i'm having problems with a query. I have two tables: country and city and i want to display the city with the highest population per country.
Here's the query:
select country.name as coname, city.name as ciname, max(city.population) as pop
from city
join country on city.countrycode=country.code
group by country.name
order by pop;`
Error
column "city.name" must appear in the GROUP BY clause or be used in an aggregate function.
I don't know how to solve this, i tried to make a subquery but it didn't work out.
How can i make it work?
You can easly get it using rank function:
select * from
(
select country.name as coname,
city.name as ciname,
city.population,
rank() over (partition by country.name order by city.population desc) as ranking
from
city
join
country
on city.countrycode=country.code
) A
where ranking = 1
Related
From this table, I'm trying to determine the nation (s) that have the highest number of teams (a nation X has a team if it has at least one athlete from that country X).
driver(id,name, team, country)
This solution restores all countries in descending order. Would it be possible to ensure that only the one (s) with the most team (s) return and not all of them? I think you should use the 'max' command but I'm not sure.
SELECT (country) ,count(distinct team)
FROM driver
GROUP BY country
order by count(distinct team) DESC;
I would use your query as a CTE and then select from it like this -
WITH t AS
(
SELECT country, count(distinct team) cnt
FROM driver
GROUP BY country
)
SELECT country, cnt FROM t
WHERE cnt = (SELECT max(cnt) FROM t);
You can combine this with a window function:
with counts as (
SELECT country,
count(distinct team) as num_teams,
dense_rank() over (order by count(distinct team) desc) as rnk
FROM driver
GROUP BY country
)
select country, num_teams
from counts
where rnk = 1;
If you are using Postgres 14, you can use fetch first with the option with ties:
SELECT country,
count(distinct team) as num_teams
FROM driver
GROUP BY country
order by count(distinct team) desc
fetch first 1 rows with ties
If two countries have the same highest number of drivers, this would return both. Without the with ties option (which was introduced in Postgres 14) only one of them would be returned.
table image
I have this table that I need to sort in the following way:
need to rank Departments by Salary;
need to show if Salary = NULL - 'No data to be shown' message
need to add total salary paid to the department
need to count people in the department
SELECT RANK() OVER (
ORDER BY Salary DESC
)
,CASE
WHEN Salary IS NULL
THEN 'NO DATA TO BE SHOWN'
ELSE Salary
,Count(Fname)
,Total(Salary) FROM dbo.Employees
I get an error saying:
Column 'dbo.Employees.Salary' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.
Why so?
Column 'dbo.Employees.Salary' is invalid in the select list because it
is not contained in either an aggregate function or the GROUP BY
clause.
Why so?
The aggregate functions are returning a single value for the whole table, you can't SELECT a field alongside them it doesn't makes sense. Like say, you have a students table you apply Sum(marks) for the whole students table, and you are then also selecting student's name Select studentname in your query. Which student's name will the database engine select? Confusing
Column "invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause"
I tried this-
using inner query
SELECT RANK() OVER (ORDER BY SAL DESC) RANK,FNAME,DEPARTMENT
CASE
WHEN SAL IS NULL THEN 'NO DATA TO BE SHOWN'
ELSE SAL
END
FROM
(SELECT COUNT(FNAME) FNAME, SUM(SALARY) SAL, DEPARTMENT
FROM TESTEMPLOYEE
GROUP BY DEPARTMENT) t
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;
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.