php – 将LIMIT 1添加到COUNT个查询
发布时间:2020-12-13 21:54:47 所属栏目:PHP教程 来源:网络整理
导读:这是我用来显示页面的COUNT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 = 1UNION ALL SELECT COUNT(URL) AS num,'GW' AS MySite FROM gw_geog_pol
这是我用来显示页面的COUNT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,'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(); 我想做两个改变: 1)从每行删除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,'GW' AS MySite FROM gw_geog WHERE URL = :MyURL AND G1 = 1 UNION ALL SELECT COUNT(URL) AS num,'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',PDO::PARAM_STR); $stmt->execute(); $Total = $stmt->fetch(); 但是,我收到错误消息Unknown column X.SUM(或X.URL等). 有人能告诉我这样做的正确方法吗? 解决方法
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,'GW' AS MySite,gw_geog.`URL` as URL FROM gw_geog WHERE URL = :MyURL AND G1 = 1 UNION ALL SELECT COUNT(URL) AS num,gw_geog_political.`URL` as URL FROM gw_geog_political WHERE URL = :MyURL ) AS X WHERE X.URL like :MyURL LIMIT 1 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |