Finding the difference between 2 dates and iterating an image based on the difference - php

Question

Ive managed to loop through a table and get the difference in days between 2 dates adjacent to each other in the table.
Multiple entries have the same date, i have it now that when that date changes, it displays an image however i want it to display the image as many times as the difference in date
$sql = mysql_query("SELECT * FROM Films_Info")
or die(mysql_error());
$last_value = null;
while ($row = mysql_fetch_assoc($sql)) {
if (!is_null($last_value)) {
$a = new DateTime($row['FilmRelease']);
echo "<p>".$row['FilmName']."</p>";
$interval = $a->diff(new DateTime($last_value));
//echo $interval->format('%d days');
$i = 0;
}
$howManydays = $interval->days;
for ( $i; $howManydays; $i++) {
echo "<img src=\"day.jpg\" />";
$howManydays = 0;
}
$last_value = $row['FilmRelease'];
}

Answer 1

for ( $i = 0; $i < $howManydays; $i++) The second is a conditional statement telling when the loop should stop.
The first section in the for loop where it says $i = 0 initialized the variable $i to 0 and then tests the condition $i < $howManydays.
Let's say $howManydays equals 1. That means 0 < 1, so the loop will perform.
At the end of the loop, the third section is called ($i++), so $i is incremented and now equals 1. The second section is called again to test conditions $i < $howManydays which is asking if 1<1 which it's not, so the loop will exit.
So if $howManydays is greater than 0, the loop should happen the integer amount that is in $howManydays.
You will want to remove $howManydays = 0; within the for loop, if you don't want it to only fire once.
The for loop
for ( $i = 0; $i < $howManydays; $i++){
// ...
}
is somewhat equivalent to the while loop:
$i = 0;
while ( $i < $howManydays ){
// ...
$i++;
}
http://php.net/manual/en/control-structures.for.php for more information
In your code above, you should also check if interval was set. You should probably just do an if rather than a while, if only need one interval.

Answer 2

you could just do a simple division on the interval and print out the image that many times
$last_value_u = $last_value->format('U');
$a_u = $a->format('U');
$interval = $a_u - $last_value_u;
$image_count = intval($interval/86400);
for($i=0;$i<$image_count;$i++)
echo "<p><img src=\"day.jpg\" /></p>";
Update
An alternative option would be to loop through the interval:
for($i=$last_value_u;$i<$a_u;)
{
if(intval(($a_u - $i)/86400) == X)
{
// run special code for specific day
}
else
{
// run code for every other day
}
$i+=86400;
}