Re: echo date('Y-m-d', $mydata->timestamp);

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It is actually a generated timestamp in MySQL.
timestamp(14)
Now what? I was hoping to avoid:
|echo substr(|$mydata->timestamp|, 0, 8);

John

|Richard Lynch wrote:
On Sun, April 22, 2007 1:05 am, John Taylor-Johnston wrote:
$mydata->timestamp = "20070419162123";

echo date('Y-m-d', $mydata->timestamp);


result: 2038-01-18

?? What is wrong?? Should be 2007-04-19?

date() takes a Unix timestamp as its input.

Unix timestamps are measured as number of seconds from Jan 1, 1970,
midnight, GMT, the birth of Disco.
[that last was a joke...]

You are handing it a pre-formatted date-stamp in YYYYMMDDHHIISS format...

You could do something like:
$t = '20070419162123';
$year = substr($t, 0, 4);
$month = substr($t, 4, 2);
$day = substr($t, 6, 2);
$hour = substr($t, 8, 2);
$minutes = substr($t, 10, 2);
$seconds = substr($t, 12, 2);
echo date(mktime($month, $day, $year, $hour, $minutes, $seconds));

I suspect strtotime() *might* handle your input and give you a Unix
timestamp...

I also suspect whatever you needed a Unix timestamp for in the first
place could have been achieved easier before you got painted into this
corner...


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