Sql server 在大写字母出现或真正需要空格的地方，如何在字符串中预加空格

在大写字母出现或真正需要空格的地方，如何在字符串中预加空格

示例代码是：

DECLARE @teams TABLE (Team NVARCHAR(100))

INSERT INTO @teams
SELECT 'TataConsultencyServices'
UNION ALL
SELECT 'TataConsultencyCompany'
UNION ALL
SELECT 'CompanyHumanResource'

预期结果

Tata Consultency Services
Tata Consultency Company
Company Human Resource

---

您肯定会从以下方面获得一些帮助：-

现在可以用sql将字符串值从行连接到单列值。 我们可以只使用任何sql连接函数

declare @str nvarchar(max) = N'ThisIsATest'
SELECT LTRIM(STUFF((
  SELECT ' ' + val FROM dbo.CaseSensitiveSQLSplitFunction(@str) FOR XML PATH('')
  ), 1, 1, '')) string

set @str = N'ThisIsASqlServerCaseSensitiveSplitStringFunction'
SELECT LTRIM(STUFF((
  SELECT ' ' + val FROM dbo.CaseSensitiveSQLSplitFunction(@str) FOR XML PATH('')
  ), 1, 1, '')) string

基于集合的解决方案：

DECLARE @s NVARCHAR(100);
SET @s = 'CompanyHumanResources';

DECLARE @Idx INT = 1;
WITH CteRecursive
AS
(
    SELECT  1 AS Idx, 
            CONVERT(NVARCHAR(200), @s) AS String
    UNION ALL
    SELECT  src.Idx + src.IsUpper + 1,
            CONVERT(NVARCHAR(200),
                CASE WHEN src.IsUpper = 1 THEN STUFF(src.String, src.Idx+1, 0, ' ') ELSE src.String END
            )
    FROM
    (
        SELECT  rec.*,
                CASE WHEN SUBSTRING(rec.String, rec.Idx, 1) <> ' ' AND SUBSTRING(rec.String, rec.Idx+1, 1) LIKE '[A-Z]' AND SUBSTRING(rec.String, rec.Idx+1, 1) COLLATE Romanian_CS_AS = UPPER(SUBSTRING(rec.String, rec.Idx+1, 1)) COLLATE Romanian_CS_AS THEN 1 ELSE 0 END AS IsUpper
        FROM    CteRecursive rec
        WHERE   rec.Idx + 1 <= LEN(rec.String)
    ) src
)
SELECT  TOP(1) x.String
FROM    CteRecursive x
ORDER BY x.Idx DESC;

而1=1 开始 更新@teams SET TeamName=STUFF（TeamName，patindex（'%[a-z，.][a-z]]'，TeamName COLLATE拉丁语1\u General\u BIN）+1,0'，） 其中patindex（'%[a-z，.][a-z]'、TeamName COLLATE Latin1\u General\u BIN）>0

结束

更新@teams SET TeamName=STUFF（TeamName，patindex（'%[A-Z][A-Z]]'，右（TeamName，LEN（TeamName）-1）对照拉丁语1\u General\u BIN）+1,0'，）

---

如果patindex（'%[A-Z][A-Z]'，TeamName COLLATE Latin1\u General\u BIN）>0

您应该用区分大小写的（

…\u CS\u…

）替换我的排序规则（

）排序规则。很好，但我不得不说，我不能自己做这个查询逻辑，所以我正在搜索新的简单逻辑，如果我能得到它，就会发布它，但这对“TCSolutions”这样的参数不起作用，它会返回像TCSolutions这样的结果，这实际上是不需要的。这是缩进行为。你的问题中没有这样的要求。谢谢你，伙计，这有助于你了解更多关于区分大小写的事情。欢迎你。如果姚认为答案值得的话，请投上一票
declare @str nvarchar(max) = N'ThisIsATest'
SELECT LTRIM(STUFF((
  SELECT ' ' + val FROM dbo.CaseSensitiveSQLSplitFunction(@str) FOR XML PATH('')
  ), 1, 1, '')) string

set @str = N'ThisIsASqlServerCaseSensitiveSplitStringFunction'
SELECT LTRIM(STUFF((
  SELECT ' ' + val FROM dbo.CaseSensitiveSQLSplitFunction(@str) FOR XML PATH('')
  ), 1, 1, '')) string

DECLARE @s NVARCHAR(100);
SET @s = 'CompanyHumanResources';

DECLARE @Idx INT = 1;
WITH CteRecursive
AS
(
    SELECT  1 AS Idx, 
            CONVERT(NVARCHAR(200), @s) AS String
    UNION ALL
    SELECT  src.Idx + src.IsUpper + 1,
            CONVERT(NVARCHAR(200),
                CASE WHEN src.IsUpper = 1 THEN STUFF(src.String, src.Idx+1, 0, ' ') ELSE src.String END
            )
    FROM
    (
        SELECT  rec.*,
                CASE WHEN SUBSTRING(rec.String, rec.Idx, 1) <> ' ' AND SUBSTRING(rec.String, rec.Idx+1, 1) LIKE '[A-Z]' AND SUBSTRING(rec.String, rec.Idx+1, 1) COLLATE Romanian_CS_AS = UPPER(SUBSTRING(rec.String, rec.Idx+1, 1)) COLLATE Romanian_CS_AS THEN 1 ELSE 0 END AS IsUpper
        FROM    CteRecursive rec
        WHERE   rec.Idx + 1 <= LEN(rec.String)
    ) src
)
SELECT  TOP(1) x.String
FROM    CteRecursive x
ORDER BY x.Idx DESC;

String
-----------------------
Company Human Resources

          IF @@ROWCOUNT = 0 BREAK