php strtotime function (2)

Question

I started to learn PHP and have to find a mistake (maybe in this code):

if($newvalues["year"] != null)
   $newvalues["year"] = date("Y-m-d", strtotime($newvalues["year"]."-01-01"));

The new date has to be saved in the array "$newvalues", but when I press the save button, it doesn't save anything. Only if the textfield "year" is empty, the other items can be saved.

Can anyone help me, please? Thanks.

Have you tried adding debugging code to that, so you can see all the values that your code is working with, and that the right part of the if statement is being triggered? — andrewsi
– andrewsi, Commented Jun 25, 2012 at 16:53

Madara's Ghost · Accepted Answer · 2012-06-25 16:54:01Z

2

You're basically doing:

100 * 80 / 80

Just save it as

if($newvalues["year"] != null)
   $newvalues["year"] .= "-01-01";

Or better yet, represent it as a DateTime object:

$newvalues = array("year" => 2012);
if ($newvalues["year"] != null) {
    $newvalues["year"] = new DateTime("{$newvalues["year"]}-01-01");
}
var_dump($newvalues["year"]);

Using a DateTime object (And the DateTime family) gives you much better and more flexible control over your date/times.

answered Jun 25, 2012 at 16:54

Madara's Ghost

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