做一個自動任務,需要查詢指定期限內(nèi)數(shù)據(jù),比如按照自然月,按照自然日,按照自然周,需要支持Oracle 和 SqlServer數(shù)據(jù)庫。
數(shù)據(jù)庫中表字段receivetime,由C++負責寫入,內(nèi)容為time_t的32位值,也就是1970年到現(xiàn)在的秒數(shù)。
研究了一天,得出如下語句:
對于SqlServer,
查詢前一天的數(shù)據(jù),
select * from table_name where datediff ( day, dateadd(day, (ReceiveTime+8*3600)/60/60/24 - 1, '19700101'), getdate() ) = 2;
請關(guān)注其中的8*3600,這是因為,SqlServer中,按照UTC計算標準時間,(ReceiveTime)/60/60/24計算的標準時間,但是數(shù)據(jù)庫服務器時區(qū)是東八區(qū),故需要加上這個值。
以下同理,不解釋。
查詢前一周的數(shù)據(jù),
查詢前一周
SELECT *
FROM tablename
WHERE datediff
(day,
dateadd
(day,
(ReceiveTime + 8 * 3600) / 60 / 60 / 24,
'1970-01-01 00:00:00'),
getdate ()) >= DatePart (Weekday, getdate ()) + 1
and
datediff
(day,
dateadd
(day,
(ReceiveTime + 8 * 3600) / 60 / 60 / 24,
'1970-01-01 00:00:00'),
getdate ()) <= DatePart (Weekday, getdate ()) + 8;
查詢前一月的數(shù)據(jù),
查詢前一月
SELECT *
FROM tablename
WHERE datediff
(month,
dateadd
(day,
(ReceiveTime + 8 * 3600) / 60 / 60 / 24,
'1970-01-01 00:00:00'),
getdate ()) = 1;
對于Oracle數(shù)據(jù)庫
查詢前一天:
查詢前一天
SELECT count(*)
FROM tablename
WHERE receivetime >=
((TRUNC (SYSDATE - 1, 'dd') - TO_DATE ('19700101', 'yyyymmdd')
)
* 24
* 3600 - 8 * 3600)
AND receivetime <
( (TRUNC (SYSDATE, 'dd') - TO_DATE ('19700101', 'yyyymmdd')
) * 24 * 3600 - 8 * 3600);
查詢前一周:
查詢前一周
SELECT count(*)
FROM tablename
WHERE receivetime <
( ( TRUNC (SYSDATE - TO_NUMBER (TO_CHAR (SYSDATE, 'W')) , 'dd')
- TO_DATE ('19700101', 'yyyymmdd')
)
* 24
* 3600 - 8 * 3600 )
AND receivetime >=
(( TRUNC (SYSDATE - TO_NUMBER (TO_CHAR (SYSDATE, 'W')) - 7, 'dd')
- TO_DATE ('19700101', 'yyyymmdd')
)
* 24
* 3600 - 8 * 3600 );
查詢前一月:
查詢前一月
SELECT count(*)
FROM tablename
WHERE receivetime >=
( ((trunc(last_day( add_months(sysdate, -2 )),'dd') + 1) - TO_DATE ('19700101', 'yyyymmdd')) * 24 * 3600 - 8 * 3600 )
AND receivetime < (((trunc(last_day( add_months(sysdate, -1 )),'dd') + 1) - TO_DATE ('19700101', 'yyyymmdd')) * 24 * 3600 - 8 * 3600);