SELECT COUNT（a.id）AS id ... =>結果數組為空！ [PHP，MySQL]

Question

我從昨天起就遇到了一些PHP問題，瀏覽網頁並感到愚蠢的感覺我錯過了一些重要的東西。

通常使用mysql_fetch_object ，嘗試使用mysql_fetch_array （但沒有幫助）。 這是代碼的一部分讓我頭疼：

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用於另一個.php文件，但是使用print_r $array已經向您顯示該數組保持為empty (0) 。 我很確定錯誤出現在這段代碼中，可能“ COUNT (x) AS y ”有問題嗎？

PS：MySQL Query工作，之前通過Workbench測試過。 我很欣賞一些好的adivce！ :-)

祝你今天愉快！

Answer 1

而不是像這樣使用它

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

你應該做這個

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

因為您正在獲取1條記錄並在使用它時導致問題

Answer 2

這不是簡單的：

$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());

請注意'。$ id。'的修改。 to'“。$ id。”'。 您的查詢正在查找數據中心ID“。$ id。”。