使用where子句選擇2個字段的計數大於1的行

Question

我有一個看起來像這樣的表：

╔════╦═════════╦═════════╦═══════╗  
║ id ║ deleted ║ status  ║  ref  ║  
╠════╬═════════╬═════════╬═══════╣  
║  1 ║       0 ║ pending ║ 10001 ║  
║  2 ║       0 ║ paid    ║ 10001 ║  
║  3 ║       0 ║ paid    ║ 10001 ║  
║  4 ║       0 ║ paid    ║ 10002 ║  
║  5 ║       1 ║ pending ║ 10002 ║  
║  6 ║       1 ║ paid    ║ 10002 ║  
║  7 ║       0 ║ pending ║ 10003 ║  
║  8 ║       0 ║ paid    ║ 10003 ║  
║  9 ║       0 ║ paid    ║ 10003 ║  
║ 10 ║       0 ║ paid    ║ 10003 ║  
║ 11 ║       0 ║ pending ║ 10004 ║  
║ 12 ║       0 ║ paid    ║ 10004 ║  
║ 13 ║       1 ║ pending ║ 10005 ║  
║ 14 ║       1 ║ paid    ║ 10005 ║  
║ 15 ║       1 ║ paid    ║ 10005 ║  
║ 16 ║       0 ║ paid    ║ 10005 ║  
║ 17 ║       0 ║ pending ║ 10006 ║  
║ 18 ║       0 ║ paid    ║ 10006 ║  
║ 19 ║       0 ║ paid    ║ 10006 ║  
╚════╩═════════╩═════════╩═══════╝

我正在嘗試編寫一個MySQL查詢，它將返回已paid但未deleted的行。 但只有在paid計數大於1的情況下，結果應該是：

╔════╦═════════╦════════╦═══════╗
║ id ║ deleted ║ status ║  ref  ║
╠════╬═════════╬════════╬═══════╣
║  2 ║       0 ║ paid   ║ 10001 ║
║  3 ║       0 ║ paid   ║ 10001 ║
║  8 ║       0 ║ paid   ║ 10003 ║
║  9 ║       0 ║ paid   ║ 10003 ║
║ 10 ║       0 ║ paid   ║ 10003 ║
║ 18 ║       0 ║ paid   ║ 10006 ║
║ 19 ║       0 ║ paid   ║ 10006 ║
╚════╩═════════╩════════╩═══════╝

我一直試圖使用以下查詢得到這個，但它沒有做我需要的東西，現在我似乎在繞圈子。 有人可以幫我一把嗎？

SELECT t1.* FROM orders t1 WHERE exists (SELECT * FROM orders t2 where t1.id != t2.id and t1.ref = t2.ref and t1.deleted = 0 and t1.status = 'paid')

非常感謝！

[編輯] arrgh！ 對不起，我忘了說我只需要返回超過1個paid狀態的行...對不起......

Answer 1

請使用以下內容（沒有數據庫實例）：

SELECT t.*
FROM orders t
INNER JOIN
(
    SELECT ref, deleted, status
    FROM orders
    WHERE deleted = 0 and status = 'paid'
    GROUP BY ref
    HAVING count(ref) > 1
) d
ON t.ref = d.ref
AND t.status = d.status
AND t.deleted = d.deleted;

（基於http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=68374 ）

Answer 2

給這個打擊：

select o.*
from orders o
inner join
(
 select ref, 
 sum(case when status = 'paid' then 1 else 0 end) as paid_count,
 count(distinct status) as total_distinct_status_count
 from orders yt
 where deleted = 0
 group by ref
) t1 on t1.ref = o.ref
where o.status = 'paid'
and o.deleted = 0
and t1.paid_count > 1
and t1.total_distinct_status_count > 1;

似乎給出了你發布的數據給出的答案。

Answer 3

我認為你正在努力解決這個問題...休息一下再次啟動你的代碼。 無論如何答案已經在第一個回答中給出了...... Chilllllll