Question

我希望协调来自2个不同表的数据，我需要执行连接和substr来创建可用于执行匹配的列。以下单独的查询反映了每个表生成的select语句将INFO : Map 1: -/- Reducer 2: 0/486 INFO : Map 1: -/- Reducer 2: 0/486 INFO : Map 1: -/- Reducer 2: 0/486 INFO : Map 1: -/- Reducer 2: 0/486（例如sitenn.zonenn）反映为节点名的匹配值。

site12.zone20

我需要编写一个查询，从一个表中选择另一个表中不存在的所有节点名。

Answer 1

一种简单的方法：使用[_tableview setRowHeight:UITableViewAutomaticDimension]; [_tableview setEstimatedRowHeight:44];

MINUS

Answer 2

使用left join查找它。它比minus，not in，not exists等快。

SELECT a.nodename 
FROM   (SELECT DISTINCT( regexp_substr(B.node_name, '*site*.*') ) AS nodename 
        FROM   opc_act_messages A, 
               opc_node_names B 
        WHERE  A.message_group = 'Ebts_Status_Alarms' 
               AND A.severity <> 2 
               AND A.node_id = B.node_id
        ) a 
LEFT JOIN 
        (SELECT 'site' 
                         || site_id 
                         || '.zone' 
                         || zone_id AS nodename 
                  FROM   aw_active_alarms 
                  GROUP  BY site_id, 
                            zone_id
        ) b 
ON a.nodename = b.nodename 
WHERE  b.nodename IS NULL

Answer 3

这会有用吗？

WITH t1
     AS (SELECT DISTINCT
                (REGEXP_SUBSTR (B.NODE_NAME, '*site*.*')) AS nodename
           FROM OPC_ACT_MESSAGES A, OPC_NODE_NAMES B
          WHERE     A.MESSAGE_GROUP = 'Ebts_Status_Alarms'
                AND A.SEVERITY <> 2
                AND A.NODE_ID = B.NODE_ID),
     t2
     AS (  SELECT 'site' || site_id || '.zone' || zone_id AS nodename
             FROM aw_active_alarms
         GROUP BY site_id, zone_id)
SELECT *
  FROM t1
 WHERE t1.nodename NOT IN (SELECT nodename FROM t2)

Oracle sql join针对提取的值

3 个答案: