SQL - select 基於值存在於另一列

Question

我想創建一個 SQL 查詢來識別沒有 A3 事件的井。 我正在使用 SQL 服務器。

我嘗試了多種方法，例如檢查計數，與事件 A3 進行比較，但仍然無法得到我想要的。

從下面的示例中，所需的結果將是 W3 和 W4

地點	事件
W1	A1
W1	A2
W1	A3
W2	A1
W2	A2
W2	A3
W3	A1
W3	A4
W4	A2
W4	A4

Answer 1

NOT EXISTS是一種非常高效的方法

例子

Select * 
 From  YourTable A
 Where not exists ( select 1 
                     from  YourTable T 
                     where T.Well=A.Well and T.[Event]='A3' 
                  )

結果

Well    Event
W3      A1
W3      A4
W4      A2
W4      A4

Answer 2

聚合是一種方法：

SELECT Site
FROM yourTable
GROUP BY Site
HAVING COUNT(CASE WHEN Event = 'A3' THEN 1 END) = 0;

如果您確實還需要查看所有站點的匹配井，則將上述內容放入 CTE 並將其用作子查詢：

WITH cte AS (
    SELECT Site
    FROM yourTable
    GROUP BY Site
    HAVING COUNT(CASE WHEN Event = 'A3' THEN 1 END) = 0
)

SELECT Site, Event
FROM yourTable
WHERE Site IN (SELECT Site FROM cte);

Answer 3

你可以做

SELECT DISTINCT site 
FROM table_name 
WHERE site NOT IN (SELECT DISTINCT site 
                   FROM table_name 
                   WHERE event = 'A3')

Answer 4

sub-query將返回帶有 A3 事件的站點。 因此使用LEFT JOIN加入表並過濾子查詢中不可用的站點

SELECT DISTINCT Site 
FROM your_table as tbl
LEFT JOIN
(
  SELECT DISTINCT Site
  FROM your_table
  WHERE Event = 'A3'
) as sq
ON tbl.site=sq.site
WHERE sq.Site IS NULL

Answer 5

SELECT Well  
FROM test
where Well not in (select Well from tablename where Evt = 'A3');