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https://github.com/miroslav-reiter/sql_zoo
Kurzy SQL, MySQL, PostgreSQL, Microsoft SQL Server
https://github.com/miroslav-reiter/sql_zoo
analysis analytics data data-analysis database db2 ibm-db2 mysql mysql-database postgres postgresql postgresql-database sql sql-injection sql-query sql-server sqlserver
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Kurzy SQL, MySQL, PostgreSQL, Microsoft SQL Server
- Host: GitHub
- URL: https://github.com/miroslav-reiter/sql_zoo
- Owner: miroslav-reiter
- Created: 2019-12-01T17:54:57.000Z (almost 5 years ago)
- Default Branch: master
- Last Pushed: 2023-10-05T07:04:54.000Z (about 1 year ago)
- Last Synced: 2023-10-05T12:58:45.023Z (about 1 year ago)
- Topics: analysis, analytics, data, data-analysis, database, db2, ibm-db2, mysql, mysql-database, postgres, postgresql, postgresql-database, sql, sql-injection, sql-query, sql-server, sqlserver
- Homepage: https://www.it-academy.sk/kategoria/databazy/kurzy-sql/
- Size: 81.1 KB
- Stars: 1
- Watchers: 1
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
Awesome Lists containing this project
README
# Kurzy SQL, MySQL, PostgreSQL, Microsoft SQL Server
**Kurzy SQL** sú koncipované tak, aby ťa doviedli **od teórie databáz k praxi**. Osvojíš si základné koncepcie a **princípy jazyka SQL**. Naučíme ťa, ako definovať **databázové objekty** a ako pracovať so **základnými príkazmi** a **ďalšími funkciami**. Už ako začiatočník zvládneš tvorbu jednoduchších dotazov. Pre pokročilých je pripravený kurz, v rámci ktorého ťa zasvätíme do jazyka **DML**, **DCL** a osvojíš si tvorbu zložitejších dotazov. Bez problémov zvládneš **správu dát**, **zabezpečenie** a **implementáci**u.
### Vyriešené príklady SQL Zoo | [SQLZOO](http://sqlzoo.net) | [Cheat mode](http://sqlzoo.net/wiki/SELECT_basics?answer=1)
## Sekcie:
0. [SELECT basics](#select-basics)
1. [SELECT name](#select-name)
2. [SELECT from WORLD](#select-from-world)
3. [SELECT from NOBEL](#select-from-nobel)
4. [SELECT in SELECT](#select-in-select)
5. [SUM and COUNT](#sum-and-count)
6. [JOIN](#join)
7. [More JOIN](#more-join)
8a. [Using NULL](#using-null)
8b. [Numeric Examples](#numeric-eamples)
9a. [Self JOIN](#self-join)
9b. [Window function](#window-unction)
## SELECT basics
Niekoľko jednoduchých dopytov, ktoré ti pomôžu začať
1.
```sql
SELECT population FROM world
WHERE name = 'Germany'
```
2.
```sql
SELECT name, gdp/population FROM world
WHERE area > 5000000
```
3.
```sql
SELECT name, population FROM world
WHERE name IN ('Ireland','Iceland','Denmark');
```
4.
```sql
SELECT name, area FROM world
WHERE area BETWEEN 200000 AND 250000
```
## SELECT name
Niekoľko dopytov k vyhľadávanie podľa vzorov/pattern matching
## SELECT from WORLD
Dopytovanie v tabuľke svet/world
1.
```sql
SELECT name, continent, population FROM world
```
2.
```sql
SELECT name FROM world
WHERE population>200000000
```
3.
```sql
SELECT name, gdp/population FROM world
WHERE population > 200000000
```
4.
```sql
SELECT name, population/1000000 FROM world
WHERE continent = 'South America'
```
5.
```sql
SELECT name,population FROM world
WHERE name IN ('France','Germany','Italy')
```
6.
```sql
SELECT name FROM world
WHERE name LIKE '%United%'
```
7.
```sql
select name, population, area from world
where population > 250000000 or area > 3000000
```
8.
```sql
select name, population, area from world
where population > 250000000 xor area > 3000000
```
9.
```sql
select name, ROUND(population/1000000,2), ROUND(gdp/1000000000,2) from world
where continent = 'South America'
```
10.
```sql
select name, ROUND(gdp/population,-3) from world
where gdp > 1000000000000
```
### Ťažšie otázky:
11.
```sql
SELECT name,
CASE WHEN continent='Oceania' THEN 'Australasia'
ELSE continent END
FROM world
WHERE name LIKE 'N%'
```
12.
```sql
SELECT name,
CASE WHEN continent='Europe' or continent='Asia' THEN 'Eurasia'
WHEN continent in ('North America','South America','Caribbean') THEN 'America'
ELSE continent END
FROM world
WHERE name LIKE 'A%' or name LIKE 'B%'
```
13.
```sql
SELECT name, continent, CASE
WHEN continent = 'Oceania' THEN 'Australasia'
WHEN continent = 'Eurasia' THEN 'Europe/Asia'
WHEN name = 'Turkey' THEN 'Europe/Asia'
WHEN continent = 'Caribbean' AND name LIKE 'B%' then 'North America'
WHEN continent = 'Caribbean' THEN 'South America'
ELSE continent END
FROM world ORDER BY name
```
## SELECT from NOBEL
Dodatočné precvičenie základných príkazov a funkcií pomocou tabuľky nositeľov Nobelovej ceny
1.
```sql
SELECT yr, subject, winner
FROM nobel
WHERE yr = 1950
```
2.
```sql
SELECT winner
FROM nobel
WHERE yr = 1962
AND subject = 'Literature'
```
3.
```sql
SELECT yr, subject
FROM nobel
WHERE winner = 'Albert Einstein'
```
4.
```sql
SELECT winner
FROM nobel
WHERE subject = 'Peace' AND yr >= 2000
```
5.
```sql
SELECT yr, subject, winner
FROM nobel
WHERE subject = 'Literature' AND yr BETWEEN 1980 AND 1989
(in MySQL BETWEEN is inclusive.. this is not always the case)
```
6.
```sql
SELECT * FROM nobel
WHERE winner IN ('Theodore Roosevelt',
'Woodrow Wilson',
'Jimmy Carter')
```
7.
```sql
SELECT winner FROM nobel
WHERE winner LIKE 'JOHN %'
```
8.
```sql
SELECT * FROM nobel
WHERE yr = 1980 AND subject = 'Physics'
OR yr = 1984 AND subject = 'Chemistry'
```
9.
```sql
SELECT * FROM nobel
WHERE yr = 1980
AND subject NOT IN ('Chemistry','Medicine')
```
10.
```sql
SELECT * FROM nobel
WHERE yr < 1910 AND subject = 'Medicine'
OR yr >= 2004 AND subject = 'Literature'
```
### Ťažšie otázky
11.
```sql
SELECT * FROM nobel
WHERE winner = 'Peter Grünberg'
```
12.
```sql
SELECT * FROM nobel
WHERE winner = 'Eugene O''Neill'
```
13.
```sql
SELECT winner, yr, subject FROM nobel
WHERE winner LIKE 'Sir %'
ORDER BY yr DESC, winner
```
14.
```sql
SELECT winner, subject
FROM nobel
WHERE yr=1984
ORDER BY subject in ('Chemistry','Physics'), subject, winner
```
## SELECT in SELECT
Príklady, v ktorých tvoríme dopyty pomocou iných dopytov
1.
```sql
SELECT name FROM world
WHERE population >
(SELECT population FROM world
WHERE name='Russia')
```
2.
```sql
SELECT name FROM world
WHERE continent = 'Europe'
AND gdp/population > (SELECT gdp/population
FROM world
WHERE name = 'United Kingdom')
```
3.
```sql
SELECT name, continent FROM world
WHERE continent IN (SELECT continent FROM world
WHERE name IN ('Argentina','Australia'))
ORDER BY name
```
4.
```sql
SELECT name, population FROM world
WHERE population > (SELECT population FROM world
WHERE name = 'Canada')
AND population < (SELECT population FROM world
WHERE name = 'Poland')
```
5.
```sql
SELECT name, CONCAT(ROUND(population/(SELECT population FROM world
WHERE name = 'Germany')*100,0),'%')
FROM world WHERE continent = 'Europe'
```
6.
```sql
SELECT name FROM world
WHERE gdp > ALL(SELECT gdp FROM world
WHERE gdp > 0 AND continent = 'Europe')
```
7.
```sql
SELECT continent, name, area FROM world x
WHERE area >= ALL
(SELECT area FROM world y
WHERE y.continent=x.continent
AND area>0)
```
8.
```sql
SELECT continent, name FROM world x
WHERE name <= ALL
(SELECT name FROM world y
WHERE y.continent=x.continent)
```
### Ťažšie otázky
9.
```sql
SELECT name, continent, population FROM world x
WHERE 25000000 >= ALL(SELECT population
FROM world y
WHERE x.continent = y.continent
AND y.population>0);
```
10.
```sql
SELECT name, continent FROM world x
WHERE population >= ALL(SELECT population*3
FROM world y
WHERE x.continent = y.continent
and y.name != x.name)
```
## SUM and COUNT
Príklady na agregované funkcie, distinct, order by, group by, having
1.
```sql
SELECT SUM(population)
FROM world
```
2.
```sql
SELECT DISTINCT continent FROM world
```
3.
```sql
SELECT SUM(gdp) FROM world
WHERE continent = 'Africa'
```
4.
```sql
SELECT COUNT(name) FROM world
WHERE area >= 1000000
```
5.
```sql
SELECT SUM(population) FROM world
WHERE name IN ('France','Germany','Spain')
```
6.
```sql
SELECT continent, COUNT(name) FROM world
GROUP BY continent
```
7.
```sql
SELECT continent, COUNT(name) FROM world
WHERE population > 10000000
GROUP BY continent
```
8.
```sql
SELECT continent FROM world
GROUP BY continent
HAVING SUM(population) > 100000000
```
## JOIN
Príklady na spájanie viacerých tabuliek hra, góly
1.
```sql
SELECT matchid, player FROM goal
WHERE teamid = 'GER'
```
2.
```sql
SELECT id,stadium,team1,team2
FROM game
WHERE id = 1012
```
3.
```sql
SELECT player, teamid, stadium, mdate
FROM game JOIN goal ON (id=matchid)
WHERE teamid = 'GER'
```
4.
```sql
SELECT team1, team2, player FROM game
JOIN goal ON (id=matchid)
WHERE player LIKE 'Mario%'
```
5.
```sql
SELECT player, teamid, coach, gtime
FROM goal
JOIN eteam on (teamid=id)
WHERE gtime<=10
```
6.
```sql
SELECT mdate,teamname FROM game
JOIN eteam ON (team1 = eteam.id)
WHERE coach = 'Fernando Santos'
```
7.
```sql
SELECT player FROM goal
JOIN game ON (matchid = id)
WHERE stadium = 'National Stadium, Warsaw'
```
### Ťažšie otázky
8.
```sql
SELECT DISTINCT player
FROM game JOIN goal ON matchid = id
WHERE (team1= 'GER' OR team2='GER')
AND teamid != 'GER'
```
9.
```sql
SELECT teamname, COUNT(*)
FROM eteam JOIN goal ON id=teamid
GROUP BY teamname
```
10.
```sql
SELECT stadium, COUNT(*) FROM goal
JOIN game ON (matchid = id)
GROUP BY stadium
```
11.
```sql
SELECT matchid, mdate, COUNT(*)
FROM game JOIN goal ON matchid = id
WHERE (team1 = 'POL' OR team2 = 'POL')
GROUP BY mdate,matchid
```
12.
```sql
SELECT matchid, mdate, COUNT(*) FROM goal
JOIN game ON (matchid=id)
WHERE teamid = 'GER'
GROUP BY matchid, mdate
```
13.
```sql
SELECT DISTINCT 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 game.id = goal.matchid
GROUP BY id, mdate, team1, team2
ORDER BY mdate, matchid, team1, team2
```
## More JOIN
Príklady na spájanie viacerých tabuliek herci, filmy, obsadenie
1.
```sql
SELECT id, title
FROM movie
WHERE yr=1962
```
2.
```sql
SELECT yr
FROM movie
WHERE title = 'Citizen Kane'
```
3.
```sql
SELECT id, title, yr FROM movie
WHERE title LIKE '%Star Trek%'
ORDER BY yr
```
4.
```sql
SELECT title FROM movie
WHERE id IN (11768, 11955, 21191)
```
5.
```sql
SELECT id FROM actor
WHERE name = 'Glenn Close'
```
6.
```sql
SELECT id FROM movie
WHERE title = 'Casablanca'
```
7.
```sql
SELECT name FROM casting JOIN actor ON (id=actorid)
WHERE movieid=11768
```
8.
```sql
SELECT name FROM casting
JOIN actor ON (actor.id=actorid)
JOIN movie ON (movie.id=movieid)
WHERE title = 'Alien'
```
9.
```sql
SELECT title FROM casting
JOIN movie ON (movie.id = movieid)
JOIN actor ON (actor.id = actorid)
WHERE name = 'Harrison Ford'
```
10.
```sql
SELECT title FROM casting
JOIN movie ON (movie.id = movieid)
JOIN actor ON (actor.id = actorid)
WHERE name = 'Harrison Ford' AND ord > 1
```
11.
```sql
SELECT title, name FROM casting
JOIN movie ON (movie.id = movieid)
JOIN actor ON (actor.id = actorid)
WHERE yr = 1962 and ord = 1
```
### Ťažšie otázky
12.
```sql
SELECT yr,COUNT(title) FROM
movie JOIN casting ON movie.id=movieid
JOIN actor ON actorid=actor.id
WHERE name='John Travolta'
GROUP BY yr
HAVING COUNT(title)=(SELECT MAX(c) FROM
(SELECT yr,COUNT(title) AS c FROM
movie JOIN casting ON movie.id=movieid
JOIN actor ON actorid=actor.id
WHERE name='John Travolta'
GROUP BY yr) AS t
)
```
13.
```sql
SELECT title, name FROM casting
JOIN movie ON movie.id = movieid
JOIN actor ON actor.id = actorid
WHERE ord = 1
AND movie.id IN
(SELECT movie.id FROM movie
JOIN casting ON movie.id = movieid
JOIN actor ON actor.id = actorid
WHERE actor.name = 'Julie Andrews')
```
14.
```sql
SELECT DISTINCT name FROM casting
JOIN movie ON movie.id = movieid
JOIN actor ON actor.id = actorid
WHERE actorid IN (
SELECT actorid FROM casting
WHERE ord = 1
GROUP BY actorid
HAVING COUNT(actorid) >= 30)
ORDER BY name
```
15.
```sql
SELECT title, COUNT(actorid) FROM casting
JOIN movie ON movieid = movie.id
WHERE yr = 1978
GROUP BY movieid, title
ORDER BY COUNT(actorid) DESC
```
16.
```sql
SELECT DISTINCT name FROM casting
JOIN actor ON actorid = actor.id
WHERE name != 'Art Garfunkel'
AND movieid IN (
SELECT movieid
FROM movie
JOIN casting ON movieid = movie.id
JOIN actor ON actorid = actor.id
WHERE actor.name = 'Art Garfunkel'
)
```
## Using NULL
Používanie stavu NULL, tabuľky učitelia, oddelenia
1.
```sql
SELECT name FROM teacher
WHERE dept IS NULL
```
2.
```sql
SELECT teacher.name, dept.name
FROM teacher INNER JOIN dept
ON (teacher.dept=dept.id)
```
3.
```sql
SELECT teacher.name, dept.name
FROM teacher LEFT JOIN dept
ON (teacher.dept=dept.id)
```
4.
```sql
SELECT teacher.name, dept.name
FROM teacher RIGHT JOIN dept
ON (teacher.dept=dept.id)
```
5.
```sql
SELECT teacher.name, COALESCE(teacher.mobile,'07986 444 2266') FROM teacher
```
6.
```sql
SELECT teacher.name, COALESCE(dept.name,'None') FROM teacher
LEFT JOIN dept ON teacher.dept = dept.id
```
7.
```sql
SELECT COUNT(name), COUNT(mobile) FROM teacher
```
8.
```sql
SELECT dept.name, COUNT(teacher.dept) FROM teacher
RIGHT JOIN dept ON dept.id = teacher.dept
GROUP BY dept.name
```
9.
```sql
SELECT name, CASE WHEN dept IN (1,2) THEN 'Sci'
ELSE 'Art'
END
FROM teacher
```
10.
```sql
SELECT name, CASE WHEN dept IN (1,2) THEN 'Sci'
WHEN dept = 3 THEN 'Art'
ELSE 'None'
END
FROM teacher
```
## Self JOIN
Príklady na spájanie tabuliek autobusové zastávky, cesty
1.
```sql
SELECT DISTINCT COUNT(*) FROM stops
```
2.
```sql
SELECT id FROM stops
WHERE name = 'Craiglockhart'
```
3.
```sql
SELECT id, name FROM stops JOIN route ON (stops.id = route.stop)
WHERE num = 4
```
4.
```sql
SELECT company, num, COUNT(*)
FROM route WHERE stop=149 OR stop=53
GROUP BY company, num
HAVING COUNT(*) = 2
```
5.
```sql
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 = (SELECT id FROM stops WHERE name = 'London Road')
```
6.
```sql
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'
```
7.
```sql
SELECT a.company, a.num
FROM route a, route b
WHERE a.num = b.num AND (a.stop = 115 AND b.stop = 137)
GROUP BY num;
```
8.
```sql
SELECT a.company, a.num
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 = 'Tollcross';
```
9.
```sql
SELECT DISTINCT name, a.company, a.num
FROM route a
JOIN route b ON (a.company = b.company AND a.num = b.num)
JOIN stops ON a.stop = stops.id
WHERE b.stop = 53;
```
10.
```sql
SELECT a.num, a.company, stopb.name, c.num, c.company
FROM route a
JOIN route b ON (a.company = b.company AND a.num = b.num)
JOIN (route c JOIN route d ON (c.company = d.company AND c.num = d.num))
JOIN stops stopa ON a.stop = stopa.id
JOIN stops stopb ON b.stop = stopb.id
JOIN stops stopc ON c.stop = stopc.id
JOIN stops stopd ON d.stop = stopd.id
WHERE stopa.name = 'Craiglockhart'
AND stopd.name = 'Sighthill'
AND stopb.name = stopc.name
ORDER BY LENGTH(a.num), b.num, stopb.name, LENGTH(c.num), d.num;
```
## Window function
Skúmame výsledky všeobecných volieb vo Veľkej Británii