I am starting to use the jquery $.ajax() but I can't get back what I want to...I send this:
$(function(){
$.ajax({
url: "graph_data.php",
type: "POST",
data: "casi=56&nada=48&nuevo=98&perfecto=100&vales=50&apenas=70&yeah=60",
dataType: "json",
error:
function (xhr, desc, exceptionobj) {
document.writeln("El error de XMLHTTPRequest dice: " + xhr.responseText);
},
success:
function (json) {
if (json.error) { alert(json.error); return; }
var output = "";
for (p in json) {
output += p + " : " + json[p] + "\n";
}
document.writeln("Results: \n\n" + output);
}
});
});
and my php is:
<?php
$data = $_POST['data'];
function array2json($data){
$json = $data;
return json_encode($json);
}
?>
and when I execute this I come out with:
Results:
just like that I used to have in the php a echo array2json statement but it just gave back gibberish...I really don't know what am I doing wrong and I've googled for about 3 hours just getting basically the same stuff. Also I don't know how to pass parameters to the "data:" in the $.ajax function in another way like getting info from the web page, can anyone please help me?
Edit
I did what you suggested and it prints the data now thank you very much =) however, I was wondering, how can I send the data to the "data:" part in jQuery so it takes it from let's say user input, also I was checking the php documentation and it says I'm allowed to write something like:
json_encode($a,JSON_HEX_TAG|JSON_HEX_APOS|JSON_HEX_QUOT|JSON_HEX_AMP)
however, if I do that I get an error saying that json_encode accepts 1 parameter and I'm giving 2...any idea why? I'm using php 5.2
