PHP not able to pass JSON array to jQuery - php

Question

Pardon me if I am making a silly mistake here... But I am little new to PHP...
I have been stuck at this for long time.
I have tried several methods but I am only able to pass ONe table row , not multiple rows
I've even tried with the following approach, but I landed with the same error.
http://www.electrictoolbox.com/json-data-jquery-php-mysql/
Here's what I am trying to do :
1) Make Ajax call using JQuery, to fetch data from server.
$.ajax({
url: 'test.php', //the script to call to get data
data: "",
dataType: 'json', //data format
success: function(data) //on recieve of reply
{
//use "data"
}
2) fetch table data from MySql database, using PHP.
$result = mysql_query("SELECT username,characterType FROM usertable");
3) pass ALL of the rows as JSON data, back to the calling function.
//This works , Returns one row and I am able to get the result back at AJAX end
$array = mysql_fetch_row($result);
echo json_encode($array);
mysql_fetch_assoc($result) FAILS with the following error :
XML Parsing Error: no element found Location: moz-nullprincipal:{6b1***-****somecrap numbers****-***86} Line Number 17, Column 3:
$array = mysql_fetch_assoc($result);
echo json_encode($array);
I have even tried using mysql_fetch_array($result, MYSQL_NUM) as suggested in the other relevant questions, but I am not able to workaround the problem.
Any Help is greatly appreciated.
============
UPDATE:
I may be wrong but just wanted to know, if this could be the possibility or not... Can this be a local setup related issue ? A Configuration mistake .. ?
I have done my localhost setup using "XAMPP installation" [ Apache / MySQL / Tomcat ... bundled in one package ] ...
When I run the file as "PHP application" it runs just fine ... I am able get "all rows" But when I deploy the code on my apache servers it doesn't run ... The whole php file comes as a response , with "XML parsing Error" [ I am using firebug to track the response ]
Thanks
Pranav

Answer 1

A few things to note here:
As Rup mentioned, $array = mysql_fetch_row($result); only returns a single row.
The other thing is that test.php does not currently return a JSON object; to see whats going on, it helps to comment out dataType: 'json', and use console.log(data); to see what is being returned. Your response probably looks something like this:
["someUsername","someChartype"]
The format of the object you are expecting is probably something more like:
[{"username":"foo","chartype":"admin"},
{"username":"bar","chartype":"user"},
{"username":"fum","chartype":"guest"}]
To get the format of the associative array, use mysql_fetch_assoc($result) instead; that will give
{"username":"someUsername","chracterType":"someChartype"}
To summarize, update your php code to:
Loop through your results, pushing them onto an array.
Encode the entire array at the end
$outputArray = array();
while( $row = mysql_fetch_assoc( $result ) ){
array_push($outputArray, $row);
}
echo json_encode($outputArray);
Take a look at How to build a JSON array from mysql database for more info. Also, as one of the answers mention, its worth using PDO rather than mysql_*

Answer 2

Don't start manually composing a JSON string as #Tuanderful indicated. Let json_encode() handle that -- just give it the right data. Try something like this:
$records = array();
while ( $record = mysql_fetch_assoc( $result ) ) {
$records[] = $record;
}
echo json_encode( $records );

Answer 3

Well I would to like this:
$.ajax({
url: 'test.php', //the script to call to get data
//data: "",
dataType: 'json', //data format
success: somevar
})
var somevar = function(data) //on recieve of reply
{
//use "data"
somevar2 = jQuery.parseJSON(data)
//now You can enter each row/cell from db(/php) like this
// somevar2[0] would be the first row from php
//if You want to do something with every line You can use:
//$.each(somevar2, function(key,value){
//$('#divname').html(value.cellnameoftable)
}
I hope it will help ;)