Find Grade From Student's Graduation Date

PHP
1

Can somebody teach me how to write or use a function that will find a student’s current grade based on their year of graduation?

For example I know that a student will graduate in 2012. How can I get my code to print out “11th Grade?”

(the year of graduation is entered into the database, but the current grade is not)

Thanks.

2

This isn’t too great, but should give you an idea:

function currentGrade($graduationDate)
	{
		$currentGradYear = "2011";
		
		$currentGrade = $graduationDate-$currentGradYear;
		$currentGrade = 12-$currentGrade;
		
		return $currentGrade." grade.";
	}

Basically even though it’s 2010, the years graduation is 2011, so I set a variable to 2011 (Could be done without inputting 2011).

Take the example of 2010, it takes 2012-the current year, leaving us with 1 year left to go, subtract 1 from 12 grades in total, returning 11.

3

Thanks a ton.

That is a good start and gives me something to play with.

The only question now is how do I figure out what the current graduation year would be because that value won’t be input anywhere in the the database I’m working with.

4

I don’t understand, if you’re running this script in your program, then everytime it’s called it’s using the variable set within the function, so you should have complete control over what $currentGradYear is equal to.

Anywho, a little modification (has comments to understand):


## -- Takes the graduation year and returns what
## -- grade the student is in currently.
## -- Based on a 12 grade system.
function currentGrade($graduationDate)
	{
		## -- Sets current month equal to the number of the month
		$currentMonth = date("n");
		
		## -- Sets current graduation year to the current year,
			## -- HOWEVER: If it is after june (6), it adds one to the year.
		$currentGradYear = ($currentMonth>6) ? date("Y")+1 : date("Y");
		
		## -- Formulates their current grade is the currentGradYear subtracted
			## -- from their graduation date.
                        ## -- Then subtracts that from 12 since there are only 12 grades
		$currentGrade = $graduationDate-$currentGradYear;
		$currentGrade = 12-$currentGrade;
		
		## -- Returns their current grade concatenated with " grade."
		return $currentGrade." grade.";
	}
5

Perfect.

The addition of finding the current month and adding one to the year was the missing piece.

Thanks. I really appreciate it.