Talk In Code

There are two ways to find out the number of days for a given month. The first is to use the date() function in conjunction with the mktime() function to create a date and format this value as the number of days in a given month.

$monthDays = date("t",mktime(0,0,0,12,1,2008));

The second way is to use the function cal_days_in_month(). This function takes three parameters.

• Canelday: There are a number of different forms of calendar to chose from. The one most English speaking people are interested in is the CAL_GREGORIAN calendar. Take a look at a full list of calendar constands.
• Month: An integer representing the month (1 – 12).
• Year: An integer representing the year.

So to find out the number of days in the month of December of 2008 you could use the following.

$monthDays = cal_days_in_month(CAL_GREGORIAN, 12, 2008);

Use the following function to work out how many years have passed since an event. This can be useful if you want to work out a persons age based on their birthday.

The function works by standardising the format of the date using the PHP strtotime() function. This is the first step of the function and sorts out if the date is valid or not. Once this has been done then the date is formatted into a standard form of yyyy-mm-dd, which is then split using the explode() function. The year of the inputted date is then subtracted from the current year, giving the age in years. A final check makes sure that the date hasn’t passed yet, and subtracts one from the years value to give a more accurate result. Here is the function:

function dateToAge($birthday){
 if (($cleaned = strtotime($birthday)) === false){
  return false; // Not a readable format
 };
 
 list($year, $month, $day) = explode('-', date('Y-m-d', $cleaned));
 $age = date('Y') - $year;
 
 if (mktime(0,0,0,$month,$day,date('Y')) < mktime()){
  // Birthday hasn't passed, so subtract one year from age
  $age--;
 };
 
 return $age;
}

Here is an example of the function in use:

echo '01/20/1980'.birthday('01/20/1980'); // prints 27

There is one issue with this function is you cannot be sure that the date will be in a standard format. Some cultures put the date as day/month and others put it as month/day. This can make the function return nothing, even though the date is correct. So the following:

echo '20-01-1979 '.birthday('20-01-1980');

Returns false.

The following function sorts this problem out by trying to clean up the date entered so that it can be recognised by the program if it initially isn’t a valid date.

function dateToAge($birthday){
 if (($cleaned = strtotime($birthday)) === false){
  if(strpos($birthday,'/')!==false){
   $birthday = explode('/',$birthday);
   if(count($birthday)==3){
    $birthday = $birthday[1].'/'.$birthday[0].'/'.$birthday[2];
   };
  };
  if(strpos($birthday,'-')!==false){
   $birthday = explode('-',$birthday);
   if(count($birthday)==3){
    $birthday = $birthday[1].'-'.$birthday[0].'-'.$birthday[2];
   };
  };
  if (($cleaned = strtotime($birthday)) === false){
   return false; // Not a readable format
  };
 };
 
 list($year, $month, $day) = explode('-', date('Y-m-d', $cleaned));
 $age = date('Y') - $year;
 
 if (mktime(0,0,0,$month,$day,date('Y')) < mktime()){
  // Birthday hasn't passed, so subtract one year from age
 $age--;
 };
 
 return $age;
}