Datetime in PHP Script

Question

I use a PHP script for a banlist that fetches the date and time, but I have an error and don\'t know how to convert it to \"normal\" time.

It lists me only the date

Answer 1

Based on this image (that you have given in deleted answer) you have milliseconds. Divide milliseconds with 1000 to get seconds.

Probably something like this:

$datetime = date('d.m.Y \u\m H:i \U\h\r', $row['expires'] / 1000);

But, if you get year 1970, that means that you are on 32bit machine, and your fetched INT is out of range. In this case you could just cut last 3 numbers with string function substr():

$datetime = date('d.m.Y \u\m H:i \U\h\r', substr($row['expires'], 0, -3));

Answer 2

Convert the date to UNIX timestamp first. Try with -

$datetime = date("d.m.Y \\u\\m H:i \\U\\h\\r", strtotime($row['expires']));
echo "<td>$datetime</td>";

Answer 3

It lists me only the date : 01.01.1970 um 00:00 Uhr

That simply means you are thinking of $row['expires'] incorrectly. That is not a UNIX Timestamp value and is producing an invalid date. It means the value essentially evaluates to 0, which is Jan 1st 1970 in UNIX time

date() requires you to send a valid Unix timestamp to it (INT 11), is that what you have in database for that field? or it is a date time field?

Try this

echo date("d.m.Y \\u\\m H:i \\U\\h\\r", "2014-10-12");   //invalid

echo date("d.m.Y \\u\\m H:i \\U\\h\\r", time());  //valid: current unix timestamp