BANCO DE PREGUNTAS DE BASES DE DATOS

1.- En una base de datos, crear una tabla y llamarla SIG_4SEMESTRE_CIVIL. Esta tabla debe tener los siguientes campos:

CAMPO TIPO DE DATO

Código_ID Numérico(entero)

Nombres Caracter (Texto)

Apellidos Caracter (Texto)

Teléfono Numérico (entero largo)

Capital Numérico (doble)

2.- crear un consulta que emplee la tabla Civil4semestre.

3.-Seleccione la instrucción SQL correcta..

a. INSERT INTO SIG_4SEMESTRE_CIVIL VALUES(1.5,"JUAN","PEREZ",3494949,100000.00);

b. INSERT INTO SIG_4SEMESTRE_CIVIL VALUES(20,"JUAN","PEREZ",#3494949,100000.00);

c. INSERT INTO SIG_4SEMESTRE_CIVIL VALUES(18,JUAN,PEREZ,3494949,100000.00);

d. INSERT INTO SIG_4SEMESTRE_CIVIL VALUES(23," JUAN","PEREZ",3494949,100000.00);

e. INSERT INTO SIG_4SEMESTRE_CIVIL VALUES(14," JUAN","PEREZ",3494949,100000.00).f. INSERT INTO SIG_4SEMESTRE_CIVIL VALUES(100,"JUAN",PEREZ,3494949,100000.00);

4.- Seleccione la sentencia SQL correcta, para desplegar todos los datos con la condición

de búsqueda en el campo Nombres

a. SELECT * FROM SIG_4SEMESTRE_CIVIL WHERE (((SIG_4SEMESTRE_CIVILNombres)="PEDRO"));

b. SELECT ? FROM SIG_4SEMESTRE_CIVIL WHERE (((SIG_4SEMESTRE_CIVIL.Nombres)="JUAN"));

c. SELECT * FROM SIG_4SEMESTRE_CIVIL WHERE (((SIG_4SEMESTRE_CIVIL.Nombres)="PEDRO")));

d. SELECT * FORM SIG_4SEMESTRE_CIVIL WHERE (((SIG_4SEMESTRE_CIVIL.Nombres)= JUAN ));

e. SELECT * FROM SIG_4SEMESTRE_CIVIL WHERE (((SIG_4SEMESTRE_CIVIL.Nombres)="PEDRO"));

f. SELECT + FROM SIG_4SEMESTRE_CIVIL WERE (((SIG_4SEMESTRE_CIVIL.Nombres)="JUAN"));

5.- Seleccione la sentencia SQL correcta, para desplegar todos los datos con la condición

de búsqueda en el campo Nombres y ordenados en forma Descendente por el campo

Apellidos

a. SELECT * FROM SIG_4SEMESTRE_CIVIL WHERE (((SIG_4SEMESTRE_CIVIL.Nombres)="SARAH"))

ORDER BY BORRAR.Apellidos DESC;

b. SELECT * FROM SIG_4SEMESTRE_CIVIL WHERE (((SIG_4SEMESTRE_CIVIL.Nombres)="SARAH"))

ORDER BY DESC BORRAR.Apellidos;

6.- Seleccione la sentencia SQL correcta, para desplegar la cantidad de registros de la

tabla SIG_4SEMESTRE_CIVIL

1. SELECT Count(Código_ID) AS "# de registros" FROM SIG_4SEMESTRE_CIVIL;

2. SELECT Contar(Código_ID) AS "# de registros" FROM SIG_4SEMESTRE_CIVIL;

3. SELECT Count(capital) AS "# de registros" FROM SIG_4SEMESTRE_CIVIL;

1.- En la base de datos PROYECTO VIVIENDA crear una consulta con las siguientes

tablas: T3DETALLEMANODEOBRA, T1GENERALESDELPROYECTO

Introducir las siguientes sentencias:

SELECT T3DETALLEMANODEOBRA.Id, T1GENERALESDELPROYECTO.[NOMBRE

DEL PROYECTO] AS Proyecto, T3DETALLEMANODEOBRA.DESCRIPCION AS

descripcion, T3DETALLEMANODEOBRA.PU, T3DETALLEMANODEOBRA.TOTAL

FROM T3DETALLEMANODEOBRA, T1GENERALESDELPROYECTO

WHERE ((([T3DETALLEMANODEOBRA]![Id])=[T1GENERALESDELPROYECTO]![Id])

AND ((T3DETALLEMANODEOBRA.TOTAL)>19)) OR

(((T3DETALLEMANODEOBRA.TOTAL)<10) AND

((T3DETALLEMANODEOBRA.PU)=1.25));

2 Realice las siguientes selecciones simples:

a. Seleccione de cuantas tablas despliega la información

1 2 3 4 5

b. Seleccione cuantos campos despliega

1 2 3 4 5 6 7

c. Seleccione el operador lógico que emplea

Or incluyente or excluyente

3 Practicar las siguientes sentencias (analizar estructura ,resultados y APLICACIONES)

1 SELECT Max(total) AS "Valor Maximo del Total"

FROM T3DETALLEMANODEOBRA

WHERE (((T3DETALLEMANODEOBRA.[No ITEM]) like"02,2*"));

2 SELECT AVG(total) AS "Valor Promedio del Total"

FROM T3DETALLEMANODEOBRA

WHERE (((T3DETALLEMANODEOBRA.[No ITEM]) like"02,2*"));

3 SELECT Min(total) AS "Valor Mínimo del Total"

FROM T3DETALLEMANODEOBRA

WHERE (((T3DETALLEMANODEOBRA.[No ITEM]) like"02,2*"));

4 SELECT SUM(total) AS "Suma del campo Total"

FROM T3DETALLEMANODEOBRA

WHERE (((T3DETALLEMANODEOBRA.[No ITEM]) like"01,*"));

5 SELECT SUM(total) AS "Suma del campo Total"

FROM T3DETALLEMANODEOBRA;

6 SELECT (TOTAL) AS ["Valor Maximo del Total"], ITEM,DESCRIPCION

FROM T3DETALLEMANODEOBRA

WHERE (((T3DETALLEMANODEOBRA.TOTAL) Between 5 And 20));

7 SELECT TOTAL AS ["Valor Maximo del Total"], ITEM,DESCRIPCION

FROM T3DETALLEMANODEOBRA

WHERE (((T3DETALLEMANODEOBRA.TOTAL) Between 5 And 20));

1.- Concepto.-(20%)

Defina Base de datos y explique una aplicación en ingeniería civil

Una colección arbitraria de tablas o archivos bajo el control de un sistema de gestión de base de datos. Aplicable en el manejo de todos los archivos de una construcción que podría ser de carretas, puentes, edificios, etc

Defina SQL

Structured query lenguage lenguaje de consulta estructurado. El estándar aceptado internacionalmente para sistemas relacionales, que incluye no solo la consulta, sino también la definición de datos, manipulación, seguridad y algunos aspectos de integridad referencial

2.- Selección múltiple.- (20%)

Seleccione cuál de las siguientes opciones se incluyen como aplicación del LENGUAJE DE CONSULTA ESTRUCTURADO (SQL):

Consulta

Definición de datos

Manipulación

Seguridad

Aspectos de integridad referencial3.- Desarrollo.- (60%)

De las siguientes propuestas, escriba las sentencias SQL que corresponda

a. Sentencia SQL para crear una base de datos llamada "PROYECTO_CARRETERA":

CREATE DATABASE PROYECTO_CARRETERA;

b. Utilizar la sentencia CREATE TABLE para crear una tabla llamada MATERIALES, que contenga cinco columnas: : MaterialesID, Nombre, Descripcion, Proveedor, Industria.

Todos deben ser tipo carácter (255) excepto Materiales ID que tiene que ser numero entero

CREATE TABLE Materiales(MaterialesID int,Nombre varchar(255),Descripcion varchar(255),proveedor varchar(255),Industria varchar(255));

c. Utilizar la sentencia sql para asignar dos alias uno para la columna SUELO como alias1, y otro para la columna UBICACIÓN como alias2 de la tabla PROYECTO

SELECT SUELO AS ALIAS1, UBICACION AS ALIAS2FROM PROYECTO;

d. Utilizar la sentencia sql para seleccionar de la tabla ALTURAS, todas las cotas del campo COTA, comprendidas en entre los valores 200 y 280

SELECT * FROM ALTURASWHERE COTA BETWEEN 200 AND 280;

e. Utilizar la sentencia IN para seleccionar de la tabla SUELOS, todos los datos, donde el campo LOCALIDAD sea ARENOSO o ARCILLOSO

SELECT * FROM SUELOSWHERE LOCALIDAD IN ('ARENOSO','ARCILLOSO');

f. Utilizar la sentencia LIKE para seleccionar de la tabla PROYECTO, todos los datos, donde el campo LOCALIDAD empieza con a,b,c

SELECT * FROM PROYECTOWHERE LOCALIDAD LIKE '[a-c]%';

g. Utilizar la sentencia DELETE para eliminar WARNES del campo UBICACIÓN de la tabla LOCALIDAD

DELETE FROM LOCALIDADWHERE UBICACION='WARNES';

h. Utilizar la sentencia ORDER BY para seleccionar todos los suelos de la tabla SUELOS, que estén en forma ascendente por el campo UBICACIÓN y descendente por el campo TIPO

SELECT * FROM SUELOSORDER BY UBICACION ASC, TIPO DESC;

i. Utilizar la sentencia sql para devolver la fecha actual del sistema y el tiempo de la tabla PROYECTO

SELECT NOW() FROM PROYECTO;

j. Utilizar la sentencia SUM para sumar la columna COSTO de la tabla MATERIALES y mostrarlo como TOTAL

SELECT SUM(COSTO) AS TOTAL FROM MATERIALES;

k. Utilizar la sentencia MAX para obtener el valor más grande del campo COTA de la tabla ALTURAS darle el alias de ALTURAMAYOR

SELECT MAX(COTA) AS ALTURAMAYOR FROM ALTURAS;

l. Utilizar la sentencia COUNT para devolver el número de registros de la columna COTA de la tabla ALTURAS

SELECT COUNT(COTA) FROM ALTURAS;

BANCO DE PREGUNTAS DE BASE DE DATOS

1.- Concepto.-

Base de datos

SQL

2.- Selección múltiple

EL LENGUAJE DE CONSULTA ESTRUCTURADO (SQL) INCLUYE LOS SIGUIENTE:

Consulta

Definición de datos

Manipulación

Seguridad

Aspectos de integridad referencial

3.- DESCRIPCION.- DEFINA QUE HACEN LAS SIGUIENTES SENTENCIAS SQL

SENTENCIA DESCRIPCIONCREATE TABLE Persons(P_Id int NOT NULL,LastName varchar(255) NOT NULL,FirstName varchar(255),Address varchar(255),City varchar(255),CHECK (P_Id>0))CREATE TABLE Persons(P_Id int NOT NULL CHECK (P_Id>0),LastName varchar(255) NOT NULL,FirstName varchar(255),Address varchar(255),City varchar(255))CREATE TABLE Persons(P_Id int NOT NULL,LastName varchar(255) NOT NULL,FirstName varchar(255),Address varchar(255),City varchar(255),CONSTRAINT chk_Person CHECK (P_Id>0 AND City='Sandnes'))CREATE TABLE Persons(P_Id int NOT NULL,LastName varchar(255) NOT NULL,FirstName varchar(255),Address varchar(255),City varchar(255) DEFAULT 'Sandnes')CREATE TABLE Orders(O_Id int NOT NULL,OrderNo int NOT NULL,P_Id int,OrderDate date DEFAULT GETDATE())ALTER TABLE PersonsALTER COLUMN City SET DEFAULT 'SANDNES'

ALTER TABLE PersonsALTER COLUMN City DROP DEFAULT

CREATE INDEX index_nameON table_name (column_name)

CREATE UNIQUE INDEX index_nameON table_name (column_name)CREATE INDEX PIndexON Persons (LastName)DROP INDEX index_name ON table_nameDROP TABLE table_nameDROP DATABASE database_nameTRUNCATE TABLE table_nameALTER TABLE table_nameADD column_name datatypeALTER TABLE table_nameDROP COLUMN column_nameALTER TABLE table_nameALTER COLUMN column_name datatypeALTER TABLE PersonsALTER COLUMN DateOfBirth yearALTER TABLE PersonsDROP COLUMN DateOfBirthCREATE TABLE Persons(ID Integer PRIMARY KEY AUTOINCREMENT,LastName varchar(255) NOT NULL,FirstName varchar(255),Address varchar(255),City varchar(255))CREATE VIEW [Current Product List] ASSELECT ProductID,ProductNameFROM ProductsWHERE Discontinued=NoSELECT * FROM [Current Product List]CREATE VIEW [Products Above Average Price] ASSELECT ProductName,UnitPriceFROM ProductsWHERE UnitPrice>(SELECT AVG(UnitPrice) FROM Products)SELECT * FROM [Products Above Average Price]CREATE VIEW [Category Sales For 1997] ASSELECT DISTINCT CategoryName,Sum(ProductSales) AS CategorySalesFROM [Product Sales for 1997]GROUP BY CategoryNameSELECT * FROM [Category Sales For 1997]SELECT * FROM [Category Sales For 1997]WHERE CategoryName='Beverages'CREATE OR REPLACE VIEW view_name ASSELECT column_name(s)

FROM table_nameWHERE conditionCREATE VIEW [Current Product List] ASSELECT ProductID,ProductName,CategoryFROM ProductsWHERE Discontinued=NoSELECT LastName,FirstName,Address FROM PersonsWHERE Address IS NULLSELECT LastName,FirstName,Address FROM PersonsWHERE Address IS NOT NULLSELECT ProductName,UnitPrice*(UnitsInStock+IIF(ISNULL(UnitsOnOrder),0,UnitsOnOrder))FROM ProductsSELECT AVG(column_name) FROM table_nameSELECT AVG(Price) AS PriceAverage FROM Products;SELECT ProductName, Price FROM ProductsWHERE Price>(SELECT AVG(Price) FROM Products);SELECT COUNT(column_name) FROM table_name;SELECT COUNT(*) FROM table_name;SELECT COUNT(CustomerID) AS OrdersFromCustomerID7 FROM OrdersWHERE CustomerID=7;SELECT FIRST(column_name) FROM table_name;SELECT LAST(column_name) FROM table_name;SELECT LAST(CustomerName) AS LastCustomer FROM Customers;SELECT MAX(column_name) FROM table_name;SELECT MAX(Price) AS HighestPrice FROM Products;SELECT MIN(column_name) FROM table_name;SELECT MIN(Price) AS SmallestOrderPrice FROM Products;SELECT SUM(column_name) FROM table_name;SELECT SUM(Quantity) AS TotalItemsOrdered FROM OrderDetails;SELECT column_name, aggregate_function(column_name)FROM table_name

WHERE column_name operator valueGROUP BY column_name;SELECT Shippers.ShipperName,COUNT(Orders.OrderID) AS NumberOfOrders FROM OrdersLEFT JOIN ShippersON Orders.ShipperID=Shippers.ShipperIDGROUP BY ShipperName;SELECT Shippers.ShipperName, Employees.LastName,COUNT(Orders.OrderID) AS NumberOfOrdersFROM ((OrdersINNER JOIN ShippersON Orders.ShipperID=Shippers.ShipperID)INNER JOIN EmployeesON Orders.EmployeeID=Employees.EmployeeID)GROUP BY ShipperName,LastName;SELECT column_name, aggregate_function(column_name)FROM table_nameWHERE column_name operator valueGROUP BY column_nameHAVING aggregate_function(column_name) operator value;SELECT Employees.LastName, COUNT(Orders.OrderID) AS NumberOfOrders FROM (OrdersINNER JOIN EmployeesON Orders.EmployeeID=Employees.EmployeeID)GROUP BY LastNameHAVING COUNT(Orders.OrderID) > 10;SELECT Employees.LastName, COUNT(Orders.OrderID) AS NumberOfOrders FROM OrdersINNER JOIN EmployeesON Orders.EmployeeID=Employees.EmployeeIDWHERE LastName='Davolio' OR LastName='Fuller'GROUP BY LastNameHAVING COUNT(Orders.OrderID) > 25;

SELECT UCASE(column_name) FROM table_name;SELECT UCASE(CustomerName) AS Customer, CityFROM Customers;SELECT LCASE(column_name) FROM table_name;SELECT LCASE(CustomerName) AS Customer, CityFROM Customers;SELECT MID(column_name,start[,length]) AS some_name FROM table_name;SELECT MID(City,1,4) AS ShortCityFROM Customers;SELECT LEN(column_name) FROM table_name;SELECT CustomerName,LEN(Address) as LengthOfAddressFROM Customers;SELECT ROUND(column_name,decimals) FROM table_name;SELECT ProductName, ROUND(Price,0) AS RoundedPriceFROM Products;SELECT NOW() FROM table_name;SELECT ProductName, Price, Now() AS PerDateFROM Products;SELECT FORMAT(column_name,format) FROM table_name;SELECT ProductName, Price, FORMAT(Now(),'YYYY-MM-DD') AS PerDateFROM Products;SELECT column_name,column_nameFROM table_name;SELECT * FROM Customers;SELECT DISTINCT column_name,column_nameFROM table_name;SELECT column_name,column_nameFROM table_nameWHERE column_name operator value;SELECT * FROM CustomersWHERE Country='Mexico';SELECT * FROM CustomersWHERE Country='Germany'AND City='Berlin';SELECT * FROM CustomersWHERE Country='Germany'

AND (City='Berlin' OR City='München');SELECT column_name, column_nameFROM table_nameORDER BY column_name ASC|DESC, column_name ASC|DESC;SELECT * FROM CustomersORDER BY Country;SELECT * FROM CustomersORDER BY Country DESC;SELECT * FROM CustomersORDER BY Country ASC, CustomerName DESC;INSERT INTO table_nameVALUES (value1,value2,value3,...);INSERT INTO Customers (CustomerName, ContactName, Address, City, PostalCode, Country)VALUES ('Cardinal','Tom B. Erichsen','Skagen 21','Stavanger','4006','Norway');UPDATE table_nameSET column1=value1,column2=value2,...WHERE some_column=some_value;UPDATE CustomersSET ContactName='Alfred Schmidt', City='Hamburg'WHERE CustomerName='Alfreds Futterkiste';DELETE FROM table_nameWHERE some_column=some_value;DELETE FROM CustomersWHERE CustomerName='Alfreds Futterkiste' AND ContactName='Maria Anders';SELECT TOP number|percent column_name(s)FROM table_name;

SELECT TOP 2 * FROM Customers;

SELECT TOP 50 PERCENT * FROM Customers;

SELECT column_name(s)FROM table_nameWHERE column_name LIKE pattern;

SELECT * FROM CustomersWHERE City LIKE 's%';

SELECT * FROM CustomersWHERE City LIKE '%s';

SELECT * FROM CustomersWHERE Country LIKE '%land%';

SELECT * FROM CustomersWHERE Country NOT LIKE '%land%';

SELECT * FROM CustomersWHERE City LIKE 'ber%';

SELECT * FROM CustomersWHERE City LIKE '_erlin';

SELECT * FROM CustomersWHERE City LIKE 'L_n_on';

SELECT * FROM CustomersWHERE City LIKE '[bsp]%';

SELECT * FROM CustomersWHERE City LIKE '[a-c]%';

SELECT column_name(s)FROM table_nameWHERE column_name IN (value1,value2,...);SELECT * FROM CustomersWHERE City IN ('Paris','London');

SELECT column_name(s)FROM table_nameWHERE column_name BETWEEN value1 AND value2;

SELECT * FROM ProductsWHERE Price BETWEEN 10 AND 20;

SELECT * FROM ProductsWHERE Price NOT BETWEEN 10 AND 20;

SELECT * FROM ProductsWHERE (Price BETWEEN 10 AND 20)AND NOT CategoryID IN (1,2,3);

SELECT * FROM ProductsWHERE ProductName NOT BETWEEN 'C' AND 'M';

SELECT * FROM OrdersWHERE OrderDate BETWEEN #07/04/1996# AND #07/09/1996#;

SELECT column_name AS alias_nameFROM table_name;

SELECT column_name(s)FROM table_name AS alias_name;

SELECT CustomerName AS Customer, ContactName AS [Contact Person]FROM Customers;

SELECT CustomerName, Address+', '+City+', '+PostalCode+', '+Country AS AddressFROM Customers;SELECT CustomerName, Address+', '+City+', '+PostalCode+', '+Country AS AddressFROM Customers;SELECT o.OrderID, o.OrderDate, c.CustomerNameFROM Customers AS c, Orders AS oWHERE c.CustomerName="Around the Horn" AND c.CustomerID=o.CustomerID;SELECT Orders.OrderID, Customers.CustomerName, Orders.OrderDateFROM OrdersINNER JOIN CustomersON

Orders.CustomerID=Customers.CustomerID;SELECT column_name(s)FROM table1INNER JOIN table2ON table1.column_name=table2.column_name;SELECT column_name(s)FROM table1JOIN table2ON table1.column_name=table2.column_name;SELECT Customers.CustomerName, Orders.OrderIDFROM CustomersINNER JOIN OrdersON Customers.CustomerID=Orders.CustomerIDORDER BY Customers.CustomerName;SELECT column_name(s)FROM table1LEFT JOIN table2ON table1.column_name=table2.column_name;SELECT column_name(s)FROM table1LEFT OUTER JOIN table2ON table1.column_name=table2.column_name;SELECT Customers.CustomerName, Orders.OrderIDFROM CustomersLEFT JOIN OrdersON Customers.CustomerID=Orders.CustomerIDORDER BY Customers.CustomerName;SELECT column_name(s)FROM table1RIGHT JOIN table2ON table1.column_name=table2.column_name;SELECT column_name(s)FROM table1RIGHT OUTER JOIN table2ON table1.column_name=table2.column_name;SELECT Orders.OrderID, Employees.FirstName

FROM OrdersRIGHT JOIN EmployeesON Orders.EmployeeID=Employees.EmployeeIDORDER BY Orders.OrderID;SELECT column_name(s)FROM table1FULL OUTER JOIN table2ON table1.column_name=table2.column_name;SELECT Customers.CustomerName, Orders.OrderIDFROM CustomersFULL OUTER JOIN OrdersON Customers.CustomerID=Orders.CustomerIDORDER BY Customers.CustomerName;SELECT column_name(s) FROM table1UNION ALLSELECT column_name(s) FROM table2;SELECT City FROM CustomersUNIONSELECT City FROM SuppliersORDER BY City;SELECT City FROM CustomersUNION ALLSELECT City FROM SuppliersORDER BY City;SELECT City, Country FROM CustomersWHERE Country='Germany'UNION ALLSELECT City, Country FROM SuppliersWHERE Country='Germany'ORDER BY City;SELECT *INTO newtable [IN externaldb]FROM table1;SELECT *INTO CustomersBackup2013FROM Customers;SELECT CustomerName, ContactNameINTO CustomersBackup2013FROM Customers;SELECT *INTO CustomersBackup2013FROM CustomersWHERE Country='Germany';

SELECT Customers.CustomerName, Orders.OrderIDINTO CustomersOrderBackup2013FROM CustomersLEFT JOIN OrdersON Customers.CustomerID=Orders.CustomerID;INSERT INTO table2SELECT * FROM table1;

INSERT INTO table2(column_name(s))SELECT column_name(s)FROM table1;INSERT INTO Customers (CustomerName, Country)SELECT SupplierName, Country FROM Suppliers;INSERT INTO Customers (CustomerName, Country)SELECT SupplierName, Country FROM SuppliersWHERE Country='Germany';CREATE DATABASE dbname;

CREATE DATABASE my_db;

CREATE TABLE table_name(column_name1 data_type(size),column_name2 data_type(size),column_name3 data_type(size),....);CREATE TABLE Persons(PersonID int,LastName varchar(255),FirstName varchar(255),Address varchar(255),City varchar(255));

Data type Description Storage

Text

Memo

Byte 1 byte

Integer 2 bytes

Long 4 bytes

Single 4 bytes

Double 8 bytes

Currency 8 bytes

AutoNumber

4 bytes

Date/Time 8 bytes

Yes/No 1 bit

Ole Object up to 1GB

Hyperlink

Lookup Wizard

4 bytes