How to pull only specific json object field value in php?

Question

I am working on a php code as shown below:

echo "<pre>";print_r($episode);echo "</pre>";

The above php code displays the following o/p:

stdClass Object
(
    [air_date] => 2019-04-15 07:15:00
    [air_duration] => 9900
    [schedule_id] => 4220986
    [program_id] => 23
)

Now on doing print_r($episode->air_date); it displays the following o/p:

2019-04-15 07:15:00

Problem Statement:

I am wondering what changes I need to make here print_r($episode->air_date); so that it displays only date, not the time.

Şafak Çıplak · Answer 1 · 2019-04-15T21:59:23.137

1

you can wrapper

echo date('Y-m-d', strtotime ($episode->air_date));
print_r(date('Y-m-d', strtotime ($episode->air_date)));

edited Apr 15 '19 at 21:59

answered Apr 15 '19 at 21:41

Şafak Çıplak

I believe that this one won't work – Oguzcan Apr 15 '19 at 21:55
sory for mistake this one i correct ```echo date('Y-m-d', strtotime('2019-04-15 07:15:00'));``` – Şafak Çıplak Apr 15 '19 at 21:56

score 1 · Answer 2 · answered Apr 15 '19 at 21:43

1

If you're just looking to print, you could just print the substring from the beginning to the position of the first space:

print(substr($episode->air_date, 0, strpos($episode->air_date, ' ')));

answered Apr 15 '19 at 21:43

Chris White

score 0 · Answer 3 · answered Apr 15 '19 at 21:58

0

Running the following code you get:

$date = "2019-04-15 07:15:00";
echo date('Y-m-d', strtotime($date));

output: 2019-04-15

This will return a false value:

$date = "2019-04-15 07:15:00";
echo date('Y-m-d', $date);

output: 1970-01-01

answered Apr 15 '19 at 21:58

Oguzcan

3 Answers3