Question

在这个网站的帮助下，我现在有一个显示来自多个数据库表的页面的查询，其中$ MyURL等于页面URL（例如，MySite / Crazy_Horse匹配表人员中的Crazy_Horse值，字段URL）。

唯一剩下的问题是静态值（MySite2）。我无法在显示页面上回应它。但它必须有效，因为如果我注释掉这一行 - $ MySite2 = $ row ['MySite2']; - 我收到一条错误消息，说明$ MySite2尚未定义。但是，当我恢复该行时，错误消息消失，但echo $ MySite2不显示任何内容。

谁能看到我做错了什么？

$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"

如果您想要显示MySite2，您必须将此值组合为您的聚合函数。

您无法显示，因为您的查询的可见性是由较高级别提供的，因此我已在您的字段列表中添加了MySite2。

告诉我它是否正常

静态值不显示（COUNT，UNION ALL查询）

1 个答案: