10 Answers
New AnswerINSERT 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;
Vaibhav Singh Yah I really did solved it. It worked for me. it was only that letter from “Giraffe”.
Try to search the forum with 'sql zoo' search term, you might get some clues, or even solution to the problem 👍
Could you please give me a clue or hint on where the issue is ; since you answered me, it will be greate.and I asked, I believe I'm wanting an answer or discussion around the problem.dont get me wrong I'm thankful for your first replay.
Of course, --------------------- INSERT INTO Animals(name,type, country_id)VALUES('Slim','Girraffe',1); SELECT Animals.name As name, Animals.type As type, Countries.country As country FROM Animals INNER JOIN Countries ON Animals.country_id=Countries.id ORDER BY country; -------------------------- That's my code
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 country; Run the below code... 🤠🤠
INSERT INTO Animals VALUES ('Slim', 'Giraffe', 1); SELECT Animals.name, Animals.type, Countries.country FROM Animals JOIN Countries ON animals.country_id = Countries.id ORDER BY country;