錯誤與一個SUM得到的Json在查詢

Question

我有以下問題，我不能讓JSON的回報怎麼一回事，因爲我有以下錯誤代碼：

注意：未定義的變量：PRECIO在的/ var/www/html等/網吧/ Consumido.php上線

這是我的代碼

<?php 
// Include confi.php 
include_once('confi.php'); 
$fecha = "'" . date('Y-m-d')."%"."'"; 
$uid = isset($_GET['uid']) ? mysql_real_escape_string($_GET['uid']) : ""; 
    if(!empty($uid)){ 
    $qur = mysql_query("SELECT Sum(comprasdetalle.Precio) FROM compras INNER JOIN comprasdetalle ON comprasdetalle.idCompras = compras.idCompras WHERE compras.CafeEmpleadoCodigo = $uid AND compras.Date LIKE $fecha 
     GROUP BY 
     comprasdetalle.Precio;"); 
    $result =array(); 
    while($r = mysql_fetch_array($qur)) 
    { 
     extract($r); 
     $result = array("Precio" => $Precio); 
    } 
    $json = array($result); 
}else{ 
    $json = array("'status'" => 0, "msg" => "'User ID not define'"); 
} 
@mysql_close($conn); 

/* Output header */ 
header('Content-type: application/json'); 
echo json_encode($json);?> 

我已經做了這個沒有SUM，它的工作原理，但如果我加上總和它給上述錯誤

你能幫我。

感謝

Answer 1

剛剛發現錯誤並修復它是這樣的：

<?php 
// Include confi.php 
include_once('confi.php'); 
$fecha = "'" . date('Y-m-d')."%"."'"; 
$uid = isset($_GET['uid']) ? mysql_real_escape_string($_GET['uid']) : ""; 
    if(!empty($uid)){ 
    $qur = mysql_query("SELECT Sum(comprasdetalle.Precio) as Precio FROM compras INNER JOIN comprasdetalle ON comprasdetalle.idCompras = compras.idCompras WHERE compras.CafeEmpleadoCodigo = $uid AND compras.Date LIKE $fecha 
     GROUP BY 
     comprasdetalle.Precio;"); 
    $result =array(); 
    while($r = mysql_fetch_array($qur)) 
    { 
     extract($r); 
     $result = array("Precio" => $Precio); 
    } 
    $json = array($result); 
}else{ 
    $json = array("'status'" => 0, "msg" => "'User ID not define'"); 
} 
@mysql_close($conn); 

/* Output header */ 
header('Content-type: application/json'); 
echo json_encode($json);?> 

只是需要一個名字添加到字段

感謝