
SQL SAKILA Gruop by
Quiz
•
Computers
•
12th Grade
•
Practice Problem
•
Easy
Jolita Kuncienė
Used 5+ times
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8 questions
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1.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Koks SQL sakinys naudojamas duomenims iš lentelės ištraukti?
SELECT
INSERT
UPDATE
DELETE
2.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Kaip iš lentelės film pasirinkti tik unikalius filmų pavadinimus?
SELECT UNIQUE title FROM film;
SELECT title FROM film;
SELECT DISTINCT title FROM film;
SELECT DIFFERENT title FROM film;
3.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Kuri funkcija SQL naudojama tekstui paversti didžiosiomis raidėmis?
CAPITALIZE()
UPPER()
TEXT_UPPER()
TO_UPPER()
4.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Kaip gauti brangiausiai kainuojančio filmo nuomos kainą (rental_rate) iš film lentelės?
SELECT HIGHEST(rental_rate) FROM film;
SELECT BIGGEST(rental_rate) FROM film;
SELECT MAX(rental_rate) FROM film;
SELECT rental_rate FROM film ORDER BY rental_rate DESC LIMIT 1;
5.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Kuris sakinys yra teisingas norint sugrupuoti filmus pagal rating ir suskaičiuoti kiek filmų turi kiekvienas reitingas?
SELECT rating, COUNT(*) FROM film SORT BY rating;
SELECT rating, COUNT(*) FROM film GROUP BY rating;
SELECT rating, COUNT(*) FROM film ORDER BY rating;
SELECT rating FROM film GROUP COUNT;
6.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Kaip surūšiuoti customer lentelės klientus pagal pavardę (last_name) abėcėlės tvarka?
SELECT * FROM customer SORT BY last_name;
SELECT * FROM customer ORDER BY last_name ASC;
SELECT * FROM customer GROUP BY last_name;
SELECT * FROM customer ORDER last_name;
7.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Kokia užklausa ištraukia visus filmus su nuomos kaina (rental_rate) tarp 2 ir 5?
SELECT * FROM film WHERE rental_rate = (2 TO 5);
SELECT * FROM film WHERE rental_rate BETWEEN 2 AND 5;
SELECT * FROM film WHERE rental_rate >= 2 AND rental_rate <= 5;
SELECT * FROM film HAVING rental_rate >= 2 AND rental_rate <= 5;
8.
MULTIPLE CHOICE QUESTION
1 min • 1 pt
Kokia užklausa parodytų visų klientų, kurie darė užsakymus, skaičių?
SELECT COUNT(*) FROM customer WHERE EXISTS (SELECT * FROM rental);
SELECT COUNT(DISTINCT customer_id) FROM customer;
SELECT COUNT(DISTINCT customer_id) FROM rental;
SELECT customer_id FROM rental GROUP BY customer_id;
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