具有包含以下行的交易表:
Id UserId PlatformId TransactionTypeId
-------------------------------------------------
0 1 3 1
1 1 1 2
2 2 3 2
3 3 2 1
4 2 3 1
如何编写一个存储过程,该存储过程可以将行聚合为具有以下格式的新表?
Id UserId Platforms TransactionTypeId
-------------------------------------------------
0 1 {"p3":1,"p1":1} {"t1":1,"t2":1}
1 2 {"p3":2} {"t2":1,"t1":1}
3 3 {"p2":1} {"t1":1}
因此,这些行将由用户汇总,计算每个平台/ transactionType并存储为键/值json字符串。
答案 0 :(得分:1)
您可以使用GROUP BY
和FOR JSON
:
SELECT MIN(ID) AS ID, UserId, MIN(sub.x) AS Platforms, MIN(sub2.x) AS Transactions
FROM tab t
OUTER APPLY (SELECT CONCAT('p', platformId) AS platform, cnt
FROM (SELECT PlatformId, COUNT(*) AS cnt
FROM tab t2 WHERE t2.UserId = t.UserId
GROUP BY PlatformId) s
FOR JSON AUTO) sub(x)
OUTER APPLY (SELECT CONCAT('t', TransactiontypeId) AS Transactions, cnt
FROM (SELECT TransactiontypeId, COUNT(*) AS cnt
FROM tab t2 WHERE t2.UserId = t.UserId
GROUP BY TransactiontypeId) s
FOR JSON AUTO) sub2(x)
GROUP BY UserId;
结果有点不同(键值数组),但请以起点为准。
答案 1 :(得分:1)
您的示例JSON并不是真正的json,但由于您希望这样做:
SELECT u.UserId, plt.pValue, ttyp.ttValue
FROM Users AS [u]
CROSS APPLY (
SELECT '{'+STUFF( (SELECT ',"'+pn.pName+'":'+LTRIM(STR(pn.pCount))
FROM (SELECT p.Name AS pName, COUNT(*) AS pCount
FROM transactions t
left JOIN Platforms p ON p.PlatformID = t.PlatformId
WHERE t.UserId = u.UserId
GROUP BY p.PlatformId, p.Name
) pn
FOR XML PATH('')),1,1,'')+'}'
) plt(pValue)
CROSS APPLY (
SELECT '{'+STUFF( (SELECT ',"'+tty.ttName+'":'+LTRIM(STR(tty.ttCount))
FROM (SELECT tt.Name AS ttName, COUNT(*) AS ttCount
FROM transactions t
left JOIN dbo.TransactionType tt ON tt.TransactionTypeId = t.TransactionTypeID
WHERE t.UserId = u.UserId
GROUP BY tt.TransactionTypeId, tt.Name
) tty
FOR XML PATH('')),1,1,'')+'}'
) ttyp(ttValue)
WHERE EXISTS (SELECT * FROM transactions t WHERE u.UserId = t.UserId)
ORDER BY UserId;