Question

我想从'num'中选择不同的值，因为它的状态仅为'YES'，一次不包括'NO'？

表：

+--------+-----+--------+
| id     | num | status |
+--------+-----+--------+
| 364519 | AAA | YES    |
| 364522 | BBB | YES    |
| 364525 | AAA | NO     |
| 364527 | CCC | YES    |
| 364530 | DDD | YES    |
| 364532 | EEE | YES    |
| 364535 | FFF | YES    |
| 364537 | DDD | YES    |
| 364538 | DDD | NO     |
| 364539 | FFF | NO     |
+--------+-----+--------+

需要结果：

+--------+-----+--------+
| id     | num | status |
+--------+-----+--------+
| 364522 | BBB | YES    |
| 364527 | CCC | YES    |
| 364532 | EEE | YES    |
+--------+-----+--------+

Answer 1

您可以使用HAVING：

SELECT MIN(id) AS id, num,'yes' AS status
FROM tab
GROUP BY num
HAVING SUM(status='NO') = 0;

<强> DBFiddle Demo

Answer 2

其他几种方法。（看起来最干净）

将数据分成两组，并确保＆＃34; A＆＃34; set在＆＃34; B＆＃34;中没有数据。组使用外连接的第一种方法仅返回＆＃34; A＆＃34;设置在＆＃34; B＆＃34;中找不到相关数量的地方集。

SELECT Min(A.ID) as ID, A.num, A.status
FROM table A
LEFT JOIN table B
  on A.Num = B.Num
 and B.Status = 'NO'
WHERE A.Status = 'YES'
  and B.Num is null
GROUP BY A.Num, A.Status

使用不存在的第二种方法。在＆＃34; A＆＃34;中只有类似的num记录不在＆＃34; B＆＃34;集。

SELECT min(A.ID) as ID, A.Num, A.Status
FROM table A
WHERE NOT EXISTS (SELECT 1 
                  FROM Table B
                  WHERE A.Num = B.Num
                    and B.Status = 'NO')
GROUP BY A.Num, A.Status

Answer 3

SELECT * FROM table
WHERE status='YES' AND num NOT IN (SELECT num FROM table WHERE status='NO')

或者这个

SELECT * FROM table t
WHERE t.status='YES' AND NOT EXISTS (SELECT * FROM table WHERE status='NO' AND num=t.num)

Answer 4

从表中选择不同（num），id，状态=＆＃39;是＆＃39;

Answer 5

我只会使用group by cluase：

select min(id) as id, num, 'YES' as status
from table t
group by num
having min(status) = max(status) and min(status) = 'YES';

在DISTINCT条件中选择DISTINCT列

5 个答案: