PHP: Warning: json_decode() expects parameter 1 to be string, object given in

Question

I am trying for get one field value from json response but I am unable to get it. My PHP code is like below

$servers    = $sp->server_list();
print_r ($servers);

Its giving me result like below

stdClass Object ( [data] => Array ( [0] => stdClass Object ( [id] => S9rr4Un0SYlPGR6E [name] => localhost [autoupdates] => 1 [firewall] => [deny_unknown_domains] => [lastconn] => 1544645254 [lastaddress] => 50.116.20.23 [datecreated] => 1511884238 [plan] => grandfathered_coach [available_runtimes] => Array ( [0] => php5.4 [1] => php5.5 [2] => php5.6 [3] => php7.0 [4] => php7.1 [5] => php7.2 [6] => php7.3 ) ) ) )

I want get value called id from it and want use it in my php. I have tried json decode method like below

$decoded_data = json_decode($servers,TRUE); 
$myid = $decoded_data['data']['id'];

its giving me error like below

Warning: json_decode() expects parameter 1 to be string, object given in C:\xampp\htdocs\code\new2.php on line 11

I have searched many questions and answer here but I am unable to get it resolved. Let me know if someone can help me. Thanks a lot.

There may be a JSON response at some point, but if so, something in the server_list() method has already decoded it for you. — Don't Panic
– Don't Panic, Commented Dec 12, 2018 at 20:36
@Don'tPanic I have got solved using answer but I am getting error called Fatal error: Uncaught Error: Cannot use object of type stdClass as array in when I use it with $systemUser = $sp->sysuser_create($server_id, $system_user, $system_user_password); Let me know if you can help me for it — Raju Bhatt
– Raju Bhatt, Commented Dec 12, 2018 at 21:20

Mohammad C · Accepted Answer · 2018-12-12 21:01:13Z

1

$servers is an object with key data which is an array. You can access the first element in the array using index 0 then you can get the id.

Json decode is not needed as $servers is already an object. You use json decode to go from json string to object. You can learn more about PHP JSON decode

$servers = $sp->server_list();
echo $myId = $servers->data[0]->id;

edited Dec 12, 2018 at 21:01

answered Dec 12, 2018 at 20:19

Mohammad C

1,3411 gold badge8 silver badges12 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

Raju Bhatt Over a year ago

its giving me error like Warning: json_decode() expects parameter 1 to be string, object given in C:\xampp\htdocs\code\new2.php on line 11

Raju Bhatt Over a year ago

this error in $decoded_data = json_decode($servers,TRUE); this line

Mohammad C Over a year ago

You dont need to use json decode. As servers in not a json string. It is an object.

Mohammad C Over a year ago

@AbraCadaver what if its array casted. Do u know if the nested object will need to be array casted separately. Either way i have removed it as an alternative.

AbraCadaver Over a year ago

You would have to cast both (array)$servers; and (array)$servers['data'][0]->id;

|

Collectives™ on Stack Overflow

PHP: Warning: json_decode() expects parameter 1 to be string, object given in

1 Answer 1

8 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

8 Comments

Your Answer

Sign up or log in

Post as a guest

Related