這是SQL Server的解決方案。這個想法是使用一個相關的子查詢來確定 每個年份和月份的最小值RecordTime
。將這個想法轉化爲DBMS的風格應該很容易。
WITH Data AS (
SELECT CONVERT(DATETIME, '1-May-2011 01:00:00') AS RecordTime, 1.23 AS FormattedValue
UNION ALL SELECT CONVERT(DATETIME, '1-May-2011 01:12:34'), 0.99
UNION ALL SELECT CONVERT(DATETIME, '1-May-2011 23:59:59'), 2.34
UNION ALL SELECT CONVERT(DATETIME, '2-May-2011 00:11:22'), 4.56
)
SELECT D1.*
FROM Data D1
WHERE D1.RecordTime = (SELECT MIN(RecordTime)
FROM Data D2
WHERE YEAR(D2.RecordTime) = YEAR(D1.RecordTime)
AND MONTH(D2.RecordTime) = MONTH(D1.RecordTime))
編輯:如果數據集較大,則加入了關於子查詢要快得多,但可能較難翻譯成你的DBMS的味道:
SELECT D1.*
FROM Data D1
INNER JOIN (
SELECT MIN(D2.RecordTime) AS FirstMonthlyRecordTime
FROM Data D2
GROUP BY YEAR(D2.RecordTime), MONTH(D2.RecordTime)
) AS SubQuery
ON D1.RecordTime = SubQuery.FirstMonthlyRecordTime
編輯2:以下代碼是我上面第一個查詢的標準SQL-92版本(在mySQL上測試過):
SELECT T1.RecordTime, T1.FormattedValue
FROM CDBHistoric T1
WHERE T1.RecordTime = (SELECT MIN(T2.RecordTime)
FROM CDBHistoric T2
WHERE EXTRACT(YEAR FROM T1.RecordTime) = EXTRACT(YEAR FROM T2.RecordTime)
AND EXTRACT(MONTH FROM T1.RecordTime) = EXTRACT(MONTH FROM T2.RecordTime))
你使用Oracle?你只是想每天獲得第一個記錄嗎? – 2011-05-06 02:10:46
什麼DBMS? SQL Server中的答案與Oracle中的答案大不相同,它與MySQL中的答案大不相同......「SQL」不足以提供答案。 – 2011-05-06 02:13:23
@Ken White:代碼是標準SQL-92,所以我認爲'sql'標記就足夠了(即爲什麼不使用標準SQL-92發佈答案?) – onedaywhen 2011-05-06 07:09:17