SQL zoo

Question

You manage a zoo. Each animal in the zoo comes from a different country. Here are the tables you have:

Answer 1

INSERT INTO Animals (name, type, country_id)
VALUES ('Slim', 'Giraffe', 1);

SELECT animals.name, animals.type, countries.country FROM animals, countries
WHERE animals.country_id=countries.id order by animals.country_id desc;

Answer 2

........is working.........
INSERT INTO Animals (name, type, country_id)
VALUES ('Slim', 'Giraffe', 1);

SELECT animals.name, animals.type, countries.country FROM animals, countries
WHERE animals.country_id=countries.id order by animals.country_id desc;

Answer 3

You don't have to quote 1 in VALUES.

While writing SELECT name, type country_id, you have to mention table name also, like Animals.name, Animals.type.

In the last ORDER BY Countries.country;

Hope this is helpful.

Answer 4

INSERT INTO Animals (name, type, country_id)
VALUES ('Slim', 'Giraffe', 1);

SELECT animals.name, animals.type, countries.country FROM animals, countries
WHERE animals.country_id=countries.id order by animals.country_id desc;

Answer 5

INSERT INTO Animals (name, type, country_id)
VALUES ('Slim', 'Giraffe', 1);

SELECT animals.name, animals.type, countries.country FROM animals, countries
WHERE animals.country_id=countries.id order by animals.country_id desc;

It`s working

Answer 6

INSERT INTO Animals (name, type, country_id)
VALUES ('Slim', 'Giraffe', 1);

SELECT animals.name, animals.type, countries.country FROM animals, countries
WHERE animals.country_id=countries.id order by animals.country_id desc;

Answer 7

Thank you Pallavi It works and I'll work hard to understand every tiny points for this SQL codes

Answer 8

thanks Islem M'HALLA

Answer 9

INSERT INTO Animals(name, type,country_id)
VALUES('Slim','Giraffe',1);

SELECT Animals.name, Animals.type, Countries.country
FROM Animals
INNER JOIN Countries
ON Animals.country_id = Countries.id
ORDER BY Countries.country;

use this ,it will help you

Answer 10

INSERT INTO Animals(name, type,country_id)
VALUES('Slim','Giraffe',1);

SELECT Animals.name, Animals.type, Countries.country
FROM Animals
INNER JOIN Countries
ON Animals.country_id = Countries.id
ORDER BY Countries.country;

Answer 11

coding zoo please

Answer 12

INSERT INTO Animals (name, type, country_id)
VALUES ('Slim', 'Giraffe', 1);

SELECT animals.name, animals.type, countries.country FROM animals, countries
WHERE animals.country_id=countries.id order by animals.country_id desc;

Answer 13

Working code, that system accepts:
/* name - "Slim", type - "Giraffe", country_id - 1 */
INSERT into animals
VALUES ('Slim','Giraffe',1);
SELECT a.name, a.type, c.country
from animals a
INNER JOIN Countries c
on a.country_id=c.id
ORDER by c.country;

Answer 14

/* name - "Slim", type - "Giraffe", country_id - 1 */

INSERT INTO animals (name, type, Country_id)
VALUES ('Slim', 'Giraffe', '1');
SELECT name type country_id
FROM animals
INNER JOIN countries
ON animal.country_id = countries.country;

I'm getting error in "type"

Answer 15

THANK YOU BHUMIKKUMAR KALOLA

Answer 16

Select … from

Answer 17

INSERT INTO Animals (name, type, country_id)
VALUES ('Slim', 'Giraffe', 1);

SELECT animals.name, animals.type, countries.country FROM animals, countries
WHERE animals.country_id=countries.id order by animals.country_id desc;

Answer 18

INSERT INTO Animals(name, type,country_id)
VALUES('Slim','Giraffe',1);

SELECT Animals.name, Animals.type, Countries.country
FROM Animals
INNER JOIN Countries
ON Animals.country_id = Countries.id
ORDER BY Countries.country;

Answer 19

INSERT INTO Animals (name, type, country_id)
VALUES ('Slim', 'Giraffe', 1);

SELECT animals.name, animals.type, countries.country FROM animals, countries
WHERE animals.country_id=countries.id order by animals.country_id desc;

Answer 20

Hi! you can use one of these options below:

NOTE:
Check capital letters on values!

Option 1:
-- name - "Slim", type - "Giraffe", country_id - 1 into table

insert into Animals (name, type, country_id) values ('Slim', 'Giraffe',1 );

-- a query to output a new table with each animal's name, type and country fields, sorted by countries.

select animals.name, animals.type, countries.country into newtable from animals inner join countries
on animals.country_id = countries.id order by countries.country;

select* from newtable;

Option 2:
-- name - "Slim", type - "Giraffe", country_id - 1 into table

insert into Animals (name, type, country_id) values ('Slim', 'Giraffe',1 );

-- a query to output a new table with each animal's name, type and country fields, sorted by countries.

create table newtable as (select animals.name, animals.type, countries.country from animals inner join countries
on animals.country_id = countries.id order by countries.country);

select* from newtable;