Changing date format, php echo from postgresql always return "Date(1970, 01, 01)"

Question

    <?php


header('Content-Type: application/json; charset=utf-8');
$db = pg_connect("host=localhost port=5432 dbname=postgres user=postgres password=blabla");

$result = pg_query($db,"SELECT kv.ph, kv.date FROM public.kv ORDER BY date DESC LIMIT 5");

$return = array();
while ($row = pg_fetch_array($result)) {

$test = $row[1];
$date = DateTime::createFromFormat('y-m-d', $test);
$formatdate = date('Y, m, d',$date);
    $return[] = array((int)$row[0],"Date($formatdate)");

} 

pg_close($db); 


echo $data=json_encode($return);
?>

I need echo like:
[[6, "Date(2017, 10, 5)"], [9, "Date(2017, 9, 30)"], ...]
I get:
[[6,"Date(1970, 01, 01)"],[9,"Date(1970, 01, 01)"],...]

I am little stuck here, please help. Data type in postgres is date. 2017-10-28

Artem Ilchenko · Accepted Answer · 2017-10-31 13:03:13Z

1

Error with second parameter in date() function. That must be int but object DateTime given.

Use date_format() instead of date() in $formatdate = date('Y, m, d',$date); row.

So your code can be:

$formatdate = date_format($date, 'Y, m, d');

or

$formatdate = $date->format('Y, m, d');

answered Oct 31, 2017 at 13:03

Artem Ilchenko

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1 Comment

ejovrh2 Over a year ago

$formatdate = date_format($date, 'Y, m, d'); works. thanks

splash58 · Accepted Answer · 2017-10-31 13:04:23Z

1

1st, DateTime::createFromFormat returns Datetime objext

2nd, i think, from db you recieve year with 4 digits, so you need use format Y

$date = DateTime::createFromFormat('Y-m-d', '2017-10-05');
echo $date->format('Y, m, d'); .. 2017, 10, 05

answered Oct 31, 2017 at 13:04

splash58

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