Find Dates Before Specified Date and Insert Them Using PHP & SQL - php

Question

I have a date outputted from the database:
$deadline = $row['DEADLINE'];
When I print $deadline it returns: 2015-05-03 18:00:00
Now I want a way to find each day 24 hours before up until todays date and then use each of those values to insert in a new table in the database.
So in this case I want:
2015-05-03 18:00:00
2015-05-02 18:00:00
2015-05-01 18:00:00
2015-04-30 18:00:00
2015-04-29 18:00:00
To be inserted into the database 5 new rows. So I know I need todays date:
$now = date("Y-m-d H:i:s");
Then I have the date of the deadline:
$deadline = $row['DEADLINE'];
I can then find the difference in days:
$dateDiff = ($deadline - $now)/(24*60*60);
This gives me a figure (in this case it will be 5). Then could I use this figure to display the date and time for 5 days prior?
Not sure how I would insert them all though. Your help would be appreciated.

Answer 1

Use DateTime object, really easiest way :
PHP :
$now = new datetime();
$sub = new datetime();
for($i=0;$i<5;$i++)
{
$sub->sub(new DateInterval('P1D'));
echo $sub->format('Y-m-d H:i:s').'<br />';
echo $sub->diff($now)->format('%a days').'<br />';
}
OUTPUT :
2015-04-29 14:18:09
1 days
2015-04-28 14:18:09
2 days
2015-04-27 14:18:09
3 days
2015-04-26 14:18:09
4 days
2015-04-25 14:18:09
5 days

Answer 2

This isent perfect, but it will probably get you started
Try Using Datetime http://php.net/manual/en/class.datetime.php
Or Carbon/Carbon Package: https://github.com/briannesbitt/Carbon
$datetime = new DateTime("2015-04-03 18:00:00");
$now = new DateTime('NOW');
$difference = $datetime->diff($now);
echo '<pre>';
for($i = 1; $i <= $difference->days; $i++)
{
if($difference->invert){
var_dump (
$temp = $datetime
->modify('-1 day')
->format('Y-m-d H:i:s')
);
}else{
var_dump (
$temp = $datetime
->modify('+1 day')
->format('Y-m-d H:i:s')
);
}
}