我有下表:
name - catridge_type - department - date - quantity
--------------------------------------------------------
Ak - Hp 305A Cyan - IT - 2016-01-13 - 5
Ad - Hp 508A Black - Hr - 2016-02-13 - 6
我有以下select语句:
$query= "SELECT catridge_type, quantity FROM catridge_details WHERE quantity > 0 ";
我的select语句将catridge_type返回为null,但将quantity作为相应的值。
请协助。
附加代码:
$result = $db->query($query);
if(!$result) { die('Error getting Catridge types ['.$db->error.']' );}
else{
//create an array to store the data
$catridge_array = array();
while ($row = $result->fetch_assoc()) {
array_push($catridge_array, new available_catridges($row['catridges'], $row['quantity']));
}
header('Content-Type: application/json');
echo json_encode($catridge_array);
}
答案 0 :(得分:0)
行$row['catridges']
不存在。试试$row['catridge_type']