Клиент серверные технологии (Курсовая работа - Информационная служба учета времени работы сотрудника), страница 3

(укажите 4 правильных ответа)

Ответы:

1 - select model, price from printer

inner join (select max(price) maxi from printer) as max_price

on price = max_price.maxi

2 - select model, price from printer

left join (select max(price) maxi from printer) as max_price on price = max_price.maxi

3 - select model, price from printer

right join (select max(price) maxi from printer) as max_price on price = max_price.maxi

4 - select model, price from printer

full join (select max(price) maxi from printer) as max_price on price = max_price.maxi

5 - select model, price from printer

cross join (select max(price) maxi from printer) as max_price

where price = max_price.maxi

6 - select model, price from (select max(price) maxi from printer) as max_price

left join printer on price = max_price.maxi

ПРАВИЛЬНЫЕ ОТВЕТЫ:

select model, price from printer

inner join (select max(price) maxi from printer) as max_price

on price = max_price.maxi

select model, price from printer

right join (select max(price) maxi from printer) as max_price on price = max_price.maxi

select model, price from printer

cross join (select max(price) maxi from printer) as max_price

where price = max_price.maxi

select model, price from (select max(price) maxi from printer) as max_price

left join printer on price = max_price.maxi

<<< ---------------------------- >>>

Дисциплина «Базы данных» (см. схему на бумаге)

Найдите среднюю скорость ПК-блокнотов, цена которых превышает 1000 дол., причем записи могут содержать неполные сведения о ПК-блокнотох.

(укажите 5 правильных ответов)

Ответы:

1 - select avg(speed) from laptop where price > 1000

2 - select avg(speed) from laptop where price in (select price from laptop where price > 1000)

3 - select sum(speed)/count(speed) from laptop where price > 1000 and speed is not null

4 - select sum(speed)/count(speed) from laptop where price > 1000

5 - select sum(speed)/count(*) from laptop where price > 1000

6 - select sum(speed)/count(speed) from laptop where price > 1000 and speed <> null

ПРАВИЛЬНЫЕ ОТВЕТЫ:

select avg(speed) from laptop where price > 1000

select avg(speed) from laptop where price in (select price from laptop where price > 1000)

select sum(speed)/count(speed) from laptop where price > 1000 and speed is not null

select sum(speed)/count(speed) from laptop where price > 1000

select sum(speed)/count(speed) from laptop where price > 1000 and speed <> null

<<< ---------------------------- >>>

Дисциплина «Базы данных» (см. схему на бумаге)

Для каждого значения скорости найдите среднюю стоимость ПК с такой же скоростью процессора. Вывести: скорость, средняя цена

(укажите 4 правильных ответа)

Ответы:

1 - select speed, avg(price) from pc group by speed

2 - select speed, avg(price) from pc group by speed order by speed

3 - select distinct speed, (select sum(price)/count(price) from pc where pc.speed=pc1.speed) from pc as pc1

4 - select speed, avg(price) from pc

5 - select speed, avg(price) from pc group by speed having count(*) > 0

6 - select speed, price from pc group by speed having avg(price) > 0

ПРАВИЛЬНЫЕ ОТВЕТЫ:

select speed, avg(price) from pc group by speed

select speed, avg(price) from pc group by speed order by speed

select distinct speed, (select sum(price)/count(price) from pc where pc.speed=pc1.speed) from pc as pc1

select speed, avg(price) from pc group by speed having count(*) > 0

<<< ---------------------------- >>>

Дисциплина «Базы данных» (см. схему на бумаге)

Для каждого значения скорости найдите среднюю стоимость ПК с такой же скоростью процессора. Вывести: скорость, средняя цена

(укажите 4 правильных ответа)

Ответы:

1 - select speed, avg(price) from pc group by speed

2 - select distinct speed, (select avg(price) from pc where pc.speed=pc1.speed) from pc as pc1

3 - select distinct speed, (select sum(price)/count(*) from pc where pc.speed=pc1.speed) from pc as pc1

4 - select distinct speed, (select avg(price) from pc where pc.speed=pc1.speed) from pc as pc1 order by 1

5 - select speed, avg(price) from pc

6 - select speed, avg(price) from pc group by speed having count(*) > 0

ПРАВИЛЬНЫЕ ОТВЕТЫ:

select speed, avg(price) from pc group by speed

select distinct speed, (select avg(price) from pc where pc.speed=pc1.speed) from pc as pc1

select distinct speed, (select avg(price) from pc where pc.speed=pc1.speed) from pc as pc1 order by 1

<<< ---------------------------- >>>

Дисциплина «Базы данных» (см. схему на бумаге)

Найдите размеры жестких дисков, совпадающих у двух и более PC.

Вывести: HD

(укажите 3 правильных ответа)

Ответы:

1 - select hd from pc group by hd having count(*)>1

2 - select hd

from (select hd, count(*) hd_count from pc group by hd) as HDs

where hd_count>1

3 - select distinct hd from pc

where (select count(*) from pc hd_count where hd_count.hd=pc.hd)>1

4 - select hd from pc group by hd where count(*)>1

5 - select hd from pc where count(*)>1 group by hd

6 - select distinct pc1.hd from pc pc1, pc pc2 where pc1.hd=pc2.hd

ПРАВИЛЬНЫЕ ОТВЕТЫ:

select hd from pc group by hd having count(*)>1

select hd

from (select hd, count(*) hd_count from pc group by hd) as HDs

where hd_count>1

select distinct hd from pc

where (select count(*) from pc hd_count where hd_count.hd=pc.hd)>1

<<< ---------------------------- >>>

Дисциплина «Базы данных» (см. схему на бумаге)

Найдите размеры жестких дисков, совпадающих у двух и более PC.

Вывести: HD

(укажите 3 правильных ответа)

Ответы:

1 - select hd from pc group by hd having count(*)>=2

2 - select distinct hd from pc where (select sum(1) from pc hd_count where hd_count.hd=pc.hd)>1

3 - select hd from pc where count(*)>1 group by hd

4 - select hd from pc where count(*)>1

5 - select distinct pc1.hd from pc pc1, pc pc2

where pc1.hd=pc2.hd and pc1.code<>pc2.code

ПРАВИЛЬНЫЕ ОТВЕТЫ:

select hd from pc group by hd having count(*)>=2

select distinct hd from pc where (select sum(1) from pc hd_count where hd_count.hd=pc.hd)>1

select distinct pc1.hd from pc pc1, pc pc2

where pc1.hd=pc2.hd and pc1.code<>pc2.code

<<< ---------------------------- >>>

Дисциплина «Базы данных» (см. схему на бумаге)

Найдите ПК-блокноты, скорость которых меньше скорости любого ПК.

Вывести: type, model, speed

(укажите 3 правильных ответа)

Ответы:

1 - select type, product.model, speed from product, laptop

where product.model = laptop.model

and speed < (select min(speed) from pc)

2 - select 'Laptop' type, model, speed from laptop

where speed < (select min(speed) from pc)

3 - select type, product.model, speed from product, laptop

where product.model = laptop.model

and speed < (select speed from pc)

4 - select type, product.model, speed

from product inner join laptop on product.model = laptop.model

where not in speed (select speed from pc)

5 - select type, product.model, speed

from product inner join laptop on product.model = laptop.model

where not exists (select model from pc where pc.speed <= laptop.speed)

6 - select type, product.model, speed

from product inner join laptop on product.model = laptop.model

where speed < (select min(speed) from pc where pc.speed < laptop.speed)

ПРАВИЛЬНЫЕ ОТВЕТЫ:

select type, product.model, speed from product, laptop

where product.model = laptop.model

and speed < (select min(speed) from pc)

select 'Laptop' type, model, speed from laptop

where speed < (select min(speed) from pc)

select type, product.model, speed

from product inner join laptop on product.model = laptop.model

where not exists (select model from pc where pc.speed <= laptop.speed)

<<< ---------------------------- >>>

Дисциплина «Базы данных» (см. схему на бумаге)

Для каждого производителя найдите средний размер экрана выпускаемых им ПК-блокнотов. Вывести: maker, средний размер экрана.

(укажите 2 правильных ответа)

Ответы:

1 - select maker, avg(screen) from product, laptop

where laptop.model=product.model group by maker

2 - select maker, sum(screen)/2 from product, laptop

where laptop.model=product.model group by maker

3 - select maker, avg(screen) from product inner join laptop on product.model=laptop.model

group by maker

4 - select maker, (select avg(screen) from laptop where product.model=laptop.model)

from product

where type = 'laptop'

5 - select maker, (select avg(screen) from laptop) from product where type = 'laptop'

ПРАВИЛЬНЫЕ ОТВЕТЫ:

select maker, avg(screen) from product, laptop

where laptop.model=product.model group by maker

select maker, avg(screen) from product inner join laptop on product.model=laptop.model

group by maker

<<< ---------------------------- >>>

Дисциплина «Базы данных» (см. схему на бумаге)

Найдите производителей, выпускающих по меньшей мере три различных модели ПК.

Вывести: Maker, число моделей

(укажите 3 правильных ответа)

Ответы:

1 - select maker, count(model) from product pr where type = 'pc'

group by maker having count(model)>=3

2 - select maker, count (distinct product.model)

from product left join pc on (product.model = pc.model) where type='pc'

group by maker having count(distinct product.model) >= 3

3 - select Maker, count(type) Count_Model from Product

group by type, maker having count(type) > 2 and type='pc'

4 - select maker, count(model) from product pr

where type = 'pc' and count(model)>=3

5 - select maker, count (*) from product left join pc on (product.model = pc.model)

where type='pc' group by maker having count(*) >= 3

ПРАВИЛЬНЫЕ ОТВЕТЫ:

select maker, count(model) from product pr where type = 'pc'