$.ajax({
url: 'ajaxfile.php',
type: 'POST',
data: {
image: base64URL
},
success: function(data){
console.log(data);
$.notify("info", "Upload successfully");
},
error: function () {
$.notify("Error on image upload");
}
});
PHP code :
<?php
$image = $_POST['image'];
$location = "src/upload/";
$image_parts = explode(";base64,", $image);
$image_base64 = base64_decode($image_parts[1]);
$filename = "screenshot_".uniqid().'.png';
$file = $location . $filename;
file_put_contents($file, $image_base64);
return [
'status' => true
]
?>
The call is done (I saw in browser console) but on console.log I have the code php returned. Seems that nothing happen, the code php is not implemented. Have you an idea ? Thx in advance and sorry for my english
I put an image with the error
file_put_contents()
returns false
on failure, so you could assign it to a variable and use that to determine your status like so:
<?php
$image = $_POST['image'];
$location = "src/upload/";
$image_parts = explode(";base64,", $image);
$image_base64 = base64_decode($image_parts[1]);
$filename = "screenshot_".uniqid().'.png';
$file = $location . $filename;
$imageData = file_put_contents($file, $image_base64);
if ($imageData !== false) {
echo "success";
} else {
http_response_code(500);
die();
}
?>