我查看了很多页面,但找不到我想要达到的目标。我有一个select语句,我希望得到的最新日期大于例如:
如果我有2015/06/01,2015/07/01的两个日期,我的表中有一条记录的日期为2015/05/01,另一条记录的日期为2015/06/20。
对于2015/06/01,我希望它的最大日期大于2015/05/01。
对于2015/07/01,我希望它的最大日期大于,在这种情况下,它将是2015/06/20。
我只希望一个选项不返回我目前获得的所有匹配项。
其他表还有其他连接等,它会使用所有这些,但为了这个目的,我已经取出了这些信息,因为它与我想要实现的目标无关。如果您需要更多信息,请发表评论。
评论之后我创建了一个Fiddle数据库结构http://sqlfiddle.com/#!6/2738f/2/0。
所以这是我当前的查询:
With IptBandRates as (
select
ibr.ipt_band_code,
ibr.ipt_rate,
ibr.concessionary_date,
max(ibr.concessionary_date) as concessionary_date2
from product.ipt_band_rates ibr
inner join quote.quote qu on '4' = qu.quote_id
where ibr.concessionary_date <= qu.[effective_date]
group by ibr.concessionary_date, ibr.ipt_rate, ipt_band_code
)
select
IptBandRates.ipt_rate as ipt_rate
from quote.premium
inner join quote.cover on quote.premium.cover_id = quote.cover.cover_id
inner join quote.quote qu on quote.cover.quote_id = qu.quote_id
inner join product.premium_class on
quote.cover.product_code = product.premium_class.product_code and
quote.cover.product_ver_no = product.premium_class.product_ver_no and
quote.cover.class_type_code = product.premium_class.class_type_code
inner join product.ipt_band on
product.premium_class.ipt_band_code = product.ipt_band.ipt_band_code
inner join IptBandRates on
product.ipt_band.ipt_band_code = IptBandRates.ipt_band_code
inner join product.ipt_band_rates ibrs on
ibrs.ipt_band_code = IptBandRates.ipt_band_code and
ibrs.concessionary_date = IptBandRates.concessionary_date
where qu.quote_id = '4'
答案 0 :(得分:0)
x 是包含日期的表格, y 仅是指获取每月第1天的日期以提取小于...的最大日期的参考。 / p>
;with
x as (
select 1 id, '2015/05/01' d
union all
select 2 id, '2015/06/20' d
union all
select 3 id, '2015/06/15' d
union all
select 4 id, '2015/05/10' d
union all
select 5 id, '2015/07/02' d
union all
select 6 id, '2015/04/02' d
),
y as (
select '2015/05/01' ref_month
union all
select '2015/06/01'
union all
select '2015/07/01'
)
SELECT *
FROM y
outer apply (
select MAX(d) max_date
from x
where d < y.ref_month
) z
结果将是:
ref_month max_date
2015/05/01 2015/04/02
2015/06/01 2015/05/10
2015/07/01 2015/06/20
答案 1 :(得分:0)
我知道这是旧的,但这是一种做法。
IF OBJECT_ID('tempdb..#FilteringDates') IS NOT NULL
DROP TABLE #FilteringDates
CREATE TABLE #FilteringDates (filterDate DATE)
INSERT INTO #FilteringDates (
filterDate)
VALUES
('2018-01-01'),
('2018-02-15'),
('2018-03-20')
IF OBJECT_ID('tempdb..#DataDates') IS NOT NULL
DROP TABLE #DataDates
CREATE TABLE #DataDates (
dataDate DATE,
otherValue INT)
INSERT INTO #DataDates (
dataDate,
otherValue)
VALUES
('2017-06-01', 1),
('2018-01-03', 90),
('2018-01-15', 150),
('2018-01-20', 35),
('2018-02-20', 400),
('2018-03-10', 425),
('2018-04-20', 3)
;WITH DateRankings AS
(
SELECT
F.filterDate,
D.dataDate,
D.otherValue,
MostRecentRanking = ROW_NUMBER() OVER (
PARTITION BY
F.filterDate
ORDER BY
D.dataDate DESC)
FROM
#FilteringDates AS F
INNER JOIN #DataDates AS D ON D.dataDate <= F.filterDate
)
SELECT
D.filterDate,
MostRecentDate = D.dataDate,
D.otherValue
FROM
DateRankings AS D
WHERE
D.MostRecentRanking = 1