CODEWITHSUNDEEP
phphtmljson
Deadpool
Deadpool

Reputation: 204

How do I put JSON into a PHP Variable?

I´m trying to put JSON-Information into a PHP variable and I´m getting Errors all the time.

This is my Code to make the JSON readable:

header('Content-Type: application/json');
$jsondecode1 = json_decode($json1);
print_r ($jsondecode1);

This is a snippet of the $jsondecode1:

Array
(
    [0] => stdClass Object
        (
            [flightId] => ******
            [******] => stdClass Object
                (
                    [******] => ******
                    [******] => ******
                    [******] => ******
                    [******] => ******
                    [******] => ******                
                )

Notice: When I echo the jsondecode1 it outputs this:

Array to string conversion in C:\xampp......\simpletest3.php on line 48

So I used print_r().

What I tried to put (for example) [flightId] into a PHP Variable:

$jsondecode1 = json_decode($json1);
$jsondecode2 = $jsondecode1->flightId;
print_r ($jsondecode2);

Output: Notice: Trying to get property 'flightId' of non-object in C:.........\simpletest3.php on line 48

I tried some other codes too, but the Outputs were very similar and I don´t want to make my Question longer than it needs to be. So, how do I put (for example) the [flightId] into a PHP variable.

Edit:

Solution:

$jsondecode1 = json_decode($json1);
$jsondecode2 = $jsondecode1[0]->flightId;
print_r ($jsondecode2);

Or:

foreach ($jsondecode1 as $data) {
    print_r($data->flightId);
}

Upvotes: 3

Views: 118

Answers (3)

Death-is-the-real-truth
Death-is-the-real-truth

Reputation: 72269

You have to do it like below (what you shown)

$jsondecode1 = json_decode($json1);
$jsondecode2 = $jsondecode1[0]->flightId;
echo $jsondecode2;

But you have multi-dimensional-array,so do like below:-

$jsondecode1 = json_decode($json1);

foreach ($jsondecode1 as $jsondecode) {
    echo $jsondecode->flightId; // here you can use echo $flightId = $jsondecode->flightId; 
}

Upvotes: 2

Robert
Robert

Reputation: 5973

Your json contains an array of objects, not 1 object.

$jsondecode1 = json_decode($json1);

// loop through all objects in the array
foreach ($jsondecode1 as $data) {
    print_r($data->flightId);
}

Or if you only want to have the first object of the array:

$jsondecode1 = json_decode($json1);
$jsondecode2 = $jsondecode1[0]->flightId;
print_r ($jsondecode2);

But then the question is why is the json an array and not a single object?

Also you can not simply echo or print out non-scalar types (like arrays or objects), for that you will have to use print_r.

Upvotes: 2

B. Desai
B. Desai

Reputation: 16436

Your flightId is inside one array whoch index is 0. So you have to first access that array then you can get flightId

try below code:

$jsondecode1 = json_decode($json1);
$jsondecode2 = $jsondecode1[0]->flightId;
print_r ($jsondecode2);

Upvotes: 2

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