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

Question

我从昨天起就遇到了一些PHP问题，浏览网页并感到愚蠢的感觉我错过了一些重要的东西。

通常使用mysql_fetch_object ，尝试使用mysql_fetch_array （但没有帮助）。 这是代码的一部分让我头疼：

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用于另一个.php文件，但是使用print_r $array已经向您显示该数组保持为empty (0) 。 我很确定错误出现在这段代码中，可能“ COUNT (x) AS y ”有问题吗？

PS：MySQL Query工作，之前通过Workbench测试过。 我很欣赏一些好的adivce！ :-)

祝你今天愉快！

Answer 1

而不是像这样使用它

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

你应该做这个

$row = mysql_fetch_object($result)
$array[] = $row->Racks;
$array[] = $row->Devices;
$array[] = $row->Cards;

因为您正在获取1条记录并在使用它时导致问题

Answer 2

这不是简单的：

$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());

请注意'。$ id。'的修改。 to'“。$ id。”'。 您的查询正在查找数据中心ID“。$ id。”。