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Tuesday, 1 August 2017

Are you having ample sleep and still see those dark under eyes?

Are you sick of seeing yourself with those deep, dark under eyes?
Not to fret, we will help you with some tips to vanish the under eye and have that baby eyes with
sparkling eyes.



1. Eye Pads: you can easily see eye pads on various online stores to soothe your eyes and give them a
refreshing look. If you don’t trust the eye pads, we can DIY them at home itself. Soak a cotton ball in
rose water or cold milk as an eye pad. Keep the cotton balls for 15 minutes and wash your face
gently. For better results do it twice a day.



2. Oil massage: Massage is the best to relax and rejuvenate your body in the best possible way. You
can use coconut oil which is found mostly in our households. The oil is full of nutrients and
nourishes the under eye skin which will help you to vanish the darkness.
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Thursday, 27 July 2017

Types of SQL Server Joins


Continuting with the previous article. Let’s discuss the last and final step to understand every type of join.

Step-4: Last step, performing different types of joins.

Let’s check out every join one by one.

Inner Join/Join: Starting with “inner join” or simply “join” keyword which give only the matched rows specified in the ON condition. Our result set would look like this:

--Using join
select P.PassengerId as PassengerId,P.PassengerName as PassengerName,
F.FlightName as FlightName,F.FlightCode as FlightCode
from tblPassengers P join tblFlight F
on P.FlightId = F.FlightId
 
Join by imagination hunt blogs
Join


--Using Inner Join
select P.PassengerId as PassengerId,P.PassengerName as PassengerName,
F.FlightName as FlightName,F.FlightCode as FlightCode
from tblPassengers P inner join tblFlight F
on P.FlightId = F.FlightId
 
Inner Join by imagination hunt blogs
Inner Join


Left Join/ Left Outer Join: Next, “left outer join” or simply “left join” keyword which give all rows from LEFT table and only matched rows from other table specified in the ON condition. Unmatched columns cell are replaced with NULL. Our result set would look like this:

--Using Left Outer Join
select P.PassengerId as PassengerId,P.PassengerName as PassengerName,
F.FlightName as FlightName,F.FlightCode as FlightCode
from tblPassengers P left outer join tblFlight F
on P.FlightId = F.FlightId
 
Left Outer Join by imagination hunt blogs
Left Outer Join


--Using Left Join
select P.PassengerId as PassengerId,P.PassengerName as PassengerName,
F.FlightName as FlightName,F.FlightCode as FlightCode
from tblPassengers P left join tblFlight F
on P.FlightId = F.FlightId
 
Left Join by imagination hunt blogs
Left Join


Right Outer Join/ Right Join: Next, “right outer join” or simply “right join” which give all rows from RIGHT table and only matched rows from other table specified in the ON condition. Unmatched columns cell are replaced with NULL. Our result set would look like this:

--Using Right Outer Join
select P.PassengerId as PassengerId,P.PassengerName as PassengerName,
F.FlightName as FlightName,F.FlightCode as FlightCode
from tblPassengers P right outer join tblFlight F
on P.FlightId = F.FlightId
 
Right Outer Join by imagination hunt blogs
Right Outer Join


--Using Right Join
select P.PassengerId as PassengerId,P.PassengerName as PassengerName,
F.FlightName as FlightName,F.FlightCode as FlightCode
from tblPassengers P right join tblFlight F
on P.FlightId = F.FlightId
 
Right Join by imagination hunt blogs
Right Join


Full Outer Join: Next, “full outer join” gives all records from LEFT and RIGHT table specified in the ON condition. Unmatched columns cell are replaced with NULL. Our result set would look like this:

--Using Full Outer Join
select P.PassengerId as PassengerId,P.PassengerName as PassengerName,
F.FlightName as FlightName,F.FlightCode as FlightCode
from tblPassengers P full outer join tblFlight F
on P.FlightId = F.FlightId
 
Full Outer Join by imagination hunt blogs
Full Outer Join


Cross Join: At last, “cross join” gives the cross product of two table. Here, it is not reqiuired to specify the ON condition. Our result set would look like this:

--Using Cross Join
select P.PassengerId as PassengerId,P.PassengerName as PassengerName,
F.FlightName as FlightName,F.FlightCode as FlightCode
from tblPassengers P cross join tblFlight F
 
Cross Join by imagination hunt blogs
Cross Join


For any query, comment us below.

Previous – Types of SQL Server Joins #1


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Simple program


Simple program #14 (Source code to check input is a positive or negative or neutral number)

Enter a number: 64
Number is positive.

The program is written in C# programming language and will accept a number as input. The logic of the program is to check whether it is positive or negative or neutral (neither positive nor negative).
 
Simple program by imagination hunt blogs
Simple program

Note: If you are new to C# and Console Application. Try to code First C# Program

Note: Read articles on how to use Loops and Conditions.

Let’s find out a simple and easy way to code the program.

Practical Implementation:

using System;

namespace patternProblem.Simple_program    //checking input is a number (Positive or negative or neutral)
{
    class Simple14
    {
        public static void Main()
        {
            try
            {
                Console.Write("Enter a number: ");
                double num = Convert.ToDouble(Console.ReadLine());

                if (num > 0)
                {
                    Console.WriteLine("Number is positive.");
                }
                else if (num < 0)
                {
                    Console.WriteLine("Number is negative.");
                }
                else
                {
                    Console.WriteLine("Number 0 is neutral.");
                }
            }
            catch (FormatException)
            {
                Console.WriteLine("Input is not a numbers. It might be an alphabet or special character.");
            }
            catch (OverflowException)
            {
                Console.WriteLine("Input number is exceeding its maximum range.");
            }
            catch (Exception ex)
            {
                Console.WriteLine(ex.Message);
            }
        }
    }
}

Output:
The input number is 64. The program will check whether input number is positive/ negative/ neutral number. The output is shown below:

Enter a number: 64
Number is positive.
Press any key to continue . . .


For any query, comment us below.

Skip to Main Table Simple program

Previous - Simple program #13


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Wednesday, 26 July 2017

Simple program

Simple program #13 (Source code to calculate number of digits)

Enter a number: 456
Number of digits used in given number 456 is 3

The program is written in C# programming language and will accept a number as input. The logic of the program is to counter number of digits.
 
Simple program by imagination hunt blogs
Simple program

Note: If you are new to C# and Console Application. Try to code First C# Program

Note: Read articles on how to use Loops and Conditions.

Let’s find out a simple and easy way to code the program.

Practical Implementation:

using System;

namespace patternProblem.Simple_program
{
    class Simple13   //Calculating number of digits
    {
        static void Main(string[] args)
        {
            Console.Write("Enter a number: ");
            int n = Convert.ToInt32(Console.ReadLine());
            int count = 0;
            Console.WriteLine();

            //Method 1: Using string in-built function length
            Console.WriteLine("Output from Method1: Using string in-built function length");
            Console.WriteLine("Number of digits used in given number {0} is {1}", n, n.ToString().Length);
            Console.WriteLine();

            //Method 2: Using for loop
            int temp = n;
            for (; n > 0; )
            {
                n = n / 10;
                count++;
            }
            Console.WriteLine("Output from Method2: Using for loop");
            Console.WriteLine("Number of digits used in given number {0} is {1}", temp, count.ToString());
            Console.WriteLine();

            //Method 3: Using while loop
            count = 0;
            n = temp;
            while (n > 0)
            {
                n = n / 10;
                count++;
            }
            Console.WriteLine("Output from Method3: Using while loop");
            Console.WriteLine("Number of digits used in given number {0} is {1}", temp, count.ToString());
            Console.ReadKey();
        }
    }
}

Output:
The input number is 4256. The program will count number of digits used in input number. The output is shown below:

Enter a number: 4256

Output from Method1: Using string in-built function length
Number of digits used in given number 4256 is 4

Output from Method2: Using for loop
Number of digits used in given number 4256 is 4

Output from Method3: Using while loop
Number of digits used in given number 4256 is 4
Press any key to continue . . .


For any query, comment us below.

Skip to Main Table Simple program

Previous - Simple program #12

Next – Simple program #14

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