Usuario:ManuelRomero/ProgramacionWeb/INAEM2017/BasesDatos/SQL/ejercicios

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Revisión a fecha de 00:45 22 may 2018; ManuelRomero (Discusión | contribuciones)

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Ejercicios de bases de datos relacionales



Icon activity.jpg
Crea conexiones
Crea una conexión con oracle usando los parámetros facilitados en clase
Crea una conexión en línea de comandos con mysql
Crea una conexión usando phpmyadmin con el gestor de bases de datos mysql



Ejecuta los siguientes ficheros tanto en mysql como en oracle


Fichero SQL de inserción

Ejercicios Sentencias DCL

Ejercicios Sentencias Select

Ejercicios select/from/where con funciones



Icon activity.jpg
Realiza las siguientes consultas

Ej1 query.png

  • En este caso JOB_ID nos da el tipo de trabajo 
Secretari@ es el tipo ST_CLERK

mysql> SELECT *  
       FROM EMPLOYEES 
       WHERE JOB_ID ="ST_CLERK"       
             AND       
             HIRE_DATE>"1997-12-31";

Ej2 query.png 

SELECT LAST_NAME, JOB_ID,SALARY, COMMISSION_PCT 
FROM EMPLOYEES
WHERE COMMISSION_PCT IS NOT NULL 
ORDER BY LAST_NAME DESC;

Ej3 query.png 

SELECT 'El salario de '||Last_name||
       'va a ser con un incremento del 10% de '||
       round(Salary *1.10) ||
       ' dólares' AS "Nuevo Salario"
FROM EMPLOYEES
WHERE COMMISSION_pCT IS NULL

Ej4 query.png 

SELECT last_name, 
       trunc(months_between(sysdate, hire_date)/12) AS "años",
       trunc(MOD(months_between(sysdate, hire_date),12))AS "meses"
FROM employees
ORDER BY hire_date;

Ej5.png 

SELECT *
FROM empleados
WHERE substr(last_name,1,1) IN ('J', 'K', 'L', 'M')

Ej6.png 

SELECT last_name, salary, 
         decode (commission_pct,NULL, 'No' ,
                               'Yes') AS "comisión"
FROM employees

Ej7.png 

SELECT department_name, location_id,job_id, salary
FROM departments d, employees e
WHERE d.department_id = e.department_id
       AND
       location_id = '&localizacion';

Ej8.png 

SELECT COUNT(*)
FROM employees
WHERE last_name LIKE '%n';
--or
SELECT COUNT(*)
FROM employees
WHERE SUBSTR(last_name, -1) = 'n';

Ej10.png 

SELECT DISTINCT job_id
FROM employees
WHERE department_id IN (10, 20);

Ej12.png 

SELECT last_name, hire_date
FROM  employees
WHERE TO_CHAR(hire_date, 'DD') < 16;

Ej13.png 

SELECT last_name, salary, TRUNC(salary, -3)/1000
FROM  employees;

Ej21.png 

SELECT last_name, TO_CHAR(hire_date, 'Month DD') BIRTHDAY
FROM employees
ORDER BY TO_CHAR(hire_date, 'DDD');




Ejercicios Sentencias Join

Ej14.png 

SELECT e.last_name, m.last_name manager, m.salary,
j.grade_level
FROM  employees e JOIN employees m
      ON    e.manager_id = m.employee_id
     JOIN job_grades j
     ON   m.salary BETWEEN j.lowest_sal AND j.highest_sal
     AND  m.salary > 15000;

Ejercicios Sentencias Group by

Actividades de funciones y agregados

Ej func 01.png

Ej func 02.png

Ej func 03.png

Ej func 04.png

Ej func 05.png

Ej func 06.png

Ej func 07.png

Ej func 08.png

Ej func 09.png

Ej func 10.png

Ej func 11.png

Ej func 12.png 

 

Ej func 13.png 

 

Archivo:Ej func 14.png 

 

Ej func 15.png 

 

Ej func 16.png 

 

Ej func 17.png 

 

Ej func 18.png 

 

Ej func 19.png 

 

Ej func 20.png 

 

Ej func 21.png 

 

Ej func 22.png 

 

Ejercicios de agrupación



Icon activity.jpg
Ejercicios de agrupación

Ej9.png 

SELECT d.department_id, d.department_name, d.location_id, COUNT(e.employee_id)
FROM  employees e 
      RIGHT OUTER JOIN departments d
 ON
     e.department_id = d.department_id
GROUP BY d.department_id, d.department_name, d.location_id;

Ej11.png 

SELECT e.job_id, COUNT(e.job_id) FREQUENCY
FROM employees e 
     JOIN departments d
     ON e.department_id = d.department_id
WHERE   d.department_name IN ('Administration', 'Executive')
GROUP BY e.job_id
ORDER BY FREQUENCY DESC;

Ej15.png 

SELECT   d.department_id, d.department_name, COUNT(e1.employee_id) employees,  
         NVL(TO_CHAR(AVG(e1.salary), '99999.99'), 'No average') avg_sal,  
         e2.last_name, e2.salary, e2.job_id
FROM  departments d RIGHT OUTER JOIN employees e1
      ON  d.department_id = e1.department_id
      RIGHT OUTER JOIN employees e2
      ON   d.department_id = e2.department_id
      GROUP BY d.department_id, d.department_name, e2.last_name,
           e2.salary,
           e2.job_id
      ORDER BY d.department_id, employees;

Ej18.png

a) Contrate menos de tres empleados
SELECT d.department_id, d.department_name, COUNT(*)
FROM
departments d JOIN employees e
ON
d.department_id = e.department_id
GROUP BY d.department_id, d.department_name
HAVING COUNT(*) < 3;
b) Tenga el mayor número de empleados
SELECT d.department_id, d.department_name, COUNT(*)
FROM
departments d JOIN employees e
ON
d.department_id = e.department_id
GROUP BY d.department_id, d.department_name
HAVING COUNT(*) = (SELECT MAX(COUNT(*))
FROM
employees
GROUP BY department_id);
c) Tenga el menor número de empleados
SELECT d.department_id, d.department_name, COUNT(*)
FROM
departments d JOIN employees e
ON
d.department_id = e.department_id
GROUP BY d.department_id, d.department_name
HAVING COUNT(*) = (SELECT MIN(COUNT(*))
FROM
employees
GROUP BY department_id);

Ej19.png 

SELECT e.employee_id, e.last_name, e.department_id, e.salary,
AVG(s.salary)
FROM
employees e JOIN employees s
ON
e.department_id = s.department_id
GROUP BY e.employee_id, e.last_name, e.department_id,
e.salary;



Ejercicios Sentencias de subconsultas

Ejercicios con subconsultas


Prac subquery 1.png
Prac subquery 1s.png

Prac subquery 2.png
Prac subquery 2s.png

Prac subquery 3.png
Prac subquery 3s.png

Prac subquery 4.png
Prac subquery 4s.png

Prac subquery 5.png

Prac subquery 6.png

Prac subquery 7.png


Icon activity.jpg

Ejercicios de subconsultas Ej16.png 

SELECT department_id, MIN(salary)
FROM  employees
GROUP BY department_id
HAVING AVG(salary) = (SELECT MAX(AVG(salary))
                      FROM  employees
                      GROUP BY department_id);

Ej17.png 

SELECT *
FROM
departments
WHERE   department_id NOT IN(SELECT department_id
                             FROM employees
                             WHERE job_id = 'SA_REP' 
                               AND department_id IS NOT NULL);

Ej20.png 

SELECT last_name, TO_CHAR(hire_date, 'DAY') DAY
FROM  employees
WHERE TO_CHAR(hire_date, 'Day') = (SELECT TO_CHAR(hire_date, 'Day')
                                    FROM employees
                                    GROUP BY TO_CHAR(hire_date, 'Day')
                                    HAVING COUNT(*) = (SELECT MAX(COUNT(*))
                                                       FROM    employees
                                                       GROUP BY TO_CHAR(hire_date,'Day')));
{{{1}}}


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