Created
March 15, 2011 15:06
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3.1. (EMPLOYEES, DEPARTMENTS) Nazw działów (department_name), nazwisk (last_name) oraz płac (salary) pracowników, którzy zarabiają ponad 12000, posortowane od najlepiej do najgorzej zarabiających (6 wyników): | |
a) bez wykorzystania polecenia JOIN | |
b) z wykorzystaniem polecenia JOIN | |
3.2. (EMPLOYEES, DEPARTMENTS, LOCATIONS) Nazwisk pracowników (last_name), nazw ich działów (department_name) i miast (city), w których pracują (106 wyników): | |
a) bez wykorzystania polecenia JOIN | |
b) z wykorzystaniem polecenia JOIN | |
3.3. (DEPARTMENTS, EMPLOYEES) Nazw działów (department_name) i ilości pracowników, którzy w nich pracują (11 wyników). | |
*3.4. Państw (country_name) i ilości pracowników, którzy w nich pracują (4 wyniki). | |
**3.5. (EMPLOYEES) Nazwisk pracowników (last_name) i nazwisk ich kierowników (last_name), posortowane rosnąco po nazwiskach kierowników. Użyj aliasów kolumn. (106 wyników) | |
3.6. Oceń trudność pytań (skala 1-10: 1-prościutkie, 10-diabelnie trudne) + napisz komentarz z sugestią do przyszłych ćwiczeń, jeśli chciał(a)byś coś zmienić Uśmiech | |
================= | |
3.1 : | |
a) SELECT d.department_name, e.last_name, e.salary FROM employees e, departments d WHERE (d.department_id = e.department_id) AND (e.salary > 12000) ORDER BY salary DESC; | |
b) SELECT d.department_name, e.last_name, e.salary FROM employees e JOIN departments d ON d.department_id = e.department_id WHERE e.salary > 12000 ORDER BY salary DESC; | |
3.2 : | |
a) SELECT e.last_name, d.department_name, l.city FROM employees e, departments d, locations l WHERE d.department_id = e.department_id AND l.location_id = d.location_id; | |
b)SELECT e.last_name, d.department_name, l.city FROM departments d JOIN employees e ON d.department_id=e.department_id JOIN locations l ON d.location_id=l.location_id; | |
3.3 : | |
SELECT d.department_name, COUNT(*) FROM departments d JOIN employees e ON d.department_id = e.department_id GROUP BY d.department_name; | |
3.4 | |
SELECT c.country_name, COUNT(*) FROM countries c JOIN locations l ON l.country_id = c.country_id JOIN departments d ON d.location_id = l.location_id JOIN employees e ON e.department_id = d.department_id GROUP BY c.country_name; | |
3.5 | |
SELECT e.last_name, e2.last_name FROM employees e JOIN employees e2 ON e.employee_id = e2.manager_id ORDER BY e2.manager_id; | |
3.6 | |
Trochę nieklarowne polecenia, zwłaszcza do 3.5 | |
Ocena 4 |
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