Question

我有一张表格，根据数字显示评分数字，如下所示：

0-24 =新手
25-49 =实习生
50-74 =中等
75-99 = Pro
100 =超人

我需要取一个数值，并将其与上面最接近的等效等级相匹配。因此， 80 等同于 Pro ，因为它符合基本 Pro 级别，但低于基础超人级别。

我已经编写了一些代码来测试它，但TOP 1过滤器似乎在JOIN之前应用，这意味着我总是看到最低等级而不是实际等级。任何人都可以帮我提出更好的选择吗？

代码在这里：

DECLARE @tbl_levels TABLE ([schemeID] int, [calevel] numeric(5,2), [desc] nvarchar(200))
DECLARE @level numeric(5,2) = 70
DECLARE @schemeID int = 1

insert into @tbl_levels ([schemeID],[calevel],[desc]) VALUES (1, 0, 'Newbie')
insert into @tbl_levels ([schemeID],[calevel],[desc]) VALUES (1, 25, 'Trainee')
insert into @tbl_levels ([schemeID],[calevel],[desc]) VALUES (1, 50, 'Moderate')
insert into @tbl_levels ([schemeID],[calevel],[desc]) VALUES (1, 75, 'Pro')
insert into @tbl_levels ([schemeID],[calevel],[desc]) VALUES (1, 100, 'Superman')

SELECT
    *
FROM
    (SELECT
        @level as [level],
        @schemeID as [schemeID]) ca LEFT OUTER JOIN 
    (SELECT 
        TOP 1 * 
    FROM 
        @tbl_levels
    ORDER BY 
        calevel ASC
    ) lvl ON lvl.schemeID = ca.schemeID AND lvl.calevel <= ca.[level]

Answer 1

SELECT @level as Level,s.*
from @tbl_levels s
where [calevel]=(Select Max([calevel]) from @tbl_levels where [calevel]<=@level)

作为对你的评论的回答

SELECT
    t.[level],
    (Select [desc] from @tbl_levels where [calevel]=(Select Max([calevel]) from @tbl_levels where [calevel]<=t.[level])) as [Desc]
FROM
    @test t

Answer 2

您可以将问题重新定义为“匹配给定值的最高级别是多少？”然后解决方案变得明显。 70级是'中等'：

SELECT TOP(1) @level, *
FROM @tbl_levels
WHERE [calevel] <= @level
ORDER BY calevel DESC;

对于多值，只需使用CROSS APPLY：

SELECT
    t.[level],
    lvl.[desc]
FROM @test t 
cross apply (
select top(1) *
from @tbl_levels
where [calevel] <= [level]
order by calevel desc) as lvl;

Answer 3

（更新）尝试：

;with r as 
(select l.*, row_number() over (partition by [schemeID] order by [calevel]) rn
 from @tbl_levels l),
cte as
(select l.*, n.[calevel] as [nextlevel]
 from r l 
 left join r n on l.[schemeID] = n.[schemeID] and l.rn+1 = n.rn)
SELECT
    ca.*, cte.[calevel],cte.[nextlevel],cte.[desc]
FROM @test ca 
LEFT OUTER JOIN cte
ON ca.schemeID = cte.schemeID AND 
   ca.[level] >= cte.[calevel] AND
   ca.[level] < coalesce(cte.[nextlevel], ca.[level]+1)

（SQLFiddle here）

Answer 4

DECLARE @tbl_levels TABLE ([schemeID] int, [calevel] numeric(5,2), [desc] nvarchar(200))
DECLARE @level numeric(5,2) = 70
DECLARE @schemeID int = 1

insert into @tbl_levels ([schemeID],[calevel],[desc]) VALUES (1, 0, 'Newbie')
insert into @tbl_levels ([schemeID],[calevel],[desc]) VALUES (1, 25, 'Trainee')
insert into @tbl_levels ([schemeID],[calevel],[desc]) VALUES (1, 50, 'Moderate')
insert into @tbl_levels ([schemeID],[calevel],[desc]) VALUES (1, 75, 'Pro')
insert into @tbl_levels ([schemeID],[calevel],[desc]) VALUES (1, 100, 'Superman')

SELECT
    TOP 1 *
FROM
    (SELECT
        @level as [level],
        @schemeID as [schemeID]) ca LEFT OUTER JOIN 
    (SELECT *
        FROM 
        @tbl_levels
    ) lvl ON lvl.schemeID = ca.schemeID AND lvl.calevel <= ca.[level]
    ORDER BY lvl.calevel DESC

拉吉

在SQL Server中匹配单个数字到最接近范围

4 个答案: