How do I put JSON into a PHP Variable?

Question

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);
}

Death-is-the-real-truth · Accepted Answer · 2017-12-27 11:22:08Z

2

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; 
}

edited Dec 27, 2017 at 11:22

answered Dec 27, 2017 at 11:07

Death-is-the-real-truth

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2 Comments

Death-is-the-real-truth Over a year ago

@duffbeerisgood also when you used json_decode() it converts the json string into object array

Death-is-the-real-truth Over a year ago

@duffbeerisgood for the current scenario, yes, as you want to iterate over json data and try to get the values form it.

B. Desai · Accepted Answer · 2017-12-27 11:08:07Z

2

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);

answered Dec 27, 2017 at 11:08

B. Desai

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Comments

Robert · Accepted Answer · 2017-12-27 11:17:39Z

2

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.

edited Dec 27, 2017 at 11:17

answered Dec 27, 2017 at 11:07

Robert

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3 Comments

Deadpool Over a year ago

Great Solution. Thank you !

Deadpool Over a year ago

I have no Idea why the json is an array, I thought you guys could tell me that.

Robert Over a year ago

@duffbeerisgood That is your input, so I can not tell you why it is like that :) $json1 is a string containing something like [{"flightId": "******", ...}]. That is an array of objects. If it were to have {"flightId": "******", ...}, without the [ and ] around it, it's a single json object.

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