我有一个名为Online_Transaction的表，我希望显示特定月份中所有类型的事务以及每种类型（已完成，已拒绝，待批准）的数量。直到现在我才有这个问题，但我反复得到任何想法吗？
SELECT DISTINCT
TRANSACTION_TYPE_ID ,
YEAR(CREATED_ON) AS YEAR ,
MONTH(CREATED_ON) AS MONTH ,
( SELECT
Count(TRANSACTION_TYPE_ID)
FROM
ONLINE_TRANSACTION
WHERE
ONLINE_TRANSACTION.STATUS_ID = 'COMPLETED' AND MONTH(CREATED_ON) = '2' ) AS COMPLETED ,
( SELECT
Count(TRANSACTION_TYPE_ID)
FROM
ONLINE_TRANSACTION
WHERE
ONLINE_TRANSACTION.STATUS_ID = 'DECLINED' AND MONTH(CREATED_ON) = '2' ) AS DECLINED ,
( SELECT
Count(TRANSACTION_TYPE_ID)
FROM
ONLINE_TRANSACTION
WHERE
ONLINE_TRANSACTION.STATUS_ID = 'FAILED' AND MONTH(CREATED_ON) = '2' ) AS FAILED ,
( SELECT
Count(TRANSACTION_TYPE_ID)
FROM
ONLINE_TRANSACTION
WHERE
ONLINE_TRANSACTION.STATUS_ID = 'PENDING_AUTH' AND MONTH(CREATED_ON) = '2' ) AS PENDING_AUTH
--(SELECT Count(*) from )
FROM
ONLINE_TRANSACTION
WHERE
MONTH(CREATED_ON) = '2'
GROUP BY
TRANSACTION_TYPE_ID ,
ONLINE_TRANSACTION.CREATED_ON
我得到了这些结果：
TRANSACTION_TYPE_ID YEAR MONTH COMPLETED DECLINED FAILED
------------------------------------ ----------- ----------- ----------- ----------- -------
INSURANCE--TYPE 2009 2 9712 177 0
CHEQUEBOOK-TYPE 2009 2 9712 177 0
CHEQUE-STOP-YPE 2009 2 9712 177 0
PAYMENT-TRANS-TYPE 2009 2 9712 177 0
DOMESTIC-TRANSFER-TYPE 2009 2 9712 177 0
PAYMENT-TRANS-TYPE 2009 2 9712 177 0
INTRA-ACCOUNT-TRANS-TYPE 2009 2 9712 177 0
INTRA-BANK-TRANS-TYPE 2009 2 9712 177 0
STANDING-ORDER-TYPE 2009 2 9712 177 0
STATEMENT-REORDERING TYPE 2009 2 9712 177 0
PAYMENTS-TRANS-TYPE 2009 2 9712 177 0
如您所见，结果重复出现，表中应该是不同的值 有什么想法吗？
尝试了下面的答案，这就是我得到的： 这种类型的结果集 - 它为不同行中的每种交易类型显示不同的结果，所以如果说交易类型是内部转移，我有10个完成，2个被拒绝，它将显示完成一行而被拒绝在另一行？如何在每个交易类型的一行中显示它
TRANSACTION_TYPE_ID YEAR MONTH REJECTED COMPLETED POSTED
ALPHA-INSURANCE-TRANS-TYPE 2009 2 0 12 0
CHEQUEBOOK-ORDER-TRANS-TYPE 2009 2 0 0 0
CHEQUEBOOK-ORDER-TRANS-TYPE 2009 2 0 52 0
CHEQUEBOOK-ORDER-TRANS-TYPE 2009 2 2 0 0
CHEQUE-STOP-TRANS-TYPE 2009 2 0 3 0
PAYMENT-TRANS-TYPE 2009 2 0 361 0
PAYMENT-TRANS-TYPE 2009 2 1 0 0
DOMESTIC-TRANSFER-TRANS-TYPE 2009 2 0 0 0
DOMESTIC-TRANSFER-TRANS-TYPE 2009 2 0 541 0
DOMESTIC-TRANSFER-TRANS-TYPE 2009 2 6 0 0
查询如下所示：
SELECT DISTINCT
TRANSACTION_TYPE_ID ,
YEAR(CREATED_ON) AS YEAR ,
MONTH(CREATED_ON) AS MONTH ,
SUM(CASE ONLINE_TRANSACTION.STATUS_ID
WHEN 'STATUS-TRANS-REJECTED ' THEN 1
ELSE 0
END) AS REJECTED ,
SUM(CASE ONLINE_TRANSACTION.STATUS_ID
WHEN 'STATUS-TRANS-COMPLETED ' THEN 1
ELSE 0
END) AS COMPLETED ,
SUM(CASE ONLINE_TRANSACTION.STATUS_ID
WHEN 'STATUS-TRANS-DECLINDED ' THEN 1
ELSE 0
END) AS DECLINED ,
SUM(CASE ONLINE_TRANSACTION.STATUS_ID
WHEN 'STATUS-TRANS-FAILED' THEN 1
ELSE 0
END) AS FAILED ,
SUM(CASE ONLINE_TRANSACTION.STATUS_ID
WHEN 'STATUS-TRANS-PENDING-AUTH ' THEN 1
ELSE 0
END) AS PENDING_AUTH ,
SUM(CASE ONLINE_TRANSACTION.STATUS_ID
WHEN 'STATUS-TRANS-PENDING-POST ' THEN 1
ELSE 0
END) AS PENDING_POST ,
SUM(CASE ONLINE_TRANSACTION.STATUS_ID
WHEN 'STATUS-TRANS-PENDING' THEN 1
ELSE 0
END) AS PENDING ,
SUM(CASE ONLINE_TRANSACTION.STATUS_ID
WHEN 'ALPHA-STATUS-TRANS-POSTED' THEN 1
ELSE 0
END) AS POSTED
FROM
ONLINE_TRANSACTION
WHERE
MONTH(CREATED_ON) = '2'
GROUP BY
TRANSACTION_TYPE_ID ,
YEAR(CREATED_ON) ,
MONTH(CREATED_ON) ,
STATUS_ID
答案 0 :(得分：2)
我尝试了other answers只是为了我自己的启发，并认为我会分享结果，以防它帮助其他人。数据集很小，但显示了原则。
sqlite> SELECT * FROM transactions;
id type_id status_id
---------- ---------- ----------
1 insurance completed
2 insurance declined
3 cheque-stop completed
4 cheque-stop completed
sqlite> SELECT
...> type_id,
...> SUM(status_id == 'completed') AS completed,
...> SUM(status_id == 'declined') AS declined
...> FROM transactions
...> GROUP BY type_id;
type_id completed declined
----------- ---------- ----------
cheque-stop 2 0
insurance 1 1
答案 1 :(得分：0)
您的子查询中需要另一个WHERE子句链接TRANSACTION_TYPE_IDs。
SELECT DISTINCT
TRANSACTION_TYPE_ID, YEAR(CREATED_ON)AS YEAR, MONTH(CREATED_ON)AS MONTH ,
(SELECT Count(TRANSACTION_TYPE_ID) FROM ONLINE_TRANSACTION
WHERE ONLINE_TRANSACTION.STATUS_ID ='COMPLETED'AND MONTH(CREATED_ON)='2'
AND TRANSACTION_TYPE_ID=ot.TRANSACTION_TYPE_ID) AS COMPLETED,
(SELECT Count(TRANSACTION_TYPE_ID) FROM ONLINE_TRANSACTION
WHERE ONLINE_TRANSACTION.STATUS_ID ='DECLINED'AND MONTH(CREATED_ON)='2'
AND TRANSACTION_TYPE_ID=ot.TRANSACTION_TYPE_ID) AS DECLINED,
(SELECT Count(TRANSACTION_TYPE_ID) from ONLINE_TRANSACTION
WHERE ONLINE_TRANSACTION.STATUS_ID ='FAILED' AND MONTH(CREATED_ON)='2'
AND TRANSACTION_TYPE_ID=ot.TRANSACTION_TYPE_ID) AS FAILED,
(SELECT Count(TRANSACTION_TYPE_ID) from ONLINE_TRANSACTION
WHERE ONLINE_TRANSACTION.STATUS_ID ='PENDING_AUTH'AND MONTH(CREATED_ON)='2'
AND TRANSACTION_TYPE_ID=ot.TRANSACTION_TYPE_ID) AS PENDING_AUTH
FROM ONLINE_TRANSACTION ot
WHERE MONTH(CREATED_ON)='2'
GROUP BY TRANSACTION_TYPE_ID,ONLINE_TRANSACTION.CREATED_ON
答案 2 :(得分：0)
嵌套select语句中的where子句不区分不同的事务类型。数字总是总金额。
尝试查看总和和案例。
TRANSACTION_TYPE_ID, YEAR(CREATED_ON)AS YEAR, MONTH(CREATED_ON)AS MONTH ,
SUM(CASE ONLINE_TRANSACTION.STATUS_ID WHEN 'COMPLETED' THEN 1 ELSE 0 END) AS COMPLETED,
SUM(CASE ONLINE_TRANSACTION.STATUS_ID WHEN 'DECLINED' THEN 1 ELSE 0 END) AS DECLINED,
SUM(CASE ONLINE_TRANSACTION.STATUS_ID WHEN 'FAILED' THEN 1 ELSE 0 END) AS FAILED,
SUM(CASE ONLINE_TRANSACTION.STATUS_ID WHEN 'PENDING_AUTH' THEN 1 ELSE 0 END) AS PENDING_AUTH
FROM ONLINE_TRANSACTION WHERE MONTH(CREATED_ON)='2'
GROUP BY TRANSACTION_TYPE_ID
答案 3 :(得分：0)
您的主SELECT条款中的括号SELECT会回到ONLINE_TRANSACTION总表中，因此当然他们会在不考虑类型的情况下得到他们的计数，而是整体而言！
如何将SUM更改为STATUS_ID（通过相同的分组，因此不需要额外的语法）IF检查 - 取1表示相等，0表示差异（ bool到0/1 int转换在许多SQL方言中是自动的，在最坏的情况下，无论你选择何种方言，你都需要CASE...WHEN或CAST或{{1}}。< / p>
答案 4 :(得分：0)
SELECT TRANSACTION_TYPE_ID, YEAR(CREATED_ON) AS YEAR, MONTH(CREATED_ON) AS MONTH ,
SUM(CASE WHEN STATUS_ID = 'COMPLETED' THEN 1 ELSE 0 END) AS Completed,
SUM(CASE WHEN STATUS_ID = 'DECLINED' THEN 1 ELSE 0 END) AS Declined,
SUM(CASE WHEN STATUS_ID = 'FAILED' THEN 1 ELSE 0 END) AS Failed,
SUM(CASE WHEN STATUS_ID = 'PENDING_AUTH' THEN 1 ELSE 0 END) AS Pending_Auth
FROM ONLINE_TRANSACTION
WHERE MONTH(CREATED_ON) = 2
GROUP BY TRANSACTION_TYPE_ID, YEAR(CREATED_ON), MONTH(CREATED_ON)
看看这是否有意义。