PHP : Sending data using fetch()

Question

i'm trying to send some data using fetch() but in return i'm getting

SyntaxError: Unexpected token , in JSON at position 23

here's what i'm doing

fetch('/api.php', {
        method: 'POST',
        body: JSON.stringify({
            nom : "Issa",
            prenom: "Oule"}),
        headers : {"Content-Type" : "application/json"},
    })
.then(res => res.json())

.then((res) => {

    console.log(res);
})
.catch((err) =>{

    console.log(err);
})

in my php file :

    <?php
    header('Access-Control-Allow-Origin: *');
    header('Content-Type: application/json');
    header('Access-Control-Allow-Methods: GET, POST, PATCH, PUT, DELETE, OPTIONS');
    header('Access-Control-Allow-Headers: Origin, Content-Type, X-Auth-Token');

    $contentType = isset($_SERVER["CONTENT_TYPE"]) ? trim($_SERVER["CONTENT_TYPE"]) : '';

    if($contentType === "application/json") {
  
    $content = trim(file_get_contents("php://input"));
    $decoded = json_decode($content, true);

    if(! is_array($decoded)) {
        echo '{"status":"error decoding"}';
  
    } else {
      
        echo '{"status":"ok", "Nom": '. $decoded['nom'];.', "Prenom": '.$decoded['prenom'].'}';
    }
}else{

    echo '{"status":"error data"}';
}

?>

someone can help me please

thanks in advance

Do not build your json manually. Instead, create an array, then use json_encode to make sure it's in a correct format — aynber
– aynber, Commented Jan 31, 2021 at 2:08
Most likely you are missing double quotes around nom, prenom in "Nom": '. $decoded['nom'];.', "Prenom": '.$decoded['prenom']. But as mentioned, do this with json_encode() because utf characters need handling as well. — Michael Berkowski
– Michael Berkowski, Commented Jan 31, 2021 at 2:17

Tangentially Perpendicular · Accepted Answer · 2021-01-31 02:17:08Z

1

You have a spurious semi-colon in the JSON string you're creating:

echo '{"status":"ok", "Nom": '. $decoded['nom'];.', "Prenom": '.$decoded['prenom'].'}';
                                               ^ here

But that's not your only error here.

It's better to create an array and use json_encode() to create the JSON:

$jsonData = ["status"=>"ok", "Nom"=>$decoded['nom'], "Prenom"=>$decoded['prenom']];
echo json_encode($jsonData);

answered Jan 31, 2021 at 2:17

Tangentially Perpendicular

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Destiny Brotobor · Accepted Answer · 2021-01-31 02:38:12Z

0

Just Change your echo statement to return json_encode array like this.

echo json_encode([
            "status" => "ok",
            "Nom" => $decoded['nom'],
            "Prenom" => $decoded['prenom']
        ]);

answered Jan 31, 2021 at 2:38

Destiny Brotobor

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