So I have 3 tables referencing cars, assurance and accident.
I want to know the brand of vehicles who had the most accidents, compared to others.
I have tried a lot of ways to that, but mostly i only get or all the brands returned or the brand of the car that was registered the most, not the one that had most accidents
These are my tables
create table car(
n_veic bigint not null,
matric varchar(15) not null,
pais_matric text not null,
n_pess bigint not null,
tipo text not null,
cor text not null,
brand text not null,
modelo varchar(15),
primary key(n_veic),
unique(matric),
foreign key (n_pess) references pessoa(n_pess)
);
create table ensurance(
apolice bigint not null,
segurado bigint not null,
car bigint not null,
datai date not null,
dataf date not null,
cobertura numeric(10,2) not null,
primary key(apolice),
unique(segurado, veiculo),
foreign key (segurado) references pessoa(n_pess),
foreign key (car) references car(n_veic)
);
create table accident(
n_acid bigint not null,
pess_segura bigint not null,
veic_seguro bigint not null,
data date not null,
local varchar(255) not null,
descr text not null,
primary key(n_acid),
unique(n_acid, veic_seguro),
foreign key (pess_segura,veic_seguro) references ensurance(segurado, car)
This is what i tried
SELECT marca
FROM veiculo NATURAL JOIN acidente
GROUP BY marca
HAVING count (distinct n_veic)>=ALL
(SELECT count (distinct n_veic)
FROM veiculo NATURAL JOIN acidente
GROUP BY marca);
I think the logic is:
select c.marca, count(*) as num_acidentes
from acidente a join
car c
on a.veic_seguro = c.n_veic
group by c.marca
order by num_acidentes desc;
You can use fetch first 1 row only -- or whatever is appropriate for your database -- to get only one row.
Try this-
Note:
1. Try to avoid NATURAL JOIN and use specific column reference.
2. Rethink DISTINCT for count is really necessary or not.
SELECT TOP 1 marca, COUNT(DISTINCT n_veic)
FROM veiculo
NATURAL JOIN acidente
GROUP BY marca
ORDER BY COUNT(DISTINCT n_veic) DESC
Related
So I am working on a football world cup database. These are my important tables:
CREATE TABLE Countries(
Cid SERIAL PRIMARY KEY,
Name VARCHAR(256) NOT NULL UNIQUE
);
CREATE TABLE Stadiums(
Sid SERIAL PRIMARY KEY,
Name VARCHAR(256) NOT NULL UNIQUE,
Cid INT REFERENCES Countries NOT NULL
);
CREATE TABLE Groups(
Gid SERIAL PRIMARY KEY,
Name VARCHAR(64) NOT NULL,
TYear SMALLINT REFERENCES Tournaments NOT NULL
);
CREATE TABLE Teams(
Tid SERIAL PRIMARY KEY,
Cid INT REFERENCES Countries NOT NULL,
Gid INT REFERENCES Groups NOT NULL
);
CREATE TABLE Matches(
Mid INT PRIMARY KEY,
HomeTid INT REFERENCES Teams NOT NULL,
VisitTid INT REFERENCES Teams NOT NULL,
HomeScore SMALLINT NOT NULL,
VisitScore SMALLINT NOT NULL,
MatchDate DATE NOT NULL,
MatchType VARCHAR(64) NOT NULL,
Sid INT REFERENCES Stadiums NOT NULL
);
CREATE TABLE tempmatches(
year INTEGER,
host_country VARCHAR(255),
match_id INTEGER,
type VARCHAR(255),
date DATE,
location1 VARCHAR(255),
team1 VARCHAR(255),
team2 VARCHAR(255),
score1 INTEGER,
score2 INTEGER
);
so my current problem is that I need to populate the columns HomeTid and VisitId of the Matches table with the tid's from the team's table that corresponds with the country of that team from the countries table but I'm not sure how to do that. I tried a few queries but none of them seemed to work. Has anyone an idea on how to solve this?
Using JOIN keyword you can combine 2 tables data.
Here, in this case, you have to use 3-way join.
Eg:
3-way JOIN can be used in this way:
SELECT * FROM table1 JOIN table2 ON table1.col1=table2.col2 JOIN table3 ON table3.col3=table.col1;
Here the 2 tables will get joined based on the columns mentioned after the "ON" keyword.
Here is a reference link in which JOIN is explained clearly
https://www.w3schools.com/sql/sql_join.asp
Use JOIN operation. In SQL it helps to combine several tables(queries) in one
I have some problems with a Query that I can not get out, the structure of the tables is as follows:
child
CREATE TABLE CHILD(
child_id SMALLINT,
child_name VARCHAR(255) NOT NULL,
birth_date DATE NOT NULL,
gender VARCHAR(255) NOT NULL,
address VARCHAR(255),
city VARCHAR(255),
CONSTRAINT PK_CHILD PRIMARY KEY(child_id)
letter
CREATE TABLE LETTER(
letter_id SMALLINT,
arrival_date DATE NOT NULL,
number_toys INTEGER NOT NULL,
delivery_mode VARCHAR(255) NOT NULL,
child_id SMALLINT,
CONSTRAINT PK_LETTER PRIMARY KEY(letter_id),
CONSTRAINT CHILD_FK FOREIGN KEY (child_id) REFERENCES CHILD(child_id)
);
wished_toy
CREATE TABLE WISHED_TOY(
letter_id SMALLINT,
toy_id SMALLINT,
CONSTRAINT PK_WISHED_TOY PRIMARY KEY(letter_id, toy_id),
CONSTRAINT LETTER_FK FOREIGN KEY (letter_id) REFERENCES LETTER(letter_id)
CONSTRAINT LETTER_FK FOREIGN KEY (toy_id) REFERENCES TOY(toy_id)
);
toy
CREATE TABLE TOY(
toy_id SMALLINT,
toy_name VARCHAR (255) NOT NULL,
price DECIMAL NOT NULL,
toy_type VARCHAR(255) NOT NULL,
manufacturer VARCHAR(255) NOT NULL,
CONSTRAINT PK_TOY PRIMARY KEY(toy_id),
);
I have to design a query that allows to obtain for each girl born before 2010 the
total number of toys you have ordered. The result must also include
those girls who have not written any letters. Specifically we want the
identifier of the girl, her name and city and the total number of toys ordered
(such as Total_number).
in the gender column of the child table can only be "m" of masculine or "f" of feminine
Do I have to use join?
the part that I have is this
select c.child_id,c.child_name,c.city
from letter l
join child c
on c.child_id=l.child_id
where c.gender='m' and c.birth_date between <='2010/01/01';
select c.child_id,c.child_name,c.city, coalesce( number_toys,0) counted
from
Child c
left outer join letter l
on c.child_id=l.child_id
where c.gender='m' and c.birth_date between <='2010/01/01';
Or
select c.child_id,c.child_name,c.city, count(wt.letter_id) actualToys
from
Child c
left outer join letter l
on c.child_id=l.child_id
left outer join wished_toy wt on l.letter_id=wt.letter_id
where c.gender='m' and c.birth_date between <='2010/01/01'
Group by
c.child_id,c.child_name,c.city
To count the toys referenced by database rows rather than the pre defined count
I'm using SQL Server query designer to try and form an outer query that will return the full name and address of each insured with home policies and those without policies. My create statements are the following:
CREATE TABLE Address (
AddressID integer NOT NULL,
HouseNumber Integer NOT NULL,
Street varchar(20) NOT NULL,
CityCounty varchar(20) NOT NULL,
StateAbb char(2),
CountryAbb char(2) NOT NULL,
Zip char(5) NOT NULL,
LastUpdatedBy varchar(20) NOT NULL,
LastUpdated date NOT NULL,
CONSTRAINT PK_Address PRIMARY KEY (AddressID));
CREATE TABLE Insured(
InsuredID integer NOT NULL,
FirstName varchar(15) NOT NULL,
LastName varchar(15) NOT NULL,
MI char(1),
DateOfBirth date NOT NULL,
CreditScore integer NOT NULL,
AddressID integer NOT NULL,
DriversLicenseNumber varchar(35),
LastUpdatedBy varchar(20) NOT NULL,
LastUpdated date NOT NULL,
CONSTRAINT PK_Insured PRIMARY KEY (InsuredID),
CONSTRAINT FK_InsuredAddress FOREIGN KEY (AddressID) references Address);
CREATE TABLE Policy(
PolicyID integer NOT NULL,
EffectiveDate date NOT NULL,
TerminationDate date NOT NULL,
Amount Numeric (8,2) NOT NULL,
PolicyYear integer NOT NULL,
PolicyType char(1) NOT NULL,
InsuredID integer NOT NULL,
AddressID integer NOT NULL,
LastUpdatedBy varchar(20) NOT NULL,
LastUpdated date NOT NULL,
CONSTRAINT PK_Policy PRIMARY KEY (PolicyID),
CONSTRAINT FK_PolicyAddress FOREIGN KEY (AddressID) references Address,
CONSTRAINT FK_PolicyInsured FOREIGN KEY (InsuredID) references Insured);
CREATE TABLE Home(
PolicyID integer NOT NULL,
ExteriorType varchar(30) NOT NULL,
Alarm char(3) NOT NULL,
DistanceToFireStation integer NOT NULL,
LastUpdatedBy varchar(20) NOT NULL,
LastUpdated date NOT NULL,
CONSTRAINT PK_Home PRIMARY KEY (PolicyID),
CONSTRAINT FK_HomePolicy FOREIGN KEY (PolicyID) references Policy);
CREATE TABLE Auto(
PolicyID integer NOT NULL,
VinNumber varchar(30) NOT NULL,
Make varchar(15) NOT NULL,
Model varchar(20) NOT NULL,
MilesPerYear integer NOT NULL,
LastUpdatedBy varchar(20) NOT NULL,
LastUpdated date NOT NULL,
CONSTRAINT PK_Auto PRIMARY KEY (PolicyID),
CONSTRAINT FK_AutoPolicy FOREIGN KEY (PolicyID) references Policy);
I believe that the query requires tables address, insured, policy and an outer right or left join but I cant get SQL server to recognize this as it keeps forming an inner join and cross join. What do I need for a query that returns insureds with home policies and their addresses and insureds with no policy and their addresses?
What I've tried so far:
SELECT Insured.InsuredID, Insured.FirstName,
Insured.LastName, Address.HouseNumber,
Policy.PolicyID
FROM Address RIGHT JOIN Policy
ON Address.AddressID = Policy.AddressID
RIGHT JOIN Insured ON Policy.AddressID = Insured.AddressID
ORDER BY Insured.InsuredID
This is the most recent query that returns what I need for insureds with a home policy but for the insureds without a policy I get nulls in the address.
SELECT i.InsuredID, i.FirstName, i.MI, i.LastName,
a.HouseNumber, a.Street, a.CityCounty, a.StateAbb, a.CountryAbb, a.Zip
FROM INSURED i
LEFT JOIN (SELECT * FROM Policy WHERE PolicyType = 'H') HomePolicy on
i.InsuredID = HomePolicy.InsuredID
LEFT JOIN Address a on HomePolicy.AddressID = a.AddressID;
Could you try this query:
SELECT i.InsuredID,
i.FirstName,
i.LastName,
a.HouseNumber,
p.PolicyID
FROM insured i
LEFT JOIN policy p ON i.AddressID = p.AddressID AND p.PolicyType = 'H'
LEFT JOIN address a ON i.AddressID = a.AddressID
ORDER BY i.InsuredID;
I think the joins were in the wrong order. Does this give you what you need?
Update: Joining the Insured table to the Address table will show you addresses regardless of if they have a policy or not.
Database design seem good.I think we have minor doubts regarding "PolicyType" column. what value it hold and what is the purpose of this column.
Say PolicyType='H' it mean that it is home policy . Or other way of finding same query is to check if that policyid exists in home table.
Is this correct ?
Main query,
What do I need for a query that returns insureds with home policies
and their addresses and insureds with no policy and their addresses?
Check this script,
--insureds with home policies and their addresses
select i.InsuredID,
i.FirstName,
i.LastName
A.HouseNumber
,1 INDICATE
from Insured i
INNER JOIN policy p ON i.InsuredID = p.InsuredID
INNER JOIN [Address] A ON A.ADDRESSID=I.ADDRESSID
WHERE EXISTS(SELECT PolicyID FROM Home H WHERE h.PolicyID=P.PolicyID)
AND NOT EXISTS(SELECT PolicyID FROM [Auto] a WHERE A.PolicyID=P.PolicyID)
UNION ALL
--insureds with no policy and their addresses
select i.InsuredID,
i.FirstName,
i.LastName
,A.HouseNumber
,0 INDICATE
from Insured i
INNER JOIN [Address] A ON A.ADDRESSID=I.ADDRESSID
WHERE EXISTS(SELECT InsuredID FROM policy p WHERE i.InsuredID = p.InsuredID )
I have use "EXISTS clause" because that table column is not require in your output.
I would need if I could help with the following code:
CREATE TABLE `car` (
`car_id` int(10) unsigned NOT NULL auto_increment,
`car_name` varchar(25) NOT NULL,
`car_year` date NOT NULL,
PRIMARY KEY (`user_id`)
) ENGINE=InnoDB;
CREATE TABLE `people` (
`peo_id` int(10) unsigned NOT NULL auto_increment,
`peo_name` varchar(50) NOT NULL,
`peo_surname` varchar(200) NOT NULL,
`car_id` int(10) unsigned NOT NULL,
PRIMARY KEY (`fav_id`),
KEY `FK_Favorites` (`user_id`),
CONSTRAINT `FK_Favorites` FOREIGN KEY (`car_id`) REFERENCES `car` (`car_id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB;
I have questions, show all cars that are more than 3 people ordered by year of car.
Thank you very much, sorry for my bad English
SELECT
c.car_id
,c.car_year
,COUNT(p.peo_id)
FROM car c
INNER JOIN people p on c.card_id = p.car_id
GROUP BY c.car_id ,c.car_year
HAVING COUNT(p.peo_id) > 3
ORDER BY c.car_year
First, find number of cars with more than three people
select car_id
from people
group by car_id
having count(*) > 3
Now, get the car details
select * from car
where car_id in
(
select car_id
from people
group by car_id
having count(*) > 3
)
order by car_year
I have the following problem to solve in SQL :
d) A query that provides management information on take up of the various types of activities on offer. For each type of activity, the query should show the total number of individuals who took that type of activity and the average number of individuals taking each type of activity.
Here are my tables :
CREATE TABLE accommodations
(
chalet_number int PRIMARY KEY,
chalet_name varchar(40) NOT NULL,
no_it_sleeps number(2) NOT NULL,
indivppw number(4) NOT NULL
)
CREATE TABLE supervisors
(
supervisor_number int PRIMARY KEY,
supervisor_forename varchar(30) NOT NULL,
supervisor_surname varchar(30) NOT NULL,
mobile_number varchar(11) NOT NULL
)
CREATE TABLE visitors
(
visitor_ID int PRIMARY KEY,
group_ID int NOT NULL,
forename varchar(20) NOT NULL,
surname varchar(20) NOT NULL,
dob date NOT NULL,
gender varchar(1) NOT NULL
)
CREATE TABLE activities
(
activity_code varchar(10) PRIMARY KEY,
activity_title varchar(20) NOT NULL,
"type" varchar(20) NOT NULL
)
CREATE TABLE "groups"
(
group_ID int PRIMARY KEY,
group_leader varchar(20) NOT NULL,
group_name varchar(30)
number_in_group number(2) NOT NULL
)
CREATE TABLE bookings
(
group_ID int NOT NULL,
start_date date NOT NULL,
chalet_number int NOT NULL,
no_in_chalet number(2) NOT NULL,
start_date date NOT NULL,
end_date date NOT NULL,
CONSTRAINT bookings_pk PRIMARY KEY(group_ID, chalet_number));
CREATE TABLE schedule
(
schedule_ID int PRIMARY KEY,
activity_code varchar(10) NOT NULL,
time_of_activity number(4,2) NOT NULL,
am_pm varchar(2) NOT NULL,
"date" date NOT NULL
)
CREATE TABLE activity_bookings
(
visitor_ID int NOT NULL,
schedule_ID int NOT NULL,
supervisor_number int NOT NULL,
comments varchar(200),
CONSTRAINT event_booking_pk PRIMARY KEY(visitor_ID, schedule_ID));
ALTER TABLE visitors
ADD FOREIGN KEY (group_ID)
REFERENCES "groups"(group_ID)
ALTER TABLE Schedule
ADD FOREIGN KEY (activity_code)
REFERENCES activities(activity_code)
ALTER TABLE bookings
ADD FOREIGN KEY (group_ID)
REFERENCES "groups"(group_ID)
ALTER TABLE bookings
ADD FOREIGN KEY (chalet_number)
REFERENCES accommodations(chalet_number)
ALTER TABLE activity_bookings
ADD FOREIGN KEY (visitor_ID)
REFERENCES visitors(visitor_ID)
ALTER TABLE activity_bookings
ADD FOREIGN KEY (schedule_ID)
REFERENCES schedule(schedule_ID)
ALTER TABLE activity_bookings
ADD FOREIGN KEY (supervisor_number)
REFERENCES supervisors(supervisor_number)
I have the following solution:
SELECT activities."type", 'overalltotal' AS OT, ('overalltotal' / 'activities') AS AVG
FROM activities, schedule
WHERE 'overalltotal' = (SELECT SUM(COUNT(schedule_ID))
FROM activities, schedule
WHERE schedule.activity_code = activities.activity_code
GROUP BY activities."type"
)
AND 'activities' = (SELECT COUNT(DISTINCT activities."type")
FROM activities
)
AND schedule.activity_code = activities.activity_code
GROUP BY activities."type";
I have implemented sample data and code to check the variables above:
SELECT SUM(COUNT(schedule_ID))
FROM activities, schedule
WHERE schedule.activity_code = activities.activity_code
GROUP BY activities."type";
Result : 20
SELECT COUNT(DISTINCT activities."type")
FROM activities;
Result : 5
However when running the code :
ORA-01722: invalid number
01722. 00000 - "invalid number"
*Cause:
*Action:
EDIT:
Using Dave's Code i have the following output:
Snowboarding 15
sledding 19
Snowmobiling 6
Ice Skating 5
Skiing 24
How would i do the final part of the question?
and the average number of individuals taking each type of activity.
You must use double quotes around column names in Oracle, not single quotes. For example, "overalltotal". Single quotes are for text strings, which is why you're getting an invalid number error.
EDIT: This is probably the type of query you want to use:
SELECT activities."type", COUNT(*) AS total, COUNT(*)/(COUNT(*) OVER ()) AS "avg"
FROM activities a
JOIN schedule s ON a.activity_code=s.activity_code
JOIN activity_bookings ab ON s.schedule_ID=ab.schedule_ID
GROUP BY activities."type";
Basically, because each activity booking has one visitor id, we want to get all the activity bookings for each activity. We have to go through schedule to do that. They we group the rows by the activity type and count how many activity bookings we have for each type.