delete hours, minutes and secconds in age Unix PHP? - php

Question

I work on one project with PHP and Mysql and I need to calculate the ages but I need delete the hours, minutes and seconds part. Here is an example:
The date of birth 2018-03-06 17:35:00 but
I need 2018-03-06 00:00:00
Here is my code:
function FetchAgeCaduce($MaxDias=3){
$fechaInicial = time();//date("Y-m-d");
$fecha = date("d/m/Y",$FechaFinal);
$fechaInicial = strtotime("d/m/Y",$fecha);
for ($i=0; $i<$MaxDias; $i++)
{
$Segundos = $Segundos + 86400;
$caduca = time()+$Segundos;//date("D",time()+$Segundos);
$var = date("D",$caduca);
if ($var == "Sat")
{
$i--;
}
else if ($var == "Sun")
{
$i--;
}
else
{
$FechaFinal = time()+$Segundos;
}
}
return $FechaFinal;
}
How can I work on it to get the expected result?

Answer 1

The right way / The MySQL way:
Store your data as DATE in MySQL, not DATETIME. DATE itself is described here. It does not store time, so you'll always have 00:00:00 as default.
The less right way / The PHP way:
Jay, Zeus, Freaking, K. Reist. Learn to use DateTime already, stop using date! It's like every second question here and noone learns DateTime!
Use it!
$dt = new DateTime();
echo $dt->format("Y-m-d 00:00:00");
The output:
2018-03-06 00:00:00
P.S. $FechaFinal on line 5 is not defined.

Answer 2

You can format the date on either the application or database layer.
Using MySQL
You can use MySQL DATE_FORMAT Function
to parse the date into the wanted format.
SELECT DATE_FORMAT(NOW(),'%Y-%m-%d 00:00:00') dob;
See DEMO on SQL Fiddle.
Using PHP
Say you don't want to format the date in the database layer, may be because you would need to use the time part in some other calculations. Use the date functions as illustrated below:
$dt = new DateTime();
echo $dt->format("Y-m-d 00:00:00");
//getting the expected date from the formatted string
$formateDate = DateTime::createFromFormat('Y-m-d H:i:s', $dt->format("Y-m-d 00:00:00"));