SQLZOO练习题答案参考（全）

自学SQL将近一年，比较有名的SQL题库都尝试刷过，目前SQLZOO、XUESQL、牛客、Hackerrank初阶中阶、炼码基本刷完，Hackerrank高阶、Leetcode、SQL_intern仍然在刷，题目越来越难，不过水平也在逐渐提高啦！

一边刷高阶题，一边还会回头复盘基础，尝试多种解法，因此会不定时分享刷的题库的参考解法。注：SQL没有严格的格式要求，是我个人比较习惯缩进和大小写。

首先从SQLZOO开始。

SQLZOO有增删改查的讲解，也有练习题，涉及内容很基础，是个很考虑新手的题库了。

适用人群：刚入门小白、巩固基础准备进阶

优点：1.既可以学习，也可以刷题，相对全面

2.练习题中也包含讲解，情境化设置，循序渐进，难度递增，适合新手培养逻辑

3.某些部分会出现难题，对一个内容可以锻炼地比较深入

缺点：1.练习题的情境化设置导致有些题目要求查询的字段表述不清，会因为缺少答案里的字段而报错

2.部分题目有问题，莫名其妙报错

3.练习题涉及内容不全面，增删改没有对应的练习题

解法参考如下：

SELECT basics部分

-- Modify it to show the population of Germany
SELECT population
FROM world
WHERE name = 'Germany';-- Show the name and the population for 'Sweden', 'Norway' and 'Denmark'
SELECT name,population
FROM world
WHERE name in ('Sweden','Norway','Denmark');-- 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部分

-- Show the name, continent and population of all countries
SELECTname,continent,population
FROM world;-- Show the name for the countries that have a population of at least 200 million
SELECT name
FROM world
WHERE population >= 200000000;-- Give the name and the per capita GDP for those countries with a population of at least 200 million
SELECTname,gdp/population
FROM world
WHERE population >= 200000000;-- 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
SELECTname,population/1000000
FROM world
WHERE continent = 'South America';-- Show the name and population for France, Germany, Italy
SELECT name,population
FROM world
WHERE name in ('France','Germany','Italy');-- Show the countries which have a name that includes the word 'United'
SELECT name
FROM world
WHERE name like '%United%';-- Show the countries that are big by area or big by population. Show name, population and area
-- 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
SELECTname,population,area
FROM world
WHERE area >= 3000000 or population >= 250000000;-- Show the countries that are big by area (more than 3 million) or big by population (more than 250 million) but not both
-- Show name, population and area
SELECTname,population,area
FROM world
WHERE (area >= 3000000 and population < 250000000) or(area < 3000000 and population >= 250000000);-- For South America show population in millions and GDP in billions both to 2 decimal places
SELECTname,round(population/1000000,2),round(gdp/1000000000,2)
FROM world
WHERE continent = 'South America';-- Show per-capita GDP for the trillion dollar countries to the nearest $1000
SELECTname,round(gdp/population,-3)
FROM world
WHERE gdp >= 1000000000000;-- Show the name and capital where the name and the capital have the same number of characters
SELECT name,capital
FROM world
WHERE len(name)=len(capital);-- 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
SELECTname,capital
FROM world
WHERE left(name,1) = left(capital,1) AND name != capital;-- Find the country that has all the vowels and no spaces in its name
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部分

-- Change the query shown so that it displays Nobel prizes for 1950
SELECT *
FROM nobel
WHERE yr = '1950';-- Show who won the 1962 prize for literature
SELECT winner
FROM nobel
WHERE yr = '1962' and subject = 'Literature';-- Show the year and subject that won 'Albert Einstein' his prize
SELECTyr,subject
FROM nobel
WHERE winner = 'Albert Einstein';-- Give the name of the 'peace' winners since the year 2000, including 2000
SELECT winner
FROM nobel
WHERE yr >= '2000' and subject = 'Peace';-- Show all details of the literature prize winners for 1980 to 1989 inclusive
SELECT *
FROM nobel
WHERE (yr between 1980 and 1989) and subject = 'Literature';-- Show all details of the presidential winners：
-- Theodore Roosevelt、Thomas Woodrow Wilson、Jimmy Carter、Barack Obama
SELECT *
FROM nobel
WHERE winner in ('Theodore Roosevelt','Woodrow Wilson','Jimmy Carter','Barack Obama');-- Show the winners with first name John
SELECT winner
FROM nobel
WHERE winner like 'John%';-- 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'
UNION
SELECT *
FROM nobel
WHERE yr = 1984 and subject = 'Chemistry';-- Show the year, subject, and name of winners for 1980 excluding chemistry and medicine
SELECT *
FROM nobel
WHERE subject not in ('Chemistry','Medicine') and yr = 1980;-- 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 yr < 1910 and subject = 'Medicine'
UNION
SELECT *
FROM nobel
WHERE yr >= 2004 and subject = 'Literature';-- Find all details of the prize won by PETER GRÜNBERG
SELECT *
FROM nobel
WHERE winner = 'PETER GRÜNBERG';-- Find all details of the prize won by EUGENE O'NEILL
SELECT *
FROM nobel
WHERE winner = 'EUGENE O''NEILL';-- 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;-- 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
SELECTwinner,subject
FROM nobel
WHERE yr = 1984
ORDER BY CASE WHEN subject not in ('Chemistry','Physics') THEN 0ELSE 1 END,subject,winner

SELECT within SELECT部分

-- 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');-- 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 worldWHERE name = 'United Kingdom');-- List the name and continent of countries in the continents containing either Argentina or Australia
-- Order by name of the country
SELECTname,continent
FROM world
WHERE continent in (SELECT continent FROM worldWHERE name in ('Argentina','Australia'))
ORDER BY name;-- Which country has a population that is more than United Kingom but less than Germany?
-- Show the name and the population
SELECTname,population
FROM world
WHERE population > (SELECT population FROM worldWHERE name = 'United Kingdom')and population < (SELECT population FROM worldWHERE name = 'Germany');-- Show the name and the population of each country in Europe
-- Show the population as a percentage of the population of Germany
SELECTname,concat(cast(round(population/(SELECT population FROM worldWHERE name = 'Germany'),2)*100 as int),'%') as percentage
FROM world
WHERE continent = '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 CASE WHEN gdp is null THEN 0ELSE gdp END gdp FROM worldWHERE continent = 'Europe');-- Find the largest country (by area) in each continent, show the continent, the name and the area
SELECTa.continent,b.name,a.area
FROM (SELECTmax(area) as area,continentFROM worldGROUP BY continent) a
JOIN world b ON a.area = b.area;-- List each continent and the name of the country that comes first alphabetically
SELECTcontinent,name
FROM (SELECTcontinent,name,row_number() over(partition by continent order by name) as rnFROM world)
WHERE rn = 1;-- 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
SELECTname,continent,population
FROM (SELECTname,continent,population,CASE WHEN max(population) over(partition by continent) <=25000000 THEN 1ELSE 0 END labelFROM world) a
WHERE label = 1;-- Some countries have populations more than three times that of all of their neighbours (in the same continent)
-- Give the countries and continents
SELECTname,continent
FROM world a
WHERE population/3 >= all(SELECT population FROM world bWHERE a.continent = b.continent and a.name != b.name);

SUM and COUNT部分

-- Show the total population of the world
SELECT sum(population)
FROM world;-- List all the continents - just once each
SELECT distinct continent
FROM world;-- Give the total GDP of Africa
SELECT sum(gdp)
FROM world
WHERE continent = 'Africa';-- How many countries have an area of at least 1000000
SELECT count(name)
FROM world
WHERE area >= 1000000;-- What is the total population of ('Estonia', 'Latvia', 'Lithuania')
SELECT sum(population)
FROM world
WHERE name in ('Estonia', 'Latvia', 'Lithuania');-- For each continent show the continent and number of countries
SELECT continent,count(name)
FROM world
GROUP BY continent;-- For each continent show the continent and number of countries with populations of at least 10 million
SELECTcontinent,count(name)
FROM world
WHERE population >= 10000000
GROUP BY continent;-- 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部分

-- 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';-- Show id, stadium, team1, team2 for just game 1012
SELECTid,stadium,team1,team2
FROM game
WHERE id = 1012;-- Modify it to show the player, teamid, stadium and mdate for every German goal
SELECTgo.player,go.teamid,ga.stadium,ga.mdate
FROM goal go
JOIN game ga ON go.matchid = ga.id
WHERE teamid = 'GER';-- Show the team1, team2 and player for every goal scored by a player called Mario player LIKE 'Mario%'
SELECTga.team1,ga.team2,go.player
FROM game ga
JOIN goal go ON ga.id = go.matchid
WHERE go.player like 'Mario%';-- Show player, teamid, coach, gtime for all goals scored in the first 10 minutes(gtime<=10)
SELECTgo.player,go.teamid,e.coach,go.gtime
FROM goal go
JOIN eteam e  ON go.teamid = e.id
WHERE gtime <= 10;-- List the dates of the matches and the name of the team in which 'Fernando Santos' was the team1 coach
SELECTga.mdate,e.teamname
FROM game ga
JOIN eteam eON ga.team1 = e.id
WHERE e.coach = 'Fernando Santos';-- List the player for every goal scored in a game where the stadium was 'National Stadium, Warsaw'
SELECT player
FROM goal go
JOIN game ga ON go.matchid = ga.id
WHERE ga.stadium = 'National Stadium, Warsaw';-- Instead show the name of all players who scored a goal against Germany
SELECT distinct player
FROM goal go
JOIN game ga ON go.matchid = ga.id
WHERE go.teamid != 'GER' and (ga.team1 = 'GER' or ga.team2 = 'GER');-- Show teamname and the total number of goals scored
SELECTteamname,count(matchid)
FROM eteam e
JOIN goal g ON e.id = g.teamid
GROUP BY teamname;-- Show the stadium and the number of goals scored in each stadium
SELECTstadium,count(teamid)
FROM game ga
JOIN goal go ON ga.id = go.matchid
GROUP BY stadium;-- For every match involving 'POL', show the matchid, date and the number of goals scored
SELECTmatchid,mdate,count(matchid)
FROM goal go
JOIN game ga ON go.matchid = ga.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'
SELECTmatchid,mdate,count(matchid)
FROM game ga
JOIN goal go ON ga.id = go.matchid
WHERE teamid = 'GER'
GROUP BY matchid,mdate;-- List every match with the goals scored by each team as shown.
-- Sort your result by mdate, matchid, team1 and team2
SELECTmdate,team1,sum(case when teamid = team1 then 1 else 0 end),team2,sum(case when teamid = team2 then 1 else 0 end)
FROM game ga
LEFT JOIN goal go ON go.matchid = ga.id
GROUP BYmdate,matchid,team1,team2
ORDER BY mdate,matchid,team1,team2;

MORE JOIN部分

--List the films where the yr is 1962 [Show id, title]
SELECTid,title
FROM movie
WHERE yr = 1962;-- Give year of 'Citizen Kane'
SELECT yr
FROM movie
WHERE title = 'Citizen Kane';--List all of the Star Trek movies, include the id, title and yr.Order results by year
SELECTid,title,yr
FROM movie
WHERE title like '%Star Trek%'
ORDER BY yr;-- What id number does the actor 'Glenn Close' have?
SELECT id
FROM actor
WHERE name = 'Glenn Close';-- What is the id of the film 'Casablanca'
SELECT id
FROM movie
WHERE title = 'Casablanca';-- Obtain the cast list for 'Casablanca'.Use movieid=11768, or whatever value you got from the previous question
-- 题目给的11768有问题，跑不出结果
SELECT name
FROM actor a
JOIN casting cON a.id = c.actorid
JOIN movie m ON c.movieid = m.id
WHERE title = 'Casablanca';-- Obtain the cast list for the film 'Alien'
SELECT name
FROM actor a
JOIN casting cON a.id = c.actorid
JOIN movie m ON c.movieid = m.id
WHERE title = 'Alien';-- List the films in which 'Harrison Ford' has appeared
SELECT title
FROM movie m
JOIN casting c  ON m.id = c.movieid
JOIN actor a  ON c.actorid = a.id
WHERE name = 'Harrison Ford';-- 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 m
JOIN casting c  ON m.id = c.movieid
JOIN actor a  ON c.actorid = a.id
WHERE name = 'Harrison Ford' and ord != 1;-- List the films together with the leading star for all 1962 films
SELECT title,name
FROM movie m
JOIN casting c ON m.id = c.movieid
JOIN actor a ON c.actorid = a.id
WHERE yr = 1962 and ord = 1;-- Show the year and the number of movies 'Rock Hudson' made each year for any year in which he made more than 2 movies
SELECTyr,count(m.id)
FROM movie m
JOIN casting cON m.id = c.movieid
JOIN actor a ON a.id = c.actorid
WHERE name = 'Rock Hudson'
GROUP BY yr
HAVING count(m.id) > 1;-- List the film title and the leading actor for all of the films 'Julie Andrews' played in
SELECT title,name
FROM movie m
JOIN casting c ON m.id = c.movieid
JOIN actor a ON c.actorid = a.id
WHERE title in (SELECTtitleFROM movie m JOIN casting c ON m.id = c.movieidJOIN actor a ON c.actorid = a.id WHERE name = 'Julie Andrews') and ord = 1;-- Obtain a list, in alphabetical order, of actors who've had at least 15 starring roles
SELECT name
FROM actor a
JOIN casting cON a.id = c.actorid
GROUP BY name
HAVING count(case when ord = 1 then actorid end) >= 15
ORDER BY name;-- List the films released in the year 1978 ordered by the number of actors in the cast,then by title
SELECTtitle,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'
SELECTname
FROM actor a
JOIN casting c ON a.id = c.actorid
WHERE movieid in (SELECT movieidFROM casting c JOIN actor a ON c.actorid = a.idWHERE name = 'Art Garfunkel' ) and name != 'Art Garfunkel';

Using NULL部分

-- List the teachers who have NULL for their department
SELECTname
FROM teacher t
WHERE dept is null;-- Note the INNER JOIN misses the teachers with no department and the departments with no teacher
-- 非练习题
SELECT t.name,d.name
FROM teacher t
JOIN dept dON t.dept=d.id;-- Use a different JOIN so that all teachers are listed
SELECTt.name,d.name
FROM teacher t
LEFT JOIN dept d  ON t.dept = d.id; -- Use a different JOIN so that all departments are listed
SELECTt.name,d.name
FROM teacher t
RIGHT JOIN dept d  ON t.dept = d.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'
SELECTname,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
SELECTt.name,coalesce(d.name,'None')
FROM teacher t
LEFT JOIN dept d  ON t.dept = d.id;-- Use COUNT to show the number of teachers and the number of mobile phones
SELECTcount(id),count(case when mobile is not null then mobile end)
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
SELECTd.name,count(t.id)
FROM teacher t
RIGHT JOIN dept d ON t.dept = d.id
GROUP BY d.name;-- Use CASE to show the name of each teacher followed by 'Sci' if the teacher is in dept 1 or 2 and 'Art' otherwise
SELECTname,case when dept in (1,2) then 'Sci'else 'Art' end
FROM teacher t;-- 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
SELECTname,case when dept in (1,2) then 'Sci'when dept = 3 then 'Art' else 'None' end
FROM teacher t;

SELF JOIN部分

-- How many stops are in the database
SELECT count(id)
FROM stops;-- 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
-- 题目可能有问题，代码正确但报错
SELECTid,name
FROM stops s
JOIN route r ON s.id = r.stop
WHERE num = 4 and company = 'LRT';-- 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 = 149 or stop = 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(53), 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
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'
SELECTdistinct a.company,a.num
FROM route a
JOIN route b ON a.company = b.company and a.num = b.num
JOIN stops s1ON a.stop = s1.id
JOIN stops s2ON b.stop = s2.id
WHERE s1.name = 'Craiglockhart' and s2.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
SELECTdistinct s2.name,a.company,a.num
FROM route a
JOIN route b ON a.company = b.company and a.num = b.num
JOIN stops s1ON a.stop = s1.id
JOIN stops s2ON b.stop = s2.id
WHERE s1.name = 'Craiglockhart' and a.company = 'LRT';-- 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
SELECTdistinct a.num,a.company,name,b.num,b.company
FROM (SELECT a.num,a.company,b.stopFROM route a JOIN route b ON a.company = b.company and a.num = b.num and a.stop != b.stopWHERE a.stop = (SELECT id FROM stops WHERE name = 'Craiglockhart')) a
JOIN (SELECT a.num,a.company,a.stopFROM route a JOIN route b ON a.company = b.company and a.num = b.num and a.stop != b.stop  WHERE b.stop = (SELECT id FROM stops WHERE name = 'Lochend')) bON a.stop = b.stop
JOIN stops s ON a.stop = s.id
order by a.num,s.name,b.num;

以上解法是我个人觉得比较直接简单的思路，如有更佳的解答欢迎讨论~

SQLZOO练习题答案参考（全）相关推荐

2020全国计算机考试ps版本,2020年3月计算机等级Photoshop练习题及参考答案
[导语]2020年3月计算机等级考试备考工作已启动,为了方便考生及时有效的备考,那么,无忧考网为您精心整理了2020年3月计算机等级Photoshop练习题及参考答案,把握机会抓紧练习吧.如想获取更多 ...
课工场-JAVA高级特性编程及实战第1章练习题3答案参考
JAVA高级特性编程及实战第1章练习题3答案参考~ 本人菜鸟,一章章地学, 本想在网上搜一下然后对下答案的, 没找着~ 本着虔诚的心,把自己做的贴出来~ 运行结果是了出来了,过程不知道是否正确欢迎大 ...
软件测试笔试练习题与参考答案（一）
软件测试笔试练习题与参考答案(一) 测试习题一．测试相关多选题 1.对手机软件的压力测试通常可以包括(ABC) A 存储压力 B 响应能力压力 C 网络流量压力 D 并发压力 2.软件验收测试的合格 ...
c语言程序设计函数题,C语言程序设计函数练习题及参考答案
<C语言程序设计函数练习题及参考答案>由会员分享,可在线阅读,更多相关<C语言程序设计函数练习题及参考答案(60页珍藏版)>请在人人文库网上搜索. 1.C语言程序设计练习题及参 ...
c语言程序设计函数题,C语言程序设计函数练习题及参考答案.doc
C语言程序设计函数练习题及参考答案.doc C 语言程序设计练习题及参考答案 1. 定义一个函数 int funint a,int b,int c, 它的功能是若 a,b,c 能构成等边三角形函数返 ...
计量经济学计算机答案第三章,计量经济学第三章练习题及参考全部解答.doc
计量经济学第三章练习题及参考全部解答.doc 1 第三章练习题及参考解答 3.1为研究中国各地区入境旅游状况,建立了各省市旅游外汇收入(Y,百万美元) .旅行社职工人数(X1,人) .国际旅游人数( ...
计算机应用基础第三版练习题答案,计算机应用基础练习题答案
计算机应用基础练习题答案 (6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! 19.90 积分 <<计算机应用基础计算机应用基础>> ...
冯诺依曼计算机程序及其执行,第4章冯.诺依曼计算机：机器级程序与其执行练习题答案解析...
第4章冯.诺依曼计算机:机器级程序与其执行练习题答案解析 (20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! 14.9 积分第 4 章冯.诺依曼计算 ...
可以练计算机应用基础的网址,计算机应用基础(第3版)章节练习题答案
计算机应用基础(第3版)章节练习题答案第1章计算机基础知识体验与探索 1.体验 (1)C (2)C (3)B (4)37 (5)167 (6)191D 10110101B 275O 1F3H (7 ...

SQLZOO练习题答案参考（全）

SQLZOO练习题答案参考（全）相关推荐

最新文章

热门文章