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Ejemplo Constrains Oracle

Incrementar los salarios de los empleados del departamento ‘IT’ en 1%, levantar una excepción en caso de que el salario  actualizado exceda el salario máximo permitido para el puesto  DECLARE   name_department  departments.department_name% TYPE := 'IT'; incremento number := 1; nvo_salario employees.SALARY%TYPE; CURSOR CIncEmp(name_dept departments.department_name% TYPE) IS     SELECT employee_id, first_name ||' '|| last_name as NombreEmp, salary, max_salary FROM employees join jobs using(job_id) JOIN departments using(department_id) WHERE department_name = name_dept     ORDER BY employee_id     FOR UPDATE OF salary NOWAIT;   BEGIN   FOR RegEmp IN CIncEmp(name_department) LOOP     DBMS_OUTPUT.PUT_LINE('ID. Empleado: ' || ' ' || RegEmp.employee_id); DBMS_OUTPUT.PUT_LINE('Nombre Empleado: ' || ' ' || RegEmp.NombreEmp); DBMS_OUTPUT.PUT_LINE('Salario Actual: ' || ' ' || RegEmp.s...

ESET instaladores x86 - x64 ESP,ENG,PTB (Shareware)

- Español Latino ESET NOD32 Antivirus http://download.eset.com/ download/win/eav/eav_nt32_esl. msi http://download.eset.com/ download/win/eav/eav_nt64_esl. msi ESET Smart Security http://download.eset.com/ download/win/ess/ess_nt32_esl. msi http://download.eset.com/ download/win/ess/ess_nt64_esl. msi - Español España ESET NOD32 Antivirus http://download.eset.com/ download/win/eav/eav_nt32_esn. msi http://download.eset.com/ download/win/eav/eav_nt64_esn. msi ESET Smart Security http://download.eset.com/ download/win/ess/ess_nt32_esn. msi http://download.eset.com/ download/win/ess/ess_nt64_esn. msi - Ingles ESET NOD32 Antivirus http://download.eset.com/ download/win/eav/eav_nt32_enu. msi http://download.eset.com/ download/win/eav/eav_nt64_enu. msi ESET Smart Security http://download.eset.com/ download/win/ess/ess_nt32_enu. msi http://download.eset.com/ download/win/ess/ess_nt64_enu. msi - Portugues ESET NOD32 Antivirus http://download.eset.com/ download/win/eav/eav_nt32_ptb. msi http://...

10 commands you should master when working with the Cisco IOS

The Cisco IOS provides thousands of commands, and configuring it can be challenging. Here are 10 commands you need to know, inside and out, when using the Cisco IOS. #1: The “?” It may seem entirely too obvious that you should know how to type  ?  to ask for help when using the Cisco IOS. However, the Cisco IOS is completely different from other operating systems when it comes to using the question mark (help key). As the IOS is a command-line operating system with thousands of possible commands and parameters, using the ? can save your day. You can use the command in many ways. First, use it when you don’t know what command to type. For example, type  ?  at the command line for a list of all possible commands. You can also use ? when you don’t know what a command’s next parameter should be. For example, you might type show ip ?  If the router requires no other parameters for the command, the router will offer CR as the only option. Finally, use ? to se...

Biseccion C++ (codigo)

% f: funcion a la que se va a sacar el cero % a, b: limites del intervalo % delta: error absoluto maximo f = @(x)(x*sin(x)-1); a = 0; b = 2; delta = 0.00001; % ya: valor de la funcion en a % yb valor de la funcion en b % yc: valor de la funcion en c % err: error de la ultima iteracion % maxi: maximo de iteraciones ya = f(a); yb = f(b); if ya * yb > 0 disp "No existe una raiz en el intervalo" break end maxi = 1 + round((log(b-a)-log(delta))/log(2)); for k = 1:maxi c = (a+b) / 2; yc = f(c); if yc == 0 a = c; b = c; elseif yb*yc > 0 b = c; yb = yc; else a = c; ya = yc; end if b-a < delta break end end c = (a+b) / 2; err = abs(b-a); yc = f(c); % mostrar resultados printf("La aproximación al cero es: %10.6f\n",c); printf("Con un error absoluto menor a %8.6f\n",err); printf("El valor de la funcion en la aproximacion es %10.8f\n",yc);

Gauss C++ (codigo)

% A es la matriz de coeficientes % b es el vector de terminos independientes % x es el vector solucion % n es el orden del sistema [n,m] = size(A); % eliminacion for i=1:n for k=i+1:n factor = (A(k,i)/A(i,i)); A(k,i:n) = A(k,i:n) - factor*A(i,i:n); b(k) = b(k) - factor*b(i); end end %sustitucion inversa x = zeros(1,n); x(n) = b(n)/A(n,n); for k=n-1:-1:1 x(k) = (b(k) - sum(x(k+1:n).*A(k,k+1:n)))/A(k,k); end x