c# - 将其转换为 SQL 的最佳方法是什么

Question

我知道对于这里的数据库大师来说，这应该是轻而易举的事。我的数据库中有一个字段，格式为 ' A/B/C/D/E/F '

格式是无关紧要的，我通常需要最后两部分，所以上面的部分是

'EF'

但如果我有另一个字符串

AB/CD/EF/GH == EFGH

我希望让最后两部分像这样“EFGH”返回

有谁知道我可以使用的 SQL 函数来拆分它

我正在使用 Microsoft SQL Server 2012 - 希望这对您有所帮助，

这是 C# 代码。

var myText = "A/B/C/D/E/F";
var identificationArray = myText.Split('/');

if(identificationArray.Length >= 2)
{
    var friendlyId = identificationArray[identificationArray.Length - 2] + identificationArray[identificationArray.Length - 1];

    return friendlyId;
}
return "";

Answer 1

这是一个答案，它以相反的顺序在字符串中搜索第二个正斜杠并返回删除正斜杠的子字符串:

declare @s varchar(20)
set @s = 'A/B/C/D/E/F'

-- result: 'EF'
select reverse(replace(left(reverse(@s), charindex('/', reverse(@s), charindex('/', reverse(@s)) + 1)), '/', ''))

set @s = 'AB/CD/EF/GH'

-- result: 'EFGH'
select reverse(replace(left(reverse(@s), charindex('/', reverse(@s), charindex('/', reverse(@s)) + 1)), '/', ''))

用其他几个输入测试这个:

set @s = '/AB/CD' -- result: 'ABCD'
set @s = 'AB/CD'  -- result: an empty string '' -- you may not want this result
set @s = 'AB'     -- result: an empty string ''

这是一种复杂得离谱的方法，可以用一系列公用表表达式 (CTE) 来做同样的事情。归功于Itzik Ben-Gan对于使用交叉连接生成计数表的 CTE 技术:

declare @s varchar(50)
set @s = 'A/B/C/D/E/F/G'
--set @s = 'AB/CD/EF/GH'
--set @s = 'AB/CD'
--set @s = 'ABCD/EFGH/IJKL'
--set @s = 'A/B'
-- set @s = 'A'

declare @result varchar(50)
set @result = ''

;with
-- cross-join a meaningless set of data together to create a lot of rows
Nbrs_2    (n) AS (SELECT 1 UNION SELECT 0 ),
Nbrs_4    (n) AS (SELECT 1 FROM Nbrs_2     n1 CROSS JOIN Nbrs_2     n2),
Nbrs_16   (n) AS (SELECT 1 FROM Nbrs_4     n1 CROSS JOIN Nbrs_4     n2),
Nbrs_256  (n) AS (SELECT 1 FROM Nbrs_16    n1 CROSS JOIN Nbrs_16    n2),
Nbrs_65536(n) AS (SELECT 1 FROM Nbrs_256   n1 CROSS JOIN Nbrs_256   n2),
Nbrs      (n) AS (SELECT 1 FROM Nbrs_65536 n1 CROSS JOIN Nbrs_65536 n2),
-- build a table of numbers from the data above; this is insanely fast
nums(n) as
(
   select row_number() over(order by n) from Nbrs
),
-- split the string into separate rows per letter
letters(n, c) as
(
    select n, substring(@s, n, 1)
    from nums
    where n < len(@s) + 1
),
-- count the slashes from the rows in descending order
-- the important slash is the second one from the end
slashes(n, num) as
(
    select n, ROW_NUMBER() over (order by n desc)
    from letters
    where c = '/'
)
select @result = @result + c
from letters
where n > (select n from slashes where num = 2) -- get everything after the second slash
and c <> '/' -- and drop out the other slash

select @result