Name of array in result of json_decode()

Question

If I do a json_decode like this:

$json = json_decode($jsondata);
echo var_dump($json);

I get something like this:

object(stdClass)#1 (1) 
{ ["QRY_JISGesch"]=> array(3) 
    { [0]=> object(stdClass)#2 (8) 
        { ["JISGeschID"]=> int(7) ["StandorteID"]=> int(0) ["FSKID"]=> int(23) 
        } 
    [1]=> object(stdClass)#3 (8) 
        { ["JISGeschID"]=> int(8) ["StandorteID"]=> int(0) ["FSKID"]=> int(22) 
        } 
    [2]=> object(stdClass)#4 (8) 
        { ["JISGeschID"]=> int(9) ["StandorteID"]=> int(0) ["FSKID"]=> int(1)
        } 
    } 
}

How do I find out "QRY_JISGesch" in code?

use "find" from your editor. – tanaydin Apr 25 '18 at 11:20 — tanaydin, Apr 25 '18 at 11:20

score 1 · Accepted Answer · answered Apr 25 '18 at 11:38

1

You could use reset() to get the first member of an object (or an array).

$json = json_decode($jsondata);
$first = reset($json);

If you only want the get the first "key", you can use key();

$json = json_decode($jsondata);
$key = key($json); // QRY_JISGesch

answered Apr 25 '18 at 11:38

Syscall

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Thanks, that was the answer I was looking for. key($json) did the trick. – klammerauf Apr 25 '18 at 11:44

score 0 · Answer 2 · answered Apr 25 '18 at 11:26

0

Try this. It will help you :-

$json = json_decode($jsondata, true);
// true means objects will be converted into associative arrays

and you can access QRY_JISGesch like this way :- $json['QRY_JISGesch']

answered Apr 25 '18 at 11:26

Yash Parekh

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Don't think that this solves the problem - if `QRY_JISGesch` is unknown, it is unknown – Nico Haase Apr 25 '18 at 11:28
I'm aware of that second param in json_decode, but this doesn't help me here. And I also know that I can access the var with $json['QRY_JISGesch']. But I need to do a function that can handle different values, other than QRY_JISGesch. – klammerauf Apr 25 '18 at 11:29
@klammerauf You want to check that `QRY_JISGesch` exists or not. correct?? – Yash Parekh Apr 25 '18 at 11:30

score 0 · Answer 3 · answered Apr 25 '18 at 11:30

0

Is there only one item given in the array, and you are not interested in the name itself, but the values? Then array_values could help.

Otherwise, you could use array_keys to read all keys and iterate over them? Or use a foreach loop which does not care about the array key neither.

answered Apr 25 '18 at 11:30

Nico Haase

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array_values($json) gives a warning "array_values() expects parameter 1 to be array, object given". array_values($json[0]) gives a fatal error "Cannot use object of type stdClass as array" – klammerauf Apr 25 '18 at 11:38
array_keys($json) gives the same warning as array_values. – klammerauf Apr 25 '18 at 11:41
Ah, yeah - this could have worked with `$json = json_decode($jsondata, true);` – Nico Haase Apr 25 '18 at 11:44

score 0 · Answer 4 · answered Apr 25 '18 at 11:33

0

get_object_vars (object $object) will give you a list of all accessible properties of the returned object.

answered Apr 25 '18 at 11:33

Cuagau

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get_object_vars($json) result is "Array" – klammerauf Apr 25 '18 at 11:42

Name of array in result of json_decode()

4 Answers4