Strip parts of one field

[Anon]

Hi,

I have a field in the database called pictime, which looks like this:

20010506223300

The first digits are the year (2001) then the month (05) the day (06) and then the hour (22) minute (33) and seconds (00)

What I want is to separate that date into $year $month and $day

It is stored in a varchar.

How can I do this???

[Anon]

Two ways to solve this problem come to mind.

The first is to just treat this as an array. It's a dirty solution, but it will work. All strings are really array's of characters. So if

$stamp = '20010506223300';

then you could do:
$year = $stamp[0] . $stamp[1] . $stamp[2] . $stamp[3];
$month = $stamp[4] . $stamp[5];
$day = $stamp[6] . $stamp[7];
$hour = $stamp[8] . $stamp[9];
$minute = $stamp[10] . $stamp[11];
$second = $stamp[12] . $stamp[13];

This is going to slow things down though... and is a really poor solution. My other recomendation works faster, and provides a bit of error checking. It uses a regular expression:

if (ereg("([0-9]{4})([0-9]{2})([0-9]{2})([0-9]{2})([0-9]{2})([0-9]{2})([0-9]{2})",$stamp,$out)) {
$year = $out[1];
$month = $out[2];
$day = $out[3];
$hour = $out[4];
$minute = $out[5];
$second = $out[6];
}

Be careful with this one, because the first array element (in index 0) is the full $stamp string itself.

Hope this helps,

Curts

[mithril]

Hey Harm,

You could change the field datatype from varchar to datetime. In a datatime datatype, the example you listed would look like this:

2001-05-06 22:33:00

To get only the year, month, date:

<?php

$data = "2001-05-06 22:33:00";
$data_split = explode(" ", $data);
$date_split = explode("-", $year[0]);

$year = $date[0];
$month = $date[1];
$day = $date[2];

?>

Hope this helps!!!

Cheers,

Geoff A. Virgo