SQL Server：在某些字符后检查大写或小写

Question

我有像'John is my name; Ram is my name; Adam is my name' 这样的数据。

我的规则是; 之后的每个第一个字母都应该是大写字母。

如何选择所有满足规则的值？

Answer 1

其他答案展示了如何将行转换为与您的模式匹配的内容。

如果您只想select 与您描述的模式匹配的行，您可以使用patindex() 或like 与区分大小写的排序规则（或使用collate 来应用）。

这假设除了分号后面的每个字母都必须是大写字母的规则之外，第一个字母也应该是大写字母。如果不是这样，只需删除where 中的第一个子句。

select *
from t
where patindex('[ABCDEFGHIJKLMNOPQRSTUVWXYZ]%', val collate latin1_general_cs_as) = 1
  and patindex('%; [^ABCDEFGHIJKLMNOPQRSTUVWXYZ]%', val collate latin1_general_cs_as) = 0

---

select *
from t
where val collate latin1_general_cs_as like '[ABCDEFGHIJKLMNOPQRSTUVWXYZ]%'
  and val collate latin1_general_cs_as not like '%; [^ABCDEFGHIJKLMNOPQRSTUVWXYZ]%'

---

测试设置：

create table t (id int not null identity(1,1),val varchar(256))
insert into t values 
 ('John is my name; Ram is my name; Adam is my name')
,('john is my name; ram is my name; adam is my name')

rextester 演示：http://rextester.com/DBGIS10645

以上都返回：

+----+--------------------------------------------------+
| id |                       val                        |
+----+--------------------------------------------------+
|  1 | John is my name; Ram is my name; Adam is my name |
+----+--------------------------------------------------+

Answer 2

你可以用这样的 XML 技巧来拆分它

DECLARE @YourString VARCHAR(100)='John is my name; Ram is my name; Adam is my name';
WITH Splitted AS
(
    SELECT CAST('<x>' + REPLACE((SELECT REPLACE(@YourString,'; ','$$SplitHere$$') AS [*] FOR XML PATH('')),'$$SplitHere$$','</x><x>')+ '</x>' AS XML) AS Casted
)
,DerivedTable AS
(
    SELECT  ROW_NUMBER() OVER(ORDER BY (SELECT NULL)) AS PartNr
           ,x.value(N'text()[1]',N'nvarchar(max)') AS Part
    FROM Splitted
    CROSS APPLY Casted.nodes(N'/x') AS X(x)
)
SELECT PartNr
      ,Part
      ,CASE WHEN ASCII(LEFT(Part,1)) BETWEEN ASCII('A') AND ASCII('Z') THEN 1 ELSE 0 END AS FirstIsCapital
FROM DerivedTable;

结果

Nr  Part            FirstLetterIsCaptial
---------------------------------------- 
1   John is my name     1
2   Ram is my name      1
3   Adam is my name     1

我不知道你的最终目标是什么...查找零件，第一个字母不是大写的？确保您的规则得到满足？

但是：
最好的办法是使用它来纠正您的设计并将这些部件放置在与1:n 相关的边桌中。

Answer 3

有点丑陋的解决方案，但你可以试一试......

Declare @str nvarchar(max) = 'John is my name; Ram is my name; Adam is my name'

Declare @xml as xml
Set @xml = cast(('<X>'+replace(@str,';' ,'</X><X>')+'</X>') as xml)
Select * from (
    Select RowN = Row_Number() over (order by (SELECT NULL)), LTrim(RTrim(N.value('.', 'nvarchar(MAX)'))) as value FROM @xml.nodes('X') as T(N) -- this is to split if you are using sql server 2016 you can use string_Split
) a
Where unicode(substring(a.[value],1,1)) = unicode(upper(substring(a.[value],1,1)))

想法是拆分字符串并用unicode值检查它是否在上面

Answer 4

你可以创建一个这样的函数。

Create FUNCTION SPLITTER (    
    @textData NVARCHAR(MAX),
    @Delimeter NVARCHAR(MAX) ) RETURNS @RtnValue TABLE  (
    Data NVARCHAR(MAX) )  AS BEGIN 
    DECLARE @index INT  DECLARE @data nvarchar(1000)    DECLARE @firstCharacter char
    SET @index = CHARINDEX(@Delimeter,@textData)

    WHILE (@index>0)
    BEGIN
                set @data = LTRIM(RTRIM(SUBSTRING(@textData, 1, @index - 1)))       set @firstCharacter = SUBSTRING(@data,1,1);
                if UNICODE(@firstCharacter) = UNICODE(upper(@firstCharacter))       begin           INSERT INTO @RtnValue (data) SELECT @data       end;

        SET @textData = SUBSTRING(@textData, @index + DATALENGTH(@Delimeter) / 2, LEN(@textData))

        SET @index = CHARINDEX(@Delimeter, @textData)
    END
        set @data = @textData   set @firstCharacter = SUBSTRING(@data,1,1);
            if UNICODE(@firstCharacter) = UNICODE(upper(@firstCharacter))   begin       INSERT INTO @RtnValue (data) SELECT @data   end;

    RETURN END

这样使用

select * from SPLITTER ('John 是我的名字; Ram 是我的名字; Adam 是我的名字', ';')

Answer 5

您可以获取NGrams8K 的副本并执行以下操作：

-- note that I made the 3rd item start with lower-case
DECLARE @YourString VARCHAR(100)='John is my name; Ram is my name; adam is my name';

WITH D(n) AS
(
  SELECT 0 UNION ALL SELECT position 
  FROM dbo.NGrams8k(@yourstring,1) WHERE token = ';'
),
TOKEN(token) AS
(
  SELECT LTRIM(SUBSTRING(@YourString, N+1, 
           ISNULL(NULLIF(CHARINDEX(';', @YourString, N+1),0), 101)-(N+1)))
  FROM D
)
SELECT token,
       FirstLetterIsCaptial = IIF(ASCII(SUBSTRING(token,1,1)) BETWEEN 65 AND 90, 1, 0)
FROM TOKEN;

结果

token              FirstLetterIsCaptial
------------------ --------------------
John is my name    1
Ram is my name     1
adam is my name    0