Question

我需要获得第1和第1个名字的名称。姓名各部分的最后一个字母（名字，中间名，姓氏）匹配。名称可以有5个部分。如果名称是：

croc lil brob tart
cron lee bro tasr

应该结果：

croc lil brob tart

Declare @name as varchar(70) = 'cron lee bro tasr test'
    Select @name, Substring(@name,1,1) As "First letter of First Name",
    Substring(@name,charindex(' ',@name) +1 -2,1) As "Last letter of First Name",
    Substring(@name,charindex(' ',@name) +1 ,1) As "First letter of Second Name",
    Substring(@name,charindex(' ',@name, charindex(' ',@name) + 1) +1 -2 ,1) As "Last letter of Second Name",
    Substring(@name,charindex(' ',@name, charindex(' ',@name) + 1) +1 ,1) As "First letter of Third Name",
    Substring(@name,charindex(' ',@name, charindex(' ',@name,charindex(' ',@name) + 1) + 1 +1)+1 -2 ,1) AS "Last Letter of 3rd Name",
    Substring(@name,charindex(' ',@name, charindex(' ',@name,charindex(' ',@name) + 1) + 1 +1)+1 ,1) AS "First Letter of 4th Name",
    Substring(@name,charindex(' ',@name,charindex(' ',@name, charindex(' ',@name,charindex(' ',@name) + 1) + 1 +1)+1)+1 -2 ,1) AS "Last Letter of 4th Name",
    Substring(@name,charindex(' ',@name,charindex(' ',@name, charindex(' ',@name,charindex(' ',@name) + 1) + 1 +1)+1)+1 ,1) AS "first Letter of 5th Name",
    SUBSTRING(Reverse(@name),1,1) AS "Last letter of 5th Name" WHERE 
    Substring(RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),1,1) = Substring(RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1)) +1 -2,1) OR 
    Substring(RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1)) +1 ,1) = Substring(RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1), charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1)) + 1) +1 -2 ,1) OR 
    Substring(RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1), charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1)) + 1) +1 ,1) = Substring(RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1), charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1)) + 1) + 1 +1)+1 -2 ,1) OR
    Substring(RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1), charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1)) + 1) + 1 +1)+1 ,1) = 
    Substring(RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1), charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1)) + 1) + 1 +1)+1)+1 -2 ,1) OR
    Substring(RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1), charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1),charindex(' ',RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1)) + 1) + 1 +1)+1)+1 ,1) = SUBSTRING(Reverse(RIGHT(@name, CHARINDEX('.', REVERSE('.' + @name)) - 1)),1,1)

通过这个选择查询，我能够得到第一个＆amp;名称各部分的最后一个字符。但是，我对where子句感到困惑。

Answer 1

试试这个：

DECLARE @t TABLE(id INT, t VARCHAR(100))

INSERT INTO @t VALUES
(1, 'croc lil brob tart'),
(2, 'cron lee bro tasr')


;WITH cte AS(
SELECT id,
       Split.a.value('.', 'VARCHAR(100)') AS t  
FROM  
(   
    SELECT id, CAST ('<M>' + REPLACE(t, ' ', '</M><M>') + '</M>' AS XML ) AS t FROM @t
) AS A CROSS APPLY t.nodes ('/M') AS Split(a))

SELECT * FROM @t t 
WHERE NOT EXISTS(SELECT * FROM cte WHERE id = t.id AND LEFT(t, 1) <> RIGHT(t, 1))

输出：

id  t
1   croc lil brob tart

Answer 2

你可以创建一个分裂函数，然后它相对容易：

SELECT * FROM dbo.TableName t
WHERE NOT EXISTS
(
   SELECT 1 FROM  dbo.Split(t.Col, ' ')x
   WHERE LEFT(x.Item, 1) <> RIGHT(x.Item, 1)
)

Demo

这是我使用的一个：

CREATE FUNCTION [dbo].[Split]
(
    @ItemList NVARCHAR(MAX), 
    @delimiter CHAR(1)
)
RETURNS @ItemTable TABLE (Item VARCHAR(250))  
AS      

BEGIN    
    DECLARE @tempItemList NVARCHAR(MAX)
    SET @tempItemList = @ItemList

    DECLARE @i INT    
    DECLARE @Item NVARCHAR(4000)

    SET @i = CHARINDEX(@delimiter, @tempItemList)

    WHILE (LEN(@tempItemList) > 0)
    BEGIN
        IF @i = 0
            SET @Item = @tempItemList
        ELSE
            SET @Item = LEFT(@tempItemList, @i - 1)
        INSERT INTO @ItemTable(Item) VALUES(@Item)
        IF @i = 0
            SET @tempItemList = ''
        ELSE
            SET @tempItemList = RIGHT(@tempItemList, LEN(@tempItemList) - @i)
        SET @i = CHARINDEX(@delimiter, @tempItemList)
    END 
    RETURN
END

检查字符串中每个单词的第一个和最后一个字符是否相同

2 个答案: