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SQL SAKILA Gruop by

Total questions: 8

Worksheet time: 9mins

Name
Class
Date
1.

Koks SQL sakinys naudojamas duomenims iš lentelės ištraukti?

a)

SELECT

b)

INSERT

c)

UPDATE

d)

DELETE

2.

Kaip iš lentelės film pasirinkti tik unikalius filmų pavadinimus?

a)

SELECT UNIQUE title FROM film;

b)

SELECT title FROM film;

c)

SELECT DISTINCT title FROM film;

d)

SELECT DIFFERENT title FROM film;

3.

Kuri funkcija SQL naudojama tekstui paversti didžiosiomis raidėmis?

a)

CAPITALIZE()

b)

UPPER()

c)

TEXT_UPPER()

d)

TO_UPPER()

4.

Kaip gauti brangiausiai kainuojančio filmo nuomos kainą (rental_rate) iš film lentelės?

a)

SELECT HIGHEST(rental_rate) FROM film;

b)

SELECT BIGGEST(rental_rate) FROM film;

c)

SELECT MAX(rental_rate) FROM film;

d)

SELECT rental_rate FROM film ORDER BY rental_rate DESC LIMIT 1;

5.

Kuris sakinys yra teisingas norint sugrupuoti filmus pagal rating ir suskaičiuoti kiek filmų turi kiekvienas reitingas?

a)

SELECT rating, COUNT(*) FROM film SORT BY rating;

b)

SELECT rating, COUNT(*) FROM film GROUP BY rating;

c)

SELECT rating, COUNT(*) FROM film ORDER BY rating;

d)

SELECT rating FROM film GROUP COUNT;

6.

Kaip surūšiuoti customer lentelės klientus pagal pavardę (last_name) abėcėlės tvarka?

a)

SELECT * FROM customer SORT BY last_name;

b)

SELECT * FROM customer ORDER BY last_name ASC;

c)

SELECT * FROM customer GROUP BY last_name;

d)

SELECT * FROM customer ORDER last_name;

7.

Kokia užklausa ištraukia visus filmus su nuomos kaina (rental_rate) tarp 2 ir 5?

a)

SELECT * FROM film WHERE rental_rate = (2 TO 5);

b)

SELECT * FROM film WHERE rental_rate BETWEEN 2 AND 5;

c)

SELECT * FROM film WHERE rental_rate >= 2 AND rental_rate <= 5;

d)

SELECT * FROM film HAVING rental_rate >= 2 AND rental_rate <= 5;

8.

Kokia užklausa parodytų visų klientų, kurie darė užsakymus, skaičių?

a)

SELECT COUNT(*) FROM customer WHERE EXISTS (SELECT * FROM rental);

b)

SELECT COUNT(DISTINCT customer_id) FROM customer;

c)

SELECT COUNT(DISTINCT customer_id) FROM rental;

d)

SELECT customer_id FROM rental GROUP BY customer_id;