php - SELECT COUNT(a.id) AS id... => 结果数组为空! [PHP, MySQL]

Question

从昨天开始，我在使用某些 PHP 时遇到了问题，浏览了 web 并有一种愚蠢的感觉，我错过了一些重要的东西。

通常使用 mysql_fetch_object，但尝试使用 mysql_fetch_array(没有帮助)。这是让我头疼的代码部分:

public static function get_datacenter_by_id($id) {
$result = mysql_query("SELECT COUNT(rack.id) AS Racks, COUNT(device.id) AS Devices, COUNT(card.id) AS Cards, COUNT(port.id) AS Ports 
          FROM datacenter, rack, device, card, port, location, building 
          WHERE location.id = building.location_id AND
          building.id = datacenter.building_id AND
          datacenter.id = '.$id.' AND
          rack.id = device.rack_id AND
          device.id = card.device_id AND
          (card.id = port.card_id1 OR
          card.id = port.card_id2)") or die ("Error in query: ".mysql_error());

$array = array();

while($row = mysql_fetch_object($result)) {
    $array[] = array($row->Racks, $row->Devices, $row->Cards, $row->Ports);                     
}

return $array;
}

$array 在另一个 .php 文件中使用，但是使用 print_r $array 已经向您展示了该数组保持为 empty (0)。我很确定错误出现在这段代码中，“COUNT (x) AS y”是否有问题？

PS:MySQL 查询有效，之前通过 Workbench 对其进行了测试。我会很感激一些好的建议! :-)

祝你有美好的一天!

Answer 1

这不是很简单吗:

$result = mysql_query("SELECT COUNT(rack.id) AS Racks, COUNT(device.id) AS Devices,     COUNT(card.id) AS Cards, COUNT(port.id) AS Ports 
      FROM datacenter, rack, device, card, port, location, building 
      WHERE location.id = building.location_id AND
      building.id = datacenter.building_id AND
      datacenter.id = '" . $id . "' AND
      rack.id = device.rack_id AND
      device.id = card.device_id AND
      (card.id = port.card_id1 OR
      card.id = port.card_id2)") or die ("Error in query: ".mysql_error());

注意来自“.$id”的修改。到 '".$id."'。您的查询正在查找“.$id.”的数据中心 ID。