GeralexGR

August 20, 2018August 20, 2018

Move files to replicated datastore using bat script

You can use a very simple bat script in order to automatically move files from a specific windows disk to another one.

An example for this particular scenario is a case that you want to copy files from a disk to another one that resides in a datastore that is replicated. You can automatically move those files from the first disk to the second one in order to get them replicated.

All you have to do, is to create an automatic task with windows task scheduler that its action will be to run the following bat script.

xcopy /s/e/y C:\Users\USER1\Desktop\folder1 D:\Users\folder2
echo %time% %date% >> D:\Users\folder2\log.txt

1 2	xcopy /s/e/y C:\Users\USER1\Desktop\folder1 D:\Users\folder2 echo %time% %date% >> D:\Users\folder2\log.txt

July 30, 2018

Get old events from Vcenter using powerCLI

If there is a reason that you want to retrieve older events from Vcenter than the 1000 that are available, you can use powerCLI in order to get specific date ranges.

After you login in the vCenter using

Connect-VIServer SERVER_IP

1	Connect-VIServer SERVER_IP

you can use Get-VIEvent with Start or Finish attributes. Be careful also to use Maxsamples attribute in order to get more than 10 that are returned by default.

For example if I wanted all events that have been captured the weekend I could use something like:

Get-VIEvent -maxsamples 10000 -start "28/07/2018"

1	Get-VIEvent -maxsamples 10000 -start "28/07/2018"

P.S Date is a DateTime and should be used as above

March 21, 2018March 21, 2018

Find the Nth Fibonacci Number – C# Code

The Fibonacci sequence begins with Fibonacci(0) = 0 and Fibonacci(1)=1 as its respective first and second terms. After these first two elements, each subsequent element is equal to the sum of the previous two elements.

Here is the basic information you need to calculate Fibonacci(n):

Fibonacci(0) = 0
Fibonacci(1) = 1
Fibonacci(n) = Fibonacci(n-1) + Fibonacci(n-2)

Task
Given n, complete the fibonacci function so it returns Fibonacci(n).

Input Format

Nth number

Output Format

Fibonacci() Nth number.

Sample Input

3

Sample Output

2

Explanation

Consider the Fibonacci sequence:

We want to know the value of Fibonacci(3). If we look at the sequence above,Fibonacci(3) evaluates to 2. Thus, we print 2 as our answer.

  public static int Fibonacci(int n)
        {
            if (n == 0) return 0;
            else if (n == 1) return 1;
            else
            {
                return Fibonacci(n - 1) + Fibonacci(n - 2);
            }
        }

public static int Fibonacci(int n)

{

if (n == 0) return 0;

else if (n == 1) return 1;

else

{

return Fibonacci(n - 1) + Fibonacci(n - 2);

}

March 20, 2018March 20, 2018

Array Left Rotation by D – C# Code

A left rotation operation on an array of size n shifts each of the array’s elements 1 unit to the left. For example, if 2 left rotations are performed on array [1,2,3,4,5], then the array would become [3,4,5,1,2].

Given an array of n integers and a number, d , perform d left rotations on the array. Then print the updated array as a single line of space-separated integers.

Input Format

The first line contains two space-separated integers denoting the respective values of (the number of integers) and (the number of left rotations you must perform).
The second line contains space-separated integers describing the respective elements of the array’s initial state.

Output Format

Print a single line of space-separated integers denoting the final state of the array after performing left rotations.

Sample Input

5 4
1 2 3 4 5

5 4

1 2 3 4 5

Sample Output

5 1 2 3 4

5 1 2 3 4

Explanation

Code

 static int[] LeftRotationByD(int[] a,int k)
        {
            int[] b = new int[a.Length];

            int index;
            int length = a.Length;
            int place;


            for (int i = 0; i < length; i++)
            {
                index = i - k;
                place = length + index;
                if (index >= 0)
                {
                    b[index] = a[i];
                }
                else
                {
                    b[place] = a[i];
                }
            }
            return b;

        }

static int[] LeftRotationByD(int[] a,int k)

{

int[] b = new int[a.Length];

int index;

int length = a.Length;

int place;

for (int i = 0; i < length; i++)

{

index = i - k;

place = length + index;

if (index >= 0)

{

b[index] = a[i];

}

else

{

b[place] = a[i];

}

return b;

}

We can see the results for 1 and 4 rotations as follows: