
Country Region City
----------- ---------- ----------
us Midwest Chicago
If using INTERSECT is not an option OR you want a faster query you could improve the query you posted a couple different ways, such as:
Option 1: let GROUP BY handle ALL the de-duplication like this:
This is the same as what you posted but without the DISTINCTS
SELECT D.Country, D.Region, D.City
FROM
(
SELECT Country, Region, City
FROM Customers
UNION ALL
SELECT Country, Region, City
FROM Employees
) AS D
GROUP BY D.Country, D.Region, D.City
HAVING COUNT(*) = 2;
Option 2: Use ROW_NUMBER
This would be my preference and will likely be most efficient
SELECT Country, Region, City
FROM
(
SELECT
rn = row_number() over (partition by D.Country, D.Region, D.City order by (SELECT null)),
D.Country, D.Region, D.City
FROM
(
SELECT Country, Region, City
FROM Customers
UNION ALL
SELECT Country, Region, City
FROM Employees
) AS D
) uniquify
WHERE rn = 2;
I am creating some queries for my project, but I face some difficulties with the follow ones:
A SELECT statement containing a subquery to retrieve a list of Locations (location id and street_address) that have employees with higher salary than the average of their department. The list must contain the number of those employees and their total salary per location. Name these aggregates respectively "emp" and "totalsalary". The locations in the list must be ordered by location_id.
Select LOCATION_ID, STREET_ADDRESS
from HR.LOCATIONS IN
(Select Employee_id
from HR.Employees
Where Salary > round(avg(SALARY)))
order by location_id;
error: SQL command not properly ended
and the second query is the following
The JOB_HISTORY table can contain more than one entries for an employee who was hired more than once. Create a query to retrieve a list of Employees that were hired more than once. Include the columns EMPLOYEE_ID, LAST_NAME, FIRST_NAME and the aggregate "Times Hired".
SELECT FIRST_NAME,LAST_NAME,EMPLOYEE_ID,
count (*)as TIMES_HIRED
from HR.JOB_HISTORY, HR.EMPLOYEES
where EMPLOYEE_ID= LAST_NAME
having COUNT(*) >1;
error: not a single-group
Try these hope they help. I am making an assumption that employee table has Location_Id column. I am adding Employee_id to Group by to make sure you get correct TotalSalary:
Select LOCATION_ID, STREET_ADDRESS, Count(Employee_id) AS emp, SUM(salary) AS totalsalary
from HR.LOCATIONS INNER JOIN
(Select Employee_id, salary
from HR.Employees
Having Salary > round(avg(SALARY), 0)) AS Emp ON HR.LOCATION_ID = Emp.Location_ID
Group By LOCATION_ID, STREET_ADDRESS, Employee_id
order by location_id;
For the second question:
SELECT FIRST_NAME,LAST_NAME,EMPLOYEE_ID,
count(Employee_id) as TIMES_HIRED
from HR.JOB_HISTORY inner join HR.EMPLOYEES On JOB_HISTORY.Employee_id = Employees.Employee_id
Group By FIRST_NAME,LAST_NAME,EMPLOYEE_ID
Having count(Employee_id) >1;
I am selecting column used in group by and count, and query looks something like
SELECT s.country, count(*) AS posts_ct
FROM store s
JOIN store_post_map sp ON sp.store_id = s.id
GROUP BY 1;
However, I want to select some more fields, like store name or store address from store table where count is max, but I don't to include that in group by clause.
For instance, to get the stores with the highest post-count per country:
SELECT DISTINCT ON (s.country)
s.country, s.store_id, s.name, sp.post_ct
FROM store s
JOIN (
SELECT store_id, count(*) AS post_ct
FROM store_post_map
GROUP BY store_id
) sp ON sp.store_id = s.id
ORDER BY s.country, sp.post_ct DESC
Add any number of columns from store to the SELECT list.
Details about this query style in this related answer:
Select first row in each GROUP BY group?
Reply to comment
This produces the count per country and picks (one of) the store(s) with the highest post-count:
SELECT DISTINCT ON (s.country)
s.country, s.store_id, s.name
,sum(post_ct) OVER (PARTITION BY s.country) AS post_ct_for_country
FROM store s
JOIN (
SELECT store_id, count(*) AS post_ct
FROM store_post_map
GROUP BY store_id
) sp ON sp.store_id = s.id
ORDER BY s.country, sp.post_ct DESC;
This works because the window function sum() is applied before DISTINCT ON per definition.