PHP Code: Get the difference between two dates

The program consist of two files. The first file can be named according to what you like as long as you will have an extension of .php and the other file is named display.php.

The first file's task is to get the date (two dates) from the user. It has textboxes and a submit button. After clicking the submit button, display.php will be called and PHP compiler will parse the file taking only the "php part". PHP is just similar to C++ in terms of syntax. Method post is being used here for the first file. Here are the codes of the two files:

FILE 1:

< html >
< head >
head
body
< form action="display.php" method="post" >
//Starting Date
//Month:
//Day:
//Year:




Ending Date




//Month:
//Day:
//Year:




//




*Formating: Month and Day should be in numbers without
leading zero.
< /body >
< /html >

FILE 2:

< html >
< body >

$month1=$_POST[mo1];
$day1=$_POST[da1];
$year1=$_POST[ye1];

$month2=$_POST[mo2];
$day2=$_POST[da2];
$year2=$_POST[ye2];

$numdays = 0;
$initial = $month1;
$final = 12;

for ($count = $year1; $count <= $year2; $count++)
{
if ($count == $year2)
$final = $month2 - 1;
else
$final = 12;
for ($x = $initial; $x <= $final; $x++)
{
switch ($x)
{
case 1: $daycount = 31;
break;
case 2: if ($year1%4 == 0)
$daycount = 28;
else
$daycount = 29;
break;
case 3: $daycount = 31;
break;
case 4: $daycount = 30;
break;
case 5: $daycount = 31;
break;
case 6: $daycount = 30;
break;
case 7: $daycount = 31;
break;
case 8: $daycount = 31;
break;
case 9: $daycount = 30;
break;
case 10: $daycount = 31;
break;
case 11: $daycount = 30;
break;
case 12: $daycount = 31;
break;
default: echo "No such month";
}
$numdays = $numdays + $daycount;
}
$initial = 1;
//echo $numdays;
//echo " ";
}

$realnumberofdays = $numdays - $day1 + $day2 + 1;

echo "There is/are ";
echo $realnumberofdays;
echo " number of day(s) in a given date"

?>
< /body >
< /html >

In file 2 (display.php), the logic is quite simple. There are two loops here. I used for loop since for me its more readable.

The inner loop dictates the month where you will get your number of days. The outer loop dictates the year, from the initial year up to the final year.

You should also take careful observation to the inputted initial and final dates (month and year to be specific)

In the program, I first assume the initial month will carry its first day. Say, instead of Feb. 3 as the input, I initialized it to Feb. 1 but of course taking into consideration the 3 days.

In the final date, I also take into consideration the month and the date. In the FOR loop, if you notice there is a $final = $month2 - 1 when the year is the final year. Why is this so? For example if your final date is December 23, we will not include the december in the loop but taking into account the 23 days.

So if we will have the dates like Feb 3, 1999 - Dec. 23, 2000, the loop will considere Feb 1, 1999 - Nov. 30, 2000 then add the 3 days in february plus the 23 days in december. Of course we will have to take into consideration the leap year as you can see in the program.

In the last formula, I have plus 1 since if we have Jan 27, 1976 - Jan 27, 1976 (the same date) it is considered one day.

If you like to consider it zero (0) day, just take "plus 1" out.

Lastly, this program is not equipped with all the validations since I know you are an intelligent user. :)

NB: This is posted as a claim for originality.