PHP: Parsing postgre's default string time-stamp format

Question

I have problem to parse the following two time-stamp strings in a generic way?

$timestamp1 = "2013-10-15 19:05:18.756932+03";
$timestamp2 = "2013-10-15 19:05:18.756932+03:00";

I have this so far but it returns with false

$datetime = DateTime::createFromFormat('Y-m-d HH:MM:SS frac "GMT"? [+-] hh ":"? MM?', $timestamp1);
$datetime = DateTime::createFromFormat('Y-m-d HH:MM:SS frac "GMT"? [+-] hh ":"? MM?', $timestamp2);

What's wrong? Thank you.

P.S. I tried this 'Y-m-d HH:MM:SSfrac"GMT"?[+-]hh":"?MM?', this ''Y-m-d HH:MM:SS[0-9]+"GMT"?[+-]hh":"?MM?' and this 'Y-m-d HH:MM:SS [0-9] "GMT"? [+-] hh ":"? MM?' but it always fails.

@Amal Murali 19:05:18, .756932 is fractional part for a second and +03 is timezone offset. — 0xC0DEGURU
– 0xC0DEGURU, Commented Nov 21, 2013 at 12:21

Vatev · Accepted Answer · 2013-11-21 12:21:00Z

6

Assuming the last part is a timezone offset, Y-m-d H:i:s.ue should work.

If this is posgre's default format you will probably need to handle the case when there are no fractional seconds (2013-10-15 19:05:18+03) with Y-m-d H:i:se.

answered Nov 21, 2013 at 12:21

Vatev

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pogush · Accepted Answer · 2013-11-21 12:25:55Z

2

You have an error in your format string.

$datetime = DateTime::createFromFormat('Y-m-d H:i:s.uP', $timestamp1);

Here is the documentation for the format string options: PHP: DateTime::createFromFormat Manual

answered Nov 21, 2013 at 12:25

pogush

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