本文是作者在做完SQLZOO Tutorial练习后整理的答案，仅供参考。

SQLZOO目录

SELECT basics
SELECT from world
SELECT from nobel
SELECT in SELECT
SUM and COUNT
JOIN
More JOIN
Using NULL
Self JOIN

SELECT basics

Introducing the world table of countries

The example uses a WHERE clause to show the population of ‘France’. Note that strings (pieces of text that are data) should be in ‘single quotes’;

Modify it to show the population of Germany

SELECT population FROM world
WHERE name = 'Germany'

Scandinavia

Checking a list The word IN allows us to check if an item is in a list. The example shows the name and population for the countries ‘Brazil’, ‘Russia’, ‘India’ and ‘China’.

Show the name and the population for ‘Sweden’, ‘Norway’ and ‘Denmark’.

SELECT name, population FROM world
WHERE name IN ('Sweden', 'Norway', 'Denmark')

Just the right size

Which countries are not too small and not too big? BETWEEN allows range checking (range specified is inclusive of boundary values). The example below shows countries with an area of 250,000-300,000 sq. km. Modify it to show the country and the area for countries with an area between 200,000 and 250,000.

SELECT name, area FROM world
WHERE area BETWEEN 200000 AND 250000

SELECT from world

Introduction

Read the notes about this table. Observe the result of running this SQL command to show the name, continent and population of all countries.

SELECT name, continent, population FROM world

Large Countries

How to use WHERE to filter records. Show the name for the countries that have a population of at least 200 million. 200 million is 200000000, there are eight zeros.

SELECT name FROM world
WHERE population > 200000000

Per capita GDP

Give the name and the per capita GDP for those countries with a population of at least 200 million.

select name, gdp/population from world
where population> 200000000

South America In millions

Show the name and population in millions for the countries of the continent ‘South America’. Divide the population by 1000000 to get population in millions.

select name,population/1000000 from world
where continent='South America'

France, Germany, Italy

Show the name and population for France, Germany, Italy

select name,population from world
where name in ('France','Germany','Italy')

United

Show the countries which have a name that includes the word ‘United’

select name from world where name like '%United%'

Two ways to be big

Two ways to be big: A country is big if it has an area of more than 3 million sq km or it has a population of more than 250 million.

Show the countries that are big by area or big by population. Show name, population and area.

select name,population,area from world
where area>3000000 or population>250000000

One or the other (but not both)

Exclusive OR (XOR). Show the countries that are big by area or big by population but not both. Show name, population and area.

Australia has a big area but a small population, it should be included.
Indonesia has a big population but a small area, it should be included.
China has a big population and big area, it should be excluded.
United Kingdom has a small population and a small area, it should be excluded.

select name,population,area from world
where area> 3000000 xor population>250000000

Rounding

Show the name and population in millions and the GDP in billions for the countries of the continent ‘South America’. Use the ROUND function to show the values to two decimal places.

For South America show population in millions and GDP in billions both to 2 decimal places.

select name,round(population/1000000,2),round(gdp/1000000000,2) from world
where continent='South America'

Trillion dollar economies

Show the name and per-capita GDP for those countries with a GDP of at least one trillion (1000000000000; that is 12 zeros). Round this value to the nearest 1000.

Show per-capita GDP for the trillion dollar countries to the nearest $1000.

select name,round(gdp/population,-3) from world
where gdp>= 1000000000000

Name and capital have the same length

Greece has capital Athens.

Each of the strings ‘Greece’, and ‘Athens’ has 6 characters.

Show the name and capital where the name and the capital have the same number of characters.

You can use the LENGTH function to find the number of characters in a string

SELECT name,capital FROM world
WHERE LENGTH(name)=LENGTH(capital)

Matching name and capital

The capital of Sweden is Stockholm. Both words start with the letter ‘S’.

Show the name and the capital where the first letters of each match. Don’t include countries where the name and the capital are the same word.
You can use the function LEFT to isolate the first character.
You can use <> as the NOT EQUALS operator.

SELECT name, capital FROM world
where LEFT(name,1)=LEFT(capital,1) and name<>capital

All the vowels

Equatorial Guinea and Dominican Republic have all of the vowels (a e i o u) in the name. They don’t count because they have more than one word in the name.

Find the country that has all the vowels and no spaces in its name.

You can use the phrase name NOT LIKE ‘%a%’ to exclude characters from your results.
The query shown misses countries like Bahamas and Belarus because they contain at least one ‘a’

SELECT name FROM world
WHERE name LIKE '%a%'
and name LIKE '%e%'
and name LIKE '%i%'
and name LIKE '%o%'
and name LIKE '%u%'AND name NOT LIKE '% %'

SELECT from nobel

Winners from 1950

Change the query shown so that it displays Nobel prizes for 1950.

SELECT yr, subject, winner FROM nobel
WHERE yr = 1950

1962 Literature

Show who won the 1962 prize for Literature.

SELECT winner FROM nobel
WHERE yr = 1962 AND subject = 'Literature'

Albert Einstein

Show the year and subject that won ‘Albert Einstein’ his prize.

select yr,subject from nobel
where winner='Albert Einstein'

Recent Peace Prizes

Give the name of the ‘Peace’ winners since the year 2000, including 2000.

select winner from nobel
where subject='Peace' and yr>=2000

Literature in the 1980’s

Show all details (yr, subject, winner) of the Literature prize winners for 1980 to 1989 inclusive.

select*from nobel
where subject='Literature' and yr between 1980 and 1989

Only Presidents

Show all details of the presidential winners:

Theodore Roosevelt
Woodrow Wilson
Jimmy Carter
Barack Obama

SELECT * FROM nobel
WHERE winner IN ('Theodore Roosevelt','Woodrow Wilson','Jimmy Carter','Barack Obama')

John

Show the winners with first name John

select winner from nobel
where winner like 'John %'

Chemistry and Physics from different years

Show the year, subject, and name of Physics winners for 1980 together with the Chemistry winners for 1984.

select*from nobel
where (yr=1980 and subject='Physics')
or (yr=1984 and subject='Chemistry')

Exclude Chemists and Medics

Show the year, subject, and name of winners for 1980 excluding Chemistry and Medicine

select*from nobel where yr=1980
and subject not in ('Chemistry','Medicine')

Early Medicine, Late Literature

Show year, subject, and name of people who won a ‘Medicine’ prize in an early year (before 1910, not including 1910) together with winners of a ‘Literature’ prize in a later year (after 2004, including 2004)

select*from nobel
where (subject='Medicine' and yr<1910)
or (subject='Literature' and yr>=2004)

Umlaut

Find all details of the prize won by PETER GRÜNBERG

select*from nobel
where winner='PETER GRÜNBERG'

Apostrophe

Find all details of the prize won by EUGENE O’NEILL

select*from nobel
where winner='EUGENE O\'NEILL'

Knights of the realm

Knights in order

List the winners, year and subject where the winner starts with Sir. Show the the most recent first, then by name order.

select winner,yr,subject from nobel
where winner like 'Sir%' order by yr desc,winner

Chemistry and Physics last

The expression subject IN (‘Chemistry’,‘Physics’) can be used as a value - it will be 0 or 1.

Show the 1984 winners and subject ordered by subject and winner name; but list Chemistry and Physics last.

SELECT winner, subject FROM nobel
WHERE yr=1984
ORDER BY subject IN ('Physics','Chemistry'),subject,winner

SELECT in SELECT

Bigger than Russia

List each country name where the population is larger than that of ‘Russia’.

SELECT name FROM world
WHERE population >(SELECT population FROM worldWHERE name='Russia')

Richer than UK

Show the countries in Europe with a per capita GDP greater than ‘United Kingdom’.

select name from world
where continent='Europe'
and gdp/population > (select gdp/population from world where name ='United Kingdom')

Neighbours of Argentina and Australia

List the name and continent of countries in the continents containing either Argentina or Australia. Order by name of the country.

select name,continent from world
where continent in (select continent from world where name in ('Argentina','Australia'))
order by name

Between Canada and Poland

Which country has a population that is more than Canada but less than Poland? Show the name and the population.

select name,population from world
where population>(select population from world where name='Canada')
and population<(select population from world where name='Poland')

Percentages of Germany

Germany (population 80 million) has the largest population of the countries in Europe. Austria (population 8.5 million) has 11% of the population of Germany.

Show the name and the population of each country in Europe. Show the population as a percentage of the population of Germany.

select name,
concat(round(population*100/(select population from world where name='Germany')),'%') from world
where continent='Europe'

Bigger than every country in Europe

Which countries have a GDP greater than every country in Europe? [Give the name only.] (Some countries may have NULL gdp values)

select name from world
where gdp>all(select gdp from world where gdp>0 and continent='Europe')

Largest in each continent

Find the largest country (by area) in each continent, show the continent, the name and the area:

SELECT continent, name, area FROM world x
WHERE area >= ALL(SELECT area FROM world y WHERE y.continent=x.continent AND area>0)

First country of each continent (alphabetically)

List each continent and the name of the country that comes first alphabetically.

select continent,name from world x
where x.name<=all(select name from world y where x.continent=y.continent)
order by name

Difficult Questions That Utilize Techniques Not Covered In Prior Sections

Find the continents where all countries have a population <= 25000000. Then find the names of the countries associated with these continents. Show name, continent and population.

select name,continent,population from world x
where 25000000>=all(select population from world y where population>0
and x.continent=y.continent)

Some countries have populations more than three times that of any of their neighbours (in the same continent). Give the countries and continents.

select name,continent from world x
where population/3>=all(select population from world y where y.continent=x.continent
and x.name!=y.name)

SUM and COUNT

Total world population

Show the total population of the world.

SELECT SUM(population) FROM world

List of continents

List all the continents - just once each.

select distinct continent from world

GDP of Africa

Give the total GDP of Africa

select sum(gdp) from world
where continent='Africa'

Count the big countries

How many countries have an area of at least 1000000

select count(name) from world
where area>1000000

Baltic states population

What is the total population of (‘Estonia’, ‘Latvia’, ‘Lithuania’)

select sum(population) from world
where name in ('Estonia', 'Latvia', 'Lithuania')

Counting the countries of each continent

For each continent show the continent and number of countries.

select continent,count(name) from world group by continent

Counting big countries in each continent

For each continent show the continent and number of countries with populations of at least 10 million.

select continent,count(name) from world
where population>10000000 group by continent

Counting big continents

List the continents that have a total population of at least 100 million.

select continent from world group by continent
having sum(population)>100000000

JOIN

The first example shows the goal scored by a player with the last name ‘Bender’. The * says to list all the columns in the table - a shorter way of saying matchid, teamid, player, gtime

Modify it to show the matchid and player name for all goals scored by Germany. To identify German players, check for: teamid = ‘GER’

SELECT matchid,player FROM goal
WHERE teamid='GER'

From the previous query you can see that Lars Bender’s scored a goal in game 1012. Now we want to know what teams were playing in that match.

Notice in the that the column matchid in the goal table corresponds to the id column in the game table. We can look up information about game 1012 by finding that row in the game table.

Show id, stadium, team1, team2 for just game 1012

SELECT id,stadium,team1,team2 FROM game where id=1012

The FROM clause says to merge data from the goal table with that from the game table. The ON says how to figure out which rows in game go with which rows in goal - the matchid from goal must match id from game. (If we wanted to be more clear/specific we could say
ON (game.id=goal.matchid)

The code below shows the player (from the goal) and stadium name (from the game table) for every goal scored.

Modify it to show the player, teamid, stadium and mdate for every German goal.

SELECT player,teamid,stadium,mdate FROM game a
JOIN goal b ON a.id=b.matchid
where teamid='GER'

Use the same JOIN as in the previous question.

Show the team1, team2 and player for every goal scored by a player called Mario player LIKE ‘Mario%’

select team1,team2,player from game a
join goal b on a.id=b.matchid
where player like 'Mario%'

The table eteam gives details of every national team including the coach. You can JOIN goal to eteam using the phrase goal JOIN eteam on teamid=id

Show player, teamid, coach, gtime for all goals scored in the first 10 minutes gtime<=10

SELECT player, teamid, coach, gtime FROM goal a
join eteam b on a.teamid=b.id
WHERE gtime<=10

To JOIN game with eteam you could use either
game JOIN eteam ON (team1=eteam.id) or game JOIN eteam ON (team2=eteam.id)

Notice that because id is a column name in both game and eteam you must specify eteam.id instead of just id

List the the dates of the matches and the name of the team in which ‘Fernando Santos’ was the team1 coach.

select mdate,teamname from game a
join eteam b on a.team1=b.id
where coach= 'Fernando Santos'

List the player for every goal scored in a game where the stadium was ‘National Stadium, Warsaw’

select player from game a
join goal b on a.id=b.matchid
where stadium='National Stadium, Warsaw'

The example query shows all goals scored in the Germany-Greece quarterfinal.
Instead show the name of all players who scored a goal against Germany.

SELECT distinct player FROM game a
JOIN goal b ON b.matchid =a.id
WHERE teamid!='GER' and (team2='GER' or team1='GER')

Show teamname and the total number of goals scored.

SELECT teamname, count(teamname) FROM eteam
JOIN goal ON id=teamid
group by teamname
ORDER BY teamname

Show the stadium and the number of goals scored in each stadium.

select stadium,count(stadium) from game a
join goal b on a.id=b.matchid
group by stadium

For every match involving ‘POL’, show the matchid, date and the number of goals scored.

SELECT matchid,mdate,count(*) FROM game
JOIN goal ON matchid = id
WHERE (team1 = 'POL' OR team2 = 'POL')
group by matchid,mdate

For every match where ‘GER’ scored, show matchid, match date and the number of goals scored by ‘GER’

select matchid,mdate,count(matchid) from goal a
join game b on a.matchid=b.id
where teamid='GER'
group by matchid,mdate

List every match with the goals scored by each team as shown. This will use “CASE WHEN” which has not been explained in any previous exercises.

Notice in the query given every goal is listed. If it was a team1 goal then a 1 appears in score1, otherwise there is a 0. You could SUM this column to get a count of the goals scored by team1. Sort your result by mdate, matchid, team1 and team2.

SELECT mdate, team1,
sum(CASE WHEN teamid=team1 THEN 1 ELSE 0 END) score1, team2,
sum(CASE WHEN teamid=team2 THEN 1 ELSE 0 END) score2 FROM game
left JOIN goal ON matchid = id
group by mdate, team1,team2

More JOIN

1962 movies

List the films where the yr is 1962 [Show id, title]

SELECT id, title FROM movie WHERE yr=1962

When was Citizen Kane released?

Give year of ‘Citizen Kane’.

select yr from movie where title='Citizen Kane'

Star Trek movies

List all of the Star Trek movies, include the id, title and yr (all of these movies include the words Star Trek in the title). Order results by year.

select id, title, yr from movie
where title like '%Star Trek%'

id for actor Glenn Close

What id number does the actor ‘Glenn Close’ have?

select id from actor where name='Glenn Close'

id for Casablanca

What is the id of the film ‘Casablanca’

select id from movie where title= 'Casablanca'

Cast list for Casablanca

Obtain the cast list for ‘Casablanca’.

what is a cast list?
Use movieid=11768, (or whatever value you got from the previous question)

select name from actor a
join casting b on a.id=b.actorid
where b.movieid=11768

Alien cast list

Obtain the cast list for the film ‘Alien’

select name from actor a
join casting b on a.id=b.actorid
join movie c on b.movieid=c.id
where title='Alien'

Harrison Ford movies

List the films in which ‘Harrison Ford’ has appeared

select title from movie a
join casting c on a.id=c.movieid
join actor b on b.id=c.actorid
where name='Harrison Ford'

Harrison Ford as a supporting actor

List the films where ‘Harrison Ford’ has appeared - but not in the starring role. [Note: the ord field of casting gives the position of the actor. If ord=1 then this actor is in the starring role]

select title from movie a
join casting c on a.id=c.movieid
join actor b on b.id=c.actorid
where name='Harrison Ford' and ord!=1

Lead actors in 1962 movies

List the films together with the leading star for all 1962 films.

select title,name from movie a
join casting c on a.id=c.movieid
join actor b on b.id=c.actorid
where yr=1962 and ord=1

Busy years for Rock Hudson

Which were the busiest years for ‘Rock Hudson’, show the year and the number of movies he made each year for any year in which he made more than 2 movies.

SELECT yr,COUNT(title) FROM movie JOIN casting ON movie.id=movieidJOIN actor   ON actorid=actor.id
WHERE name='Rock Hudson'
GROUP BY yr
HAVING COUNT(title) > 2

Lead actor in Julie Andrews movies

List the film title and the leading actor for all of the films ‘Julie Andrews’ played in.

SELECT title,name FROM casting a
join movie b on a.movieid=b.id
join actor c on a.actorid=c.id
WHERE movieid IN (SELECT movieid FROM actor c join casting a on c.id=a.actoridWHERE name='Julie Andrews') and ord=1

Actors with 30 leading roles

Obtain a list, in alphabetical order, of actors who’ve had at least 30 starring roles.

select name from actor a
join casting c on a.id=c.actorid
where ord=1 group by name having count(name)>=30 order by name

List the films released in the year 1978 ordered by the number of actors in the cast, then by title.

select title, count(actorid) from movie m
join casting c on m.id=c.movieid
where yr=1978 group by title order by count(actorid) desc, title

List all the people who have worked with ‘Art Garfunkel’.

select f.name from (select movieid from casting c
join actor a on a.id=c.actorid where name='Art Garfunkel') as e
join (select movieid, name from casting c
join actor a on a.id=c.actorid where name!='Art Garfunkel') as f
on e.movieid=f.movieid

Using NULL

List the teachers who have NULL for their department.

select name from teacher where dept is null

Note the INNER JOIN misses the teachers with no department and the departments with no teacher.

SELECT teacher.name, dept.name FROM teacher
INNER JOIN dept ON teacher.dept=dept.id

Use a different JOIN so that all teachers are listed.

SELECT teacher.name, dept.name FROM teacher
left JOIN dept ON teacher.dept=dept.id

Use a different JOIN so that all departments are listed.

SELECT teacher.name, dept.name FROM teacher
right JOIN dept ON teacher.dept=dept.id

Use COALESCE to print the mobile number. Use the number ‘07986 444 2266’ if there is no number given. Show teacher name and mobile number or ‘07986 444 2266’

select name, COALESCE(mobile,'07986 444 2266') from teacher

Use the COALESCE function and a LEFT JOIN to print the teacher name and department name. Use the string ‘None’ where there is no department.

select teacher.name, COALESCE(dept.name,'None') from teacher
left join dept on teacher.dept=dept.id

Use COUNT to show the number of teachers and the number of mobile phones.

select count(name), count(mobile) from teacher

Use COUNT and GROUP BY dept.name to show each department and the number of staff. Use a RIGHT JOIN to ensure that the Engineering department is listed.

select dept.name, count(teacher.name) from teacher
right join dept on teacher.dept=dept.id
group by 1

Use CASE to show the name of each teacher followed by ‘Sci’ if the teacher is in dept 1 or 2 and ‘Art’ otherwise.

select name,
case when dept in (1,2) then 'Sci'
else 'Art'
end
from teacher

Use CASE to show the name of each teacher followed by ‘Sci’ if the teacher is in dept 1 or 2, show ‘Art’ if the teacher’s dept is 3 and ‘None’ otherwise.

select name,
case when dept in (1,2) then 'Sci'
when dept=3 then  'Art'
else 'None'
end
from teacher

Self JOIN

How many stops are in the database.

select count(*) from stops

select count(distinct stop) from route

Find the id value for the stop ‘Craiglockhart’

select id from stops where name='Craiglockhart'

Give the id and the name for the stops on the ‘4’ ‘LRT’ service.

select id,name from stops
join route on route.stop=stops.id
where num=4 and company='LRT'

Routes and stops
The query shown gives the number of routes that visit either London Road (149) or Craiglockhart (53). Run the query and notice the two services that link these stops have a count of 2. Add a HAVING clause to restrict the output to these two routes.

SELECT company, num, COUNT(*) FROM route
WHERE stop in (149,53)
GROUP BY company, num
having COUNT(*)=2

Execute the self join shown and observe that b.stop gives all the places you can get to from Craiglockhart, without changing routes. Change the query so that it shows the services from Craiglockhart to London Road.

SELECT a.company, a.num, a.stop, b.stop FROM route a
JOIN route b ON (a.company=b.company AND a.num=b.num)
WHERE a.stop=53 and b.stop=149

The query shown is similar to the previous one, however by joining two copies of the stops table we can refer to stops by name rather than by number. Change the query so that the services between ‘Craiglockhart’ and ‘London Road’ are shown. If you are tired of these places try ‘Fairmilehead’ against ‘Tollcross’

SELECT a.company, a.num, stopa.name, stopb.name FROM route a
JOIN route b ON (a.company=b.company AND a.num=b.num)
JOIN stops stopa ON (a.stop=stopa.id)
JOIN stops stopb ON (b.stop=stopb.id)
WHERE stopa.name='Craiglockhart' and stopb.name='London Road'

Give a list of all the services which connect stops 115 and 137 (‘Haymarket’ and ‘Leith’)

SELECT distinct a.company, a.num FROM route a
JOIN route b ON (a.company=b.company AND a.num=b.num)
WHERE a.stop=115 and b.stop=137

Give a list of the services which connect the stops ‘Craiglockhart’ and ‘Tollcross’

select a.company, a.num from route a
join route b on (a.company=b.company and a.num=b.num)
join stops x on x.id=a.stop
join stops y on y.id=b.stop
where x.name='Craiglockhart' and y.name='Tollcross'

Give a distinct list of the stops which may be reached from ‘Craiglockhart’ by taking one bus, including ‘Craiglockhart’ itself, offered by the LRT company. Include the company and bus no. of the relevant services.

select y.name, a.company, a.num from route a
join route b on (a.company=b.company and a.num=b.num)
join stops x on x.id=a.stop
join stops y on y.id=b.stop
where x.name='Craiglockhart'

Find the routes involving two buses that can go from Craiglockhart to Lochend.
Show the bus no. and company for the first bus, the name of the stop for the transfer,
and the bus no. and company for the second bus.

Hint
Self-join twice to find buses that visit Craiglockhart and Lochend, then join those on matching stops.

select bus1.num, bus1.company, name, bus2.num, bus2.company from
(select start1.num, start1.company, stop1.stop from route start1
join route stop1 on start1.num=stop1.num and start1.company=stop1.company and start1.stop!=stop1.stop and start1.stop=(select id from stops where name='Craiglockhart')) bus1
join (select start2.num, start2.company, start2.stop from route start2
join route stop2 on start2.num=stop2.num and start2.company=stop2.company and start2.stop!=stop2.stop and stop2.stop=(select id from stops where name='Lochend')) bus2
on bus1.stop=bus2.stop
join stops on stops.id=bus1.stop

sqlzoo答案参考（全）相关推荐

2020年拼多多校招面试题及答案-最全最新-持续更新中
大家好我是好好学习天天编程的天天一个整天在互联网上种菜和砍柴的程序员 2020年拼多多校招面试题及答案-最全最新-持续更新中 2020年拼多多校招面试题一面-牛客网 2020年拼多多校招面试题二面- ...
【转载，整理】Linux模拟试题及答案（全）+经典Linux系统工程师面试题（附答案）
转自:http://www.ha97.com/871.html http://www.ha97.com/870.html 2010-6 11 经典Linux系统工程师面试题(附答案) 发表于: L ...
最新前端面试题整理和答案（全）一直更新
最新前端面试题整理和答案(全) 参考地址:https://blog.csdn.net/wdlhao/article/details/79079660 javascript: JavaScript中如何 ...
广东电大计算机绘图试题,电大计算机绘图期末复习试题及答案参考小抄.doc
电大计算机绘图期末复习试题及答案参考小抄一.填空题(每小题1.5分,共30分) 1．CAD的常用图形输入设备有???鼠标??.数字化仪.图形输入板.光笔.??键盘等.2．CAD的软件可分为系统软件 ...
统计应用计算机基础,计算机基础应用试题及答案参考
计算机基础应用试题及答案参考计算机技术与通信技术的结合,使计算机网络得到发展.信息服务业的兴起使社会信息资源得到更广泛的利用.下面是小编为大家搜索整理的计算机应用基础训练题,希望能给大家带来帮助! ...
新概念C语言能力教程练习3答案,新概念C语言教程答案参考(自做)中国电力
新概念C语言教程答案参考(自做)中国电力 (138页) 本资源提供全文预览,点击全文预览即可全文预览,如果喜欢文档就下载吧,查找使用更方便哦! 39.9 积分第一篇算法与C程序结构第1单元C语言程 ...
mysql待办事项表名_SSD8-Ex4待办事项列表答案参考
[实例简介] SSD8-Ex4待办事项列表答案参考:http://wangbaiyuan.cn/mysql-database-data-released-in-java-web-service-and ...
2017计算机绘图试题及答案,2017年度__电大（精编新版）计算机绘图期末复习试题及答案参考小抄.doc...
2017年度__电大(精编新版)计算机绘图期末复习试题及答案参考小抄电大计算机绘图期末复习试题及答案参考小抄一.填空题(每小题1.5分,共30分) 1．CAD的常用图形输入设备有???鼠标??.数 ...
广东电大计算机绘图试题,2014年电大计算机绘图期末复习试题及答案参考小抄.doc...
电大计算机绘图期末复习试题及答案参考小抄一.填空题(每小题1.5分,共30分) 1．CAD的常用图形输入设备有???鼠标??.数字化仪.图形输入板.光笔.??键盘等.2．CAD的软件可分为系统软件 ...

sqlzoo答案参考（全）