método jQuery $.ajax. éxito: función usando los datos de respuesta
Frecuentes
Visto 212 veces
0
Yo tengo:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>jQuery</title>
<script src="jquery-1.9.1.js"></script>
<style>
</style>
</head>
<body>
<div id="response" style="display:none"></div>
<button id="ajaxbutton">Get it!</button>
<script>
$(document).ready(function(){
$response = $("#response");
$button = $("#ajaxbutton");
$button.click(function(eventObject){
$.ajax({
url: "ajax.php",
type: "GET",
datatype: "json",
success: function(html){
$response.fadeIn("slow").html(html);
for(property in html)
{
console.log(property) + "<br />";
}
}
});
});
});
</script>
</body>
</html>
now the jQuery part is the one I am interested in only. Especially:
success: function(html){
$response.fadeIn("slow").html(html);
for(property in html)
{
console.log(property) + "<br />";
}
}
This above snippet returns the following output and I don't know why:
event.returnValue is deprecated. Please use the standard event.preventDefault() instead. jquery-1.9.1.js:3345
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¿Alguien puede explicar por qué?
1 Respuestas
0
Unfortunately I can't explain why. I'm gonna look through it a little further and see if I can't provide something for you. :) One thing I would like to say though is your jquery source is being loaded locally. I recommend using the external jquery source to load for a faster load on your page. Not sure how many people you'll have, but why not have a faster page anyways. :)
respondido 27 nov., 13:04
No es la respuesta que estás buscando? Examinar otras preguntas etiquetadas javascript jquery ajax or haz tu propia pregunta.
Can we see what is returned by your ajax.php? - ROunofF
Agrega un
console.log(html)
- Arun P Johnywhat is ajax.php supposed to return to you? - kkemple
the php script just pints a picture - Robert