PHP JSON_DECODE ARRAY

Question

I am using following code to decode json array

$json_contacts ='[{"addr_name":"1","addr_phone":"010-1111-1111"},{"addr_name":"2","addr_phone":"010-1111-1112"},{"addr_name":"3","addr_phone":"010-1111-1113"},{"addr_name":"4","addr_phone":"010-1111-1114"}]';

$contact = json_decode($json_contacts, true);
echo $json_contacts;
echo "\n";
echo $contact;
echo "\nTHE END";

I am getting following result

[{"addr_name":"1","addr_phone":"010-1111-1111"},{"addr_name":"2","addr_phone":"010-1111-1112"},{"addr_name":"3","addr_phone":"010-1111-1113"},{"addr_name":"4","addr_phone":"010-1111-1114"}]

THE END

Why is my JSON_DECODE not decoding the array properly?

turn on error reporting on php - php.net/manual/en/function.error-reporting.php — Krish R
– Krish R, Commented Aug 8, 2015 at 2:11
I have checked your code and shows warning in line "echo $contact;" , changed into "print_r($contact);" and it prints all array values. — Krish R
– Krish R, Commented Aug 8, 2015 at 5:02

Josua M C · Accepted Answer · 2015-08-08 02:09:21Z

1

Try this:

<?php
$json_contacts ='[{"addr_name":"1","addr_phone":"010-1111-1111"},{"addr_name":"2","addr_phone":"010-1111-1112"},{"addr_name":"3","addr_phone":"010-1111-1113"},{"addr_name":"4","addr_phone":"010-1111-1114"}]';

$contact = json_decode($json_contacts, true);
echo $json_contacts;
echo "\n";
echo '<pre>';
print_r( $contact);

echo '</pre>';
echo "\nTHE END";

Because you echo your previous $json_contacts and you didn't turn on PHP notice. You cannot echo $contact because it's an array.

edited Aug 8, 2015 at 2:09

answered Aug 8, 2015 at 2:03

Josua M C

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2 Comments

Ham Dong Kyun Over a year ago

I am not that much of newbie... Ofcourse the script is written with in <?php ?>. Or else it will not have shown the result as I mentioned above.

Josua M C Over a year ago

@HamDongKyun i know you're not newbie. please just copy all of that. and do print_r($contact); not echo $contact.

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