JSON 인코딩 문제 PHP

Question

인코딩 방식이 잘못 되었나요?

<?php 
include ("../includes/db_con.php"); 

    $itemResults = mysql_query("SELECT `sold_for` FROM `items` WHERE `item_did_sell`='1'") or die(); 
    $mIResults = mysql_query("SELECT `mi_price`, `mi_total_sold` FROM `misc_items` WHERE `mi_total_sold`>'1'") or die(); 
    $donationResults = mysql_query("SELECT `amount` FROM `donations`") or die(mysql_error()); 

$total = 0; 
$itemTotal = 0; 
$mITotal = 0; 
$donationTotal = 0; 

while($row = mysql_fetch_assoc($itemResults)){ 
    $itemTotal += $row['sold_for']; 
    $total += $itemTotal; 
} 

while($row = mysql_fetch_assoc($mIResults)){ 
    $mITotal += ($row['mi_price'] * $row['mi_total_sold']); 
    $total += $mITotal; 
} 

while($row = mysql_fetch_assoc($donationResults)){ 
    $donationTotal += $row['amount']; 
    $total += $donationTotal; 
} 

header("Content-Type: application/json"); 
$arr = array('items' => $itemTotal,'mitems' => $mITotal,'donations' => $donationTotal,'total' => $total); 
$arr = json_encode($arr); 
echo $arr; 

include ("../includes/db_discon.php"); 
?> 

스크립트 출력 : {"items":1000,"mitems":0,"donations":0,"total":1000}

은 (확인 된) 내 클라이언트 측 스크립트 내 JSON 어레이 내의 parase 오류가 존재에 의해 발생.

Answer 1

http://www.jquery4u.com/json/ajaxjquery-getjson-simple/