1
function CheckNewsFeed() {
$.ajax({
url: ProcurarPorNovosDados.php,
success: function(texto) {
*Aqui que está minha dúvida, como fazer o Append de resultados dinâmicos
retornado pelo PHP?*
//$("#Resultados").html(texto);
}
})};
setTimeout(CheckNewsFeed(), 3000);
I have no idea how to do this, and I don’t know if it’s possible, but every comment helps, thank you;
independent from the results list in mysql?
– Kevin mtk
When you configure the callback Success to $.ajax method it is already understood that you are returning text/html ready to be attached to the page, mainly because you have not detailed anything else. And, jQuery knows nothing about your Mysql, all it expects there is a return, no matter what you do in your PHP. What then do you mean by result list in Mysql? As I said it is expected that in your PHP you return only text/html/JSON. So, if you get from Mysql, first do the processing in PHP and return the text.
– AlfredBaudisch
Yes, I will make the results have their <div’s> in PHP itself, thank you
– Kevin mtk