不显示静态值（COUNT、UNION ALL 查询）

Question

在此站点的大量帮助下，我现在有一个查询，它显示了来自多个数据库表的页面，其中 $MyURL 等于页面 URL（例如，MySite/Crazy_Horse 与表 people、字段 URL 中的值 Crazy_Horse 匹配）。

唯一剩下的问题是静态值（MySite2）。我无法在显示页面上回显它。然而它必须工作，因为如果我注释掉这一行 - $MySite2 = $row['MySite2']; - 我收到一条错误消息，提示 $MySite2 尚未定义。但是当我恢复该行时，错误消息消失了，但 echo $MySite2 没有显示任何内容。

谁能看出我做错了什么？

$sql = "SELECT SUM(num) as num FROM (
  SELECT 'GZ' AS MySite2, COUNT(Taxon) AS num FROM gz_life WHERE Taxon = :MyURL
  UNION ALL
  SELECT 'All' AS MySite2, COUNT(Name) AS num FROM gw_geog WHERE Name = :MyURL
  UNION ALL
  SELECT 'GS' AS MySite2, COUNT(URL) AS num FROM gs WHERE URL = :MyURL
 ) AS X";

$stmt = $pdo->prepare($sql);
$stmt->bindParam(':MyURL',$MyURL,PDO::PARAM_STR);
$stmt->execute();

while ($row = $stmt->fetch())
{
 $MySite2 = $row['MySite2'];
 $Total = $row['num'];
 switch($Total)
{
 case 1:
 require($BaseINC."/$MyPHP/inc/C/2_Child.php");
 break;
 case 0:
 require_once($BaseINC."/404.php");
 break;
 default:
 require($_SERVER['DOCUMENT_ROOT']."/Dupe.php");
 break;
  }
}

Answer 1

以这种方式更改您的查询：

$sql = "SELECT X.MySite2, SUM(x.num) as num FROM (
  SELECT 'GZ' AS MySite2, COUNT(Taxon) AS num FROM gz_life WHERE Taxon = :MyURL
  UNION ALL
  SELECT 'All' AS MySite2, COUNT(Name) AS num FROM gw_geog WHERE Name = :MyURL
  UNION ALL
  SELECT 'GS' AS MySite2, COUNT(URL) AS num FROM gs WHERE URL = :MyURL
 ) AS X group by X.MySite2"

如果你想显示 MySite2，你必须根据这个值对你的聚合函数进行分组。

您无法显示，因为您的查询的可见性已由上级提供，因此我已将 MySite2 添加到您的字段列表中。

告诉我是否可以