Using JSON returned via PHP in Javascript

Solution:

You can create a new array of the data you want to return to the JS. For example, inside your function you could change your code to look like this

$Json = json_decode($JsonContents);
$output = [];

for ($x = 1; $x <= 10; $x++) {
    $hint = [];
    $hint['brand'] = $Json->hints[$x]->{'food'}->{'brand'};
    $hint['label'] = $Json->hints[$x]->{'food'}->{'label'};
    $output[] = $hint;
}

return $output;
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You can then return this data like so

echo json_encode(getFoodFromAPI($ingr));
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Your JS would look like this:

foodSearched.forEach(function(hint) {
    console.log('Brand is ' + hint.brand + ' and label is ' + hint.label);
});
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You are close to achieving your goal, you just need to return only the 'broken down' data of the API request to the JS HTTP request. At the moment, you are doing the correct PHP logic of breaking down only the data you want to return to the JS, but you are using$JsonRec = json_encode($Json); which is sending back all of the API request rather than your PHP logic.

Hope this helps.

Here is a working example for you to just test

# Eggs
$response = json_decode(file_get_contents('https://api.edamam.com/api/food-database/parser?ingr=egg&app_id=47971bbd&app_key=15ddd5fb6a552c59762e2dc7093d5f85'));

$output = [];

for ($x = 1; $x <= 10; $x++) {
    $hint = [];
    $hint['brand'] = $response->hints[$x]->{'food'}->{'brand'};
    $hint['label'] = $response->hints[$x]->{'food'}->{'label'};
    $output[] = $hint;
}

header('Content-Type: application/json');
var_dump(json_encode($response)); # Here is what your JS would see
die();
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