Segundo parcial Base de Datos
Quiz
•
Instructional Technology
•
1st - 5th Grade
•
Practice Problem
•
Medium
Edgar Caldas
Used 9+ times
FREE Resource
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20 questions
Show all answers
1.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
Buscar salario más alto y mínimo, la suma y promedio de salarios de todos los empleados. Etiquetar las columnas Máximo, Mínimo, Suma y Promedio, respectivamente. Redondo.
select max(salary),0) as Maximo, min(salary),0 as Minimo, sum(salary) as Suma, avg(salary) as Promedio
from employees
select round(max(salary),0) as Maximo, round(min(salary),0) as Minimo, round(sum(salary),0) as Suma, round(avg(salary),0) as Promedio
from employees
select round(max(salary),0) as Maximo, round(min(salary),0) as Minimo, round(sum(salary),0) as Suma, round(avg(salary),0) as Promedio
from employes
Ninguno
2.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
Encuentre la diferencia entre los salarios más altos y más bajos. etiquete la columna DIFERENCIA.
SELECT (MAX(SALARY)-MIN(SALARY)) AS DIFFERENCE
FROM EMPLOYEES
SELECT (MIN(SALARY)-MAX(SALARY)) AS DIFFERENCE
FROM EMPLOYEES
SELECT (MAX(SALARY) DIF MIN(SALARY)) AS DIFFERENCE
FROM EMPLOYEES
Ninguno
3.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
El departamento de recursos humanos quiere encontrar la duración del empleo de cada empleado. Para cada empleado, muestre el apellido y calcule el número de meses entre hoy y la fecha en que se contrató al empleado. Etiquete la columna MONTHS_WORKED. Ordene sus resultados por el número de meses empleados. Redondea el número de meses al número entero más cercano.
select last_name, round(months_between( sysdate,hire_date )) as "Months Worked" from employees order by 3
select last_name, months_between( sysdate,hire_date ) as "Months Worked" from employees order by 2
select last_name, round(months_between( sysdate,hire_date )) as "Months Worked" from employees order by 2
Ninguno
4.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
Cree un informe que produzca lo siguiente para cada empleado:
<apellido del empleado> gana <salario> mensualmente pero quiere <2 veces el salario>. Etiquete la columna Salarios de ensueño.
Select last_name || ' gana ' || to_char(salary,'$99,999.00') || ' Quiere ' || to_char(salary*2,'$99,999.00') as "Salario de sueño" from employees
Select last_name || ' gana ' || to_char(salary,'$99,999.00') || ' Quiere ' || to_char(salary,'$99,999.00') as "Salario de sueño" from employees
Select last_name || ' gana ' || to_char(salary,'$99,999.00') || ' Quiere ' || to_char(salary+2,'$99,999.00') as "Salario de sueño" from employees
Ninguno
5.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
El departamento de recursos humanos necesita un informe de todos los empleados. Escriba una consulta para mostrar el apellido, el número de departamento y el departamento de todos los empleados.
select last_name, department_id, department_name
from employees join departments
order by department_id
select last_name, department_id, department_name
from employees, departments
order by department_id
select last_name, department_id, department_name
from employees natural join departments
select last_name, department_id, department_name
from employees natural join departments
order by department_id
6.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
Determine la cantidad de gerentes sin enumerarlos. Etiquete la columna "Número de gerentes". Sugerencia: use la columna MANAGER_ID para determinar la cantidad de administradores.
SELECT COUNT(MANAGER_ID) AS "MANAGER NUMBER" FROM EMPLOYEES
SELECT AVG(MANAGER_ID) AS "MANAGER NUMBER" FROM EMPLOYEES
SELECT AVG(MANAGER_ID) AS "MANAGER NUMBER" FROM EMPLOYEES
Ninguno
7.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
Cree un informe para mostrar el número de gerente y el salario del empleado peor pagado que el gerente. Excluya a cualquier persona cuyo gerente no sea conocido. Excluya cualquier grupo donde el salario mínimo sea: $ 8,000 o menos. Ordene la producción en orden descendente de salario.
SELECT manager_id, MIN(SALARY)
FROM employees
WHERE manager_id IS NOT NULL
HAVING MIN(SALARY) > 8000
GROUP BY manager_id
ORDER BY 2 DESC
SELECT manager_id, MIN(SALARY)
FROM employees
WHERE manager_id IS NOT NULL
HAVING MIN(SALARY) <> 8000
GROUP BY manager_id
ORDER BY 2 DESC
SELECT manager_id, MIN(SALARY)
FROM employees
WHERE manager_id IS NOT NULL
HAVING MIN(SALARY) > 8000
GROUP BY manager_id
Ninguno
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