Question

这是我用来显示网页的COUNT个查询：

$sql = "
SELECT SUM(num) as num FROM (
  SELECT
    COUNT(URL) AS num
    , 'World' AS GoSection
    , 'GW' AS MySite
  FROM gw_geog
  WHERE URL = :MyURL AND G1 = 1
UNION ALL
  SELECT
    COUNT(URL) AS num
    , 'World' AS GoSection
    , 'GW' AS MySite
  FROM gw_geog_political
  WHERE URL = :MyURL
 ) AS X";

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

我想做两处修改：

1）从每一行删除 WHERE URL =：MyURL ，并将其合并到查询末尾的单行中

2）通过添加 LIMIT 1

来重复Zap

这就是我现在正在使用的。

$sql = "SELECT SUM(num) as num FROM (
 SELECT COUNT(URL) AS num, 'World' AS GoSection, 'GW' AS MySite FROM gw_geog WHERE URL = :MyURL AND G1 = 1
 UNION ALL
 SELECT COUNT(URL) AS num, 'World' AS GoSection, 'GW' AS MySite FROM gw_geog_political WHERE URL = :MyURL
 ) AS X
WHERE X.URL LIKE :MyURL LIMIT 1";
$stmt = $pdo->prepare($sql);
$stmt->bindParam(':MyURL',$MyURL,PDO::PARAM_STR);
$stmt->execute();
$Total = $stmt->fetch();

但是，我收到错误消息未知列X.SUM （或X.URL等）。

有人能告诉我这样做的正确方法吗？

Answer 1

使用Query As Follows，您将获得所需的输出：

 SELECT
    SUM(num) AS num
FROM
(
    SELECT
        COUNT(URL) AS num,
        'World' AS GoSection,
        'GW' AS MySite,
        gw_geog.`URL` as URL  
    FROM
        gw_geog
    WHERE
        URL = :MyURL
    AND G1 = 1
    UNION ALL
        SELECT
            COUNT(URL) AS num,
            'World' AS GoSection,
            'GW' AS MySite,
             gw_geog_political.`URL` as URL
        FROM
            gw_geog_political
        WHERE
            URL = :MyURL
) AS X
WHERE
    X.URL like  :MyURL
LIMIT 1

Answer 2

如果我了解你，limit 1并不是你想要的。

这是一个小提琴：http://sqlfiddle.com/#!9/5ffbd/5

SELECT count(URL) AS num
    ,'World' AS GoSection
    ,'GW' AS MySite
FROM (
    SELECT URL
        ,G1
        ,'a' AS tbl
    FROM gw_geog

    UNION ALL

    SELECT URL
        ,G1
        ,'b' AS tbl
    FROM gw_geog
    ) a
WHERE a.tbl = 'b'
    OR (
        a.tbl = 'a'
        AND a.G1 = 1
        )
    AND a.URL = :MyURL

将LIMIT 1添加到COUNT个查询

2 个答案: