PHP show date2 if date1 passed
886
I need help on this..
I am new in php and dummy..
$displaydate = "";
$paydate1 = "23-10-2016";
$paydate2 = "23-11-2016";
$paydate3 = "23-12-2016";
$paydate4 = "23-01-2017";
$paydate5 = "23-02-2017";
$paydate6 = "23-03-2017";
if todaydate is 23-10-2016 then $displaydate = "$paydate2" until 22-11-2016. When 23-11-2016 then $displaydate = "$paydate3" and so on.
then results
if the date is 23-10-2016 until 22-11-2016 $displaydate = "$paydate2"
if the date is 23-11-2016 until 22-12-2016 $displaydate = "$paydate3"
if the date is 23-12-2016 until 22-01-2016 $displaydate = "$paydate4"
if the date is 23-01-2016 until 22-02-2016 $displaydate = "$paydate5"
if the date is 23-02-2016 until 22-03-2016 $displaydate = "$paydate6"
please help with the code...
Thanks..
IM php DUMMY
Answer
Solution:
Organize your data convenient so you can easy access it. Put your paydates it an array:
UPDATED per your comment:
Transform the dates in an easy to sort format:
Sort the array to be sure we will get the dates in ascending order while looping.
Now loop through the array to find the first date higher then today: