I want to create the process of uploading an image with jquery ajax. here's my code :
<form action="" method="post" name="formPicture" enctype="multipart/form-data">
<div class="modal-body panel-modal-scrol">
<input type="file" name="picture" class="form-control" />
</div>
<div id="info"></div>
<button class="btn btn-default" data-dismiss="modal" aria-hidden="true">Cancel</button>
<input type="submit" value="Save" id="btn-saving" class="btn btn-primary" />
</form>
<script>
$("[name=formPicture]").submit(function(e){
e.preventDefault();
var formData = $(this).serialize();
console.log(formData);
$.ajax({
type: 'POST',
url: 'upload_picture.php',
data: formData,
success: function(data){
console.log(data);
}
});
});
</script>
upload_picture.php
<?php
$path = $_FILES['picture']['tmp_name'];
$file_name = $_FILES['picture']['name'];
$newPath = "upload/";
move_uploaded_file($lokasi_file,$newPath.$file_name);
$insert="INSERT INTO `picture`(`picture`)
VALUES ('$file_name')";
$query=mysql_query($insert);
if($query){
echo '{"status":"success"}';
exit;
}else{
echo '{"status":"error"}';
exit;
}
?>
but the error said "Notice: Undefined index: picture on line 4"; there's something wrong with my code ? Thanks..