ajax json decode mysql fetch array - php

Question

$result = mysql_query ("SELECT * FROM order_list");
$test = array();
$i=0;
$myjsons = array();
while($row = mysql_fetch_assoc($result)){
echo $myjsons[] = json_encode(array($row));
}
and a AJAX javascript
function ajaxFunction() {
var ajaxRequest; // The variable that makes Ajax possible!
try {
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
} catch (e) {
// Internet Explorer Browsers
try {
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e) {
try {
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e) {
// Something went wrong
alert("Your browser is too old to run me!");
return false;
}
}
}
// Create a function that will receive data sent from the server
ajaxRequest.onreadystatechange = function() {
if (ajaxRequest.readyState == 4) {
document.getElementById("resultTXT").value = eval(" (" + ajaxRequest.responseText + ")");
}
}
ajaxRequest.open("POST", "userfind.php", true);
ajaxRequest.send(null);
}​
the javascript function wont store the array into the textbox
can anyone tell me whats the problem here?
Thanks

Answer 1

try:
while($row = mysql_fetch_assoc($result)){
$myjsons[] = json_encode(array($row));
}
echo json_encode($myjsons);
instead of
while($row = mysql_fetch_assoc($result)){
echo $myjsons[] = json_encode(array($row));
}
you can use jquery library for tha ajax call http://api.jquery.com/jQuery.post/
$.post('userfind.php', function(data) {
$("#resultTXT").val(data);// ==document.getElementById("resultTXT").value =data;
},'json'
);

Answer 2

JSON output would be invalid in your example, build the array and out put a single json encoded array like so:
$myjsons = array();
while($row = mysql_fetch_assoc($result)){
$myjsons[] = $row;
}
echo json_encode($myjsons);