Good afternoon, I have this ajax code where I use to do an update on my form, it happens that when I submit it, it goes to the php page, and I do not want it, I just want the feedback.
follow the code:
function alterProduct(obj){
var form = $(obj);
var dados = new FormData(obj);
$.ajax({
url: form.attr('action'),
type: form.attr('method'),
data: dados,
processData: false,
cache: false,
contentType: false,
success: function( data ) {
if ( data == 'OK' ) {
alert('Dados enviados com sucesso');
} else {
alert(data);
}
},
error: function (request, status, error) {
alert(request.responseText);
}
});
return false;
}
My form:
<form method="post" enctype="multipart/form-data" class="update-product" onsubmit="alterProduct(this)" action="ajax/update_product.php">
<input type="text" name="id" value="<?php echo $a->current()->id; ?>" hidden>
<div class="form-group">
<label>Nome do Produto</label>
<input type="text" name="name" value="<?php echo $a->current()->name; ?>" class="form-control" placeholder="Digite o nome do seu produto">
</div>
<div class="form-group">
<label>Preço</label>
<input type="text" name="price" value="<?php echo $a->current()->price; ?>" class="form-control money" placeholder="Digite o valor do seu produto">
</div>
<div class="form-group">
<label>Quantidade</label>
<input type="number" name="qtd" class="form-control" value="<?php echo $a->current()->qtd; ?>" placeholder="Digite a quantidade a ser cadastrada">
</div>
<input type="submit" name="update" class="btn-action" value="Alterar">
<button type="button" class="btn-outline" data-dismiss="modal">Fechar</button>
</form>