If I have a number like -7 stored in a variable how can I modify a date in php by this amount. Here's what I have so far;
<?php
$the_date = '16 Mar 2018';
$test_sum = '-7';
$date = $the_date;
$date = strtotime($date);
$date = strtotime($test_sum, $date);
echo date('d M Y', $date);
?>
But this just output's the initial date value.
strtotime accepts "-7 day" as argument. I think this may work:
$the_date = '16 Mar 2018';
$test_sum = '-7';
$date = strtotime($the_date);
$date = strtotime("$test_sum day", $date);
echo date('d M Y', $date);
Running:
date('d M Y', strtotime('16 Mar 2018 -7 days'))
would produce:
09 Mar 2018
If you want to subtract days, you can simply use '-7 days'. If you don't specify the unit strtotime() probably doesn't subtracts days.
$the_date = '16 Mar 2018';
$test_sum = -7;
$date = strtotime(sprintf("%d days", $test_sum), strtotime($the_date));
echo date('d M Y', $date);
Please try with the following code :
$the_date = '16-Mar-2018';
$test_sum = '-7';
echo date('Y-m-d', strtotime($the_date. $test_sum.' days'));
What you need to use is strtotime() what this function does is it gets the string containing an English date format and try to convert it into a Unix timestamp formate which is the number of seconds since 1 of January 1970 00:00:00.
This code should work for you:
date('d M Y', strtotime($the_date. $test_sum . ' days'));
This makes this 16 Mar 2018 to this 09 Mar 2018
Hope that this helps.
More about strtotime
strtotime