I'm trying to send an XHR request to a PHP page with several parameters (2 string, 1 blob) but I am unnable to get theses values in the PHP code..
here is my code :
JS
function initRequeteUpdImg(blob, file)
{
img = document.getElementById("id");
img.src = blob;
requeteUdpRec = new XMLHttpRequest();
requeteUdpRec.onreadystatechange = callback_ReqUpdRec;
requeteUdpRec.open("POST","function/Modify_action.php", true);
requeteUdpRec.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
function readFile(event)
{
var fileType = file.type;
blob = blobUtil.arrayBufferToBlob(event.target.result, fileType);
formData = new FormData();
formData.append("id", id);
formData.append("action", "updImg");
formData.append("file", event.target.result);
requeteUdpRec.send(formData);
}
var reader = new FileReader();
reader.addEventListener('load', readFile);
reader.readAsArrayBuffer(file)
}
PHP
if(isset($_POST["action"]) && $_POST["action"]=="updImg")`{
$requete = "select eleve_id from eleves where person_id = ?";
$req_args = array($_POST['id']);
$req_type = "i";
$res = mysqli_fetch_array(prepared_query($requete, $req_args, $req_type));
$requete = "update signature SET signature_blob = '".$_POST["file"]."' where eleve_id = ".$res["eleve_id"];
$res = unprepared_query($requete, $req_args, $req_type);
}
else
{
http_response_code (404);
}`
I'm obviously doing something wrong but where..?
I'm new to JS so if you could detail it would be very nice ! Thanks for your help !
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