php date not converting properly

Question

I have this date in php: 31/01/2013

I'm trying to convert it using the strtotime function like so

date("Y-m-d", strtotime(31/01/2013));

but it keeps displaying as 1970-01-01. Any know why this is?

31 divided by 01 divided by 2013 equals 0.0153999006458023, which is cast to integer 0; which is the unix timestamp for 1st January 1970 at 00:00 GMT — Mark Baker
– Mark Baker, Commented Aug 1, 2013 at 8:57

DevZer0 · Accepted Answer · 2013-08-01 08:46:50Z

5

you should include it inside a string, not a continuous series of dividing numbers

date("Y-m-d", strtotime("31/01/2013"));

answered Aug 1, 2013 at 8:46

DevZer0

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Comments

Companjo · Accepted Answer · 2013-08-01 09:13:21Z

3

This will work

$date = str_replace("/", "-", "31/01/2013");
echo date("Y-m-d", strtotime($date));

answered Aug 1, 2013 at 9:13

Companjo

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웃웃웃웃웃 · Accepted Answer · 2013-08-01 08:50:22Z

1

Try this

$date = "31/01/2013";
$date = date("Y-m-d", strtotime($date));

Hope it will help

edited Aug 1, 2013 at 8:50

웃웃웃웃웃

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answered Aug 1, 2013 at 8:48

Sonu Sindhu

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Comments

Dev · Accepted Answer · 2013-08-01 09:19:11Z

0

Try this

$date = "01/08/2013";

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

answered Aug 1, 2013 at 9:19

Dev

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