I spent a lot of time building this select, but Im not able to solve it. I have 2 tables. First table is called car and has PK (primary key) id_car and another columns name and so on. Second table is called rating and has colums id_rating (PK), id_car (FK) and rating_value (integer). As you suspect, one car can have more than one ranting. I want to select all cars and I want to know average rating to each car. Finally, I want to order the result by this average desc. I was trying things like this:
SELECT id_car, name, average
FROM car C, rating R
WHERE C.id_car = R.id_car
ORDER BY (average) (
SELECT AVG(rating_value) AS average
FROM rating R
WHERE C.id_car = R.id_car)
but it doesn't work.
For SQL Server; Also I suggest you to use JOIN instead of WHERE table1, table2..
SELECT C.id_car, name, AVG(rating_value) AS average
FROM car C JOIN rating R
ON C.id_car = R.id_car
GROUP By C.id_car, name
ORDER BY average DESC
This implements the aggregate function AVG() and then a GROUP BY the car id and name:
select c.id_car, c.name, avg(r.rating_value) aver
from car c
left join rating r
on c.id_car = r.id_car
group by c.id_car, c.name
order by aver desc
Using a LEFT JOIN will include all cars in the result even those which have no yet been rated.
SELECT C.name, AVG(R.rating_value)
FROM car C, rating R
WHERE C.id_car = R.id_car
GROUP BY C.name
ORDER BY AVG(R.rating_value) DESC
SELECT c.id_car, c.name,avg(r.rating_value) as rating
FROM car c
join rating r
on c.id_car = r.id_car
group by r.id_car
order by rating
Related
How do you ORDER BY number of rows found in another table? I have a table for animals (these are livestock animals) and another table for awards. When an animal wins an award, the award gets added to the awards table.
People want to be able to find the animals who have won the most awards (WHERE award type is 1), ordered from most awards to least. How do I ORDER BY how many awards they have if the awards are in a separate table each with their own row?
SELECT animals.id
FROM animals
LEFT JOIN awards ON animals.id = awards.animalid
WHERE awards.type = 1
ORDER BY...
You would seem to want GROUP BY:
SELECT a.id
FROM animals a LEFT JOIN
awards aw
ON a.id = aw.animalid AND aw.type = 1
GROUP BY a.id
ORDER BY COUNT(aw.animalid) DESC;
I have a movie table which has year and movie details like title , movie id( mid) and a table m_cast where i have all the actors in that movie.
I would like to get all the actors who have never been unemployed for more than 3 years. ( Assuming actors are unemployed between two consecutive movies)
i code i came up with is
select a.yr1 y1 , b.yr2 y2 , a.yr1 - b.yr2 diff from
(select substr(substr(trim(year),-5),0,5) yr1 , * from movie m inner join m_cast p on m.mid = p.mid order by pid , yr1) a ,
(select substr(substr(trim(year),-5),0,5) yr2 , * from movie m inner join m_cast p on m.mid = p.mid order by pid, yr2) b on a.yr1 > b.yr2
where not exists
(select count(*) from movie m inner join m_cast p on m.mid = p.mid
and cast(substr(substr(trim(year),-5),0,5) as integer) < a.yr1 and cast(substr(substr(trim(year),-5),0,5) as integer) > b.yr2)
Self join itself takes a lot of time. And lag and lead functions do not work in SQLite version i am using.
I'm assuming the movie table has a column called year, and a column to identify the actor's name. Something like : year int, actorId int
The fastest way to run your query is to filter the last 3 years from your movie table and then to group by your actors the distinct count of years.
Example after filtering
ActorId Year
1. 2018
1. 2018
1. 2017
2. 2016
2. 2017
2. 2018
Then group by and select distinct :
Select actorId from movieTable group by actorId having count (distinct (Year)) =3
And that will only return the actors who have worked in the last 3 years. Once you have your actors id's filtered out in that column do a join to the table that holds their names.
Sorry about the format of my writing - did it from my cellphone.
Regards,
Jorge D. Lopez
My school task was to get names from my movie database actors which play in movies with highest ratings
I made it this way and it works :
select name,surname
from actor
where ACTORID in(
select actorid
from actor_movie
where MOVIEID in (
select movieid
from movie
where RATINGID in (
select ratingid
from rating
where PERCENT_CSFD = (
select max(percent_csfd)
from rating
)
)
)
);
the output is :
Gary Oldman
Sigourney Weaver
...but I'd like to also add to this select mentioned movie and its rating. It accessible in inner selects but I don't know how to join it with outer select in which i can work just with rows found in Actor Table.
Thank you for your answers.
You just need to join the tables properly. Afterwards you can simply add the columns you´d like to select. The final select could be looking like this.
select ac.name, ac.surname, -- go on selecting from the different tables
from actor ac
inner join actor_movie amo
on amo.actorid = ac.actorid
inner join movie mo
on amo.movieid = mo.movieid
inner join rating ra
on ra.ratingid = mo.ratingid
where ra.PERCENT_CSFD =
(select max(percent_csfd)
from rating)
A way to get your result with a slightly different method could be something like:
select *
from
(
select name, surname, percent_csfd, row_number() over ( order by percent_csfd desc) as rank
from actor
inner join actor_movie
using (actorId)
inner join movie
using (movieId)
inner join rating
using(ratingId)
(
where rank = 1
This uses row_number to evaluate the "rank" of the movie(s) and then filter for the movie(s) with the highest rating.
I have a table of cars where each car belongs to a company. In another table I have a list of company locations by city.
I want to select all cars from the cars table whose company has locations on all cities passed into the stored procedure, otherwise exclude those cars all together even if it falls short of one city.
So, I've tried something like:
select id, cartype from cars where companyid in
(
select id from locations where cityid in
(
select id from cities
)
)
This doesn't work as it obviously satisfies the condition if ANY of the cities are in the list, not all of them.
It sounds like a group by count, but can't make it work with what I tried.
I"m using MS SQL 2005
One example:
select id, cartype from cars c
where ( select count(1) from cities where id in (...))
= ( select count(distinct cityid)
from locations
where c.companyid = locations.id and cityid in (...) )
Maybe try counting all the cities, and then select the car if the company has the same number of distinct location cities are there are total cities.
SELECT id, cartype FROM cars
WHERE
--Subquery to find the number of locations belonging to car's company
(SELECT count(distinct cities.id) FROM cities
INNER JOIN locations on locations.cityid = cities.id
WHERE locations.companyId = cars.companyId)
=
--Subquery to find the total number of locations
(SELECT count(distinct cities.id) FROM cities)
I haven't tested this, and it may not be the most efficient query, but I think this might work.
Try this
SELECT e.*
FROM cars e
WHERE NOT EXISTS (
SELECT 1
FROM Cities p
WHERE p.location = e.Location
)
Imagine you have two tables, with a one to many relationship.
For this example, I will suggest that there are two tables: Person, and Homes.
The person table holds a persons name, and gives them an ID. The homes table, holds the association of homes to a person. PID joins to "Person.ID"
And, in this tiny DB, a person can have no homes, or many homes.
I hope I drew that right.
How do I write a select, that returns everyone with every specified house type?
Let's say these are valid "Types" in the homes table:
Cottage, Main, Mansion, Spaceport.
I want to return everyone, in the Person table, who has a spaceport and a Cottage.
The best I could come up with was this:
SELECT DISTINCT( p.name ) AS name
FROM person p
INNER JOIN homes h ON h.pid = p.id
WHERE 'spaceport' in (
SELECT DISTINCT( type ) AS type
FROM homes
WHERE pid = p.id
)
AND 'cottage' in (
SELECT DISTINCT( type ) AS type
FROM homes
WHERE pid = p.id
)
When I wrote that, it works, but I'm pretty sure there has to be a better way.
The HAVING clause here will guarantee that the persons returned have both types, not just one or the other.
SELECT p.name
FROM person p
INNER JOIN homes h
ON p.id = h.pid
AND h.type IN ('spaceport', 'cottage')
GROUP BY p.name
HAVING COUNT(DISTINCT h.type) = 2
select * from homes;
home_id person_id type
--
1 1 cottage
2 1 mansion
3 2 cottage
4 3 mansion
5 4 cottage
6 4 cottage
To find the id numbers of every person who has both a cottage and a mansion, group by the id number, restrict the output to cottages and mansions, and count the distinct types.
select person_id
from homes
where type in ('cottage','mansion')
group by person_id
having count(distinct type) = 2;
person_id
--
1
You can use this query in a join to get all the columns from the person table.
select person.*
from person
inner join (select person_id
from homes
where type in ('cottage','mansion')
group by person_id
having count(distinct type) = 2) T
on person.person_id = T.person_id;
Thanks to Joe for pointing out an error in my count().
Not sure about the performance on this one, but here goes:
SELECT PID FROM (
SELECT PID, COUNT(PID) cnt FROM (
SELECT DISTINCT PID, Type FROM Homes
WHERE Type IN ('Type1', 'Type2', 'Type3')
) a
GROUP BY PID
) b
WHERE b.cnt = 3
You'd have to dynamically generate your IN clause as well as the WHERE b.CNT clause.