Question

我的数据如下

98-45.3A-22
104-44.0A-23
00983-29.1-22
01757-42.5A-22
04968-37.3A2-23

输出在SQL Server中查找输出如下

00098-BA45.3A-IN-22
00104-BA44.0A-IN-23
00983-BA29.1-IN-22
01757-BA42.5A-IN-22
04968-BA37.3A2-IN-23

Answer 1

我拆分了部件以应对棘手的数据模板。这应该适用于非破折号2位数的尾部：

WITH Src AS
(
    SELECT * FROM (VALUES
    ('98-45.3A-22'),
    ('104-44.0A-23'),
    ('00983-29.1-22'),
    ('01757-42.5A-22'),
    ('04968-37.3A2-23')
    ) T(X)
), Parts AS
(
    SELECT *,
    RIGHT('00000'+SUBSTRING(X, 1, CHARINDEX('-',X, 1)-1),5) Front,
    'BA'+SUBSTRING(X, CHARINDEX('-',X, 1)+1, 2) BA,
    SUBSTRING(X, PATINDEX('%.%',X), LEN(X)-CHARINDEX('-', REVERSE(X), 1)-PATINDEX('%.%',X)+1)  P,
    SUBSTRING(X, LEN(X)-CHARINDEX('-', REVERSE(X), 1)+1, LEN(X)) En
    FROM Src
)
SELECT Front+'-'+BA+P+'-IN'+En
FROM Parts

它返回：

00098-BA45.3A-IN-22
00104-BA44.0A-IN-23
00983-BA29.1-IN-22
01757-BA42.5A-IN-22
04968-BA37.3A2-IN-23

Answer 2

试试这个，

DECLARE @String VARCHAR(100) = '98-45.3A-22'
SELECT  ISNULL(REPLICATE('0',6 - CHARINDEX('-',@String)),'')        -- Add leading Zeros
        + STUFF(
                STUFF(@String,CHARINDEX('-',@String),1,'-BA'),      -- Add 'BA'
                CHARINDEX('-',@String,CHARINDEX('-',@String)+1)+2,  -- 2 additional for the character 'BA' 
                1,'-IN')    --  Add 'IN'

如果我在第一个连字符之前有超过6位数字并且想要删除前导零以使其为6位数，该怎么办。

DECLARE @String VARCHAR(100) = '0000098-45.3A-22'
SELECT  CASE WHEN CHARINDEX('-',@String) <= 6 
            THEN ISNULL(REPLICATE('0',6 - CHARINDEX('-',@String)),'')        -- Add leading Zeros
                + STUFF(
                        STUFF(  @String,CHARINDEX('-',@String),1,'-BA'),      -- Add 'BA'
                        CHARINDEX('-',@String,CHARINDEX('-',@String)+1)+2,  -- 2 additional for the character 'BA' 
                        1,'-IN')    --  Add 'IN'
            ELSE STUFF(
                        STUFF(
                                STUFF(@String,CHARINDEX('-',@String),1,'-BA'),      -- Add 'BA'
                                CHARINDEX('-',@String,CHARINDEX('-',@String)+1)+2,  -- 2 additional for the character 'BA' 
                                1,'-IN'),    --  Add 'IN'
                        1, CHARINDEX('-',@String) - 6, ''       -- remove extra leading Zeros
                        )
        END

Answer 3

假设格式一致（例如总是以＆＃34; - ＆＃34; + 2个字符......结尾）

DECLARE @Data TABLE (Col1 VARCHAR(100))
INSERT @Data ( Col1 )
SELECT Col1 
FROM (
    VALUES ('98-45.3A-22'), ('104-44.0A-23'),
        ('00983-29.1-22'), ('01757-42.5A-22'),
        ('04968-37.3A2-23')
) x (Col1)


SELECT RIGHT('0000' + LEFT(Col1, CHARINDEX('-', Col1) - 1), 5)
    + '-BA' + SUBSTRING(Col1, CHARINDEX('-', Col1) + 1, CHARINDEX('.', Col1) - CHARINDEX('-', Col1))
    + SUBSTRING(Col1, CHARINDEX('.', Col1) + 1, LEN(Col1) - CHARINDEX('.', Col1) -  3)
    + '-IN-' + RIGHT(Col1, 2)
FROM @Data

在SQL中始终执行此字符串操作并非理想的IMO。您可以将其移出到表示层，或将预先格式化的值存储在数据库中，以便每次都节省成本。

Answer 4

使用 REPLICATE 和 CHARINDEX ：

复制：将重复给定的字符，直到达到函数中指定的所需计数

CharIndex：查找任何字符的第一个匹配项

Declare @Data AS VARCHAR(50)='98-45.3A-22'
SELECT REPLICATE('0',6-CHARINDEX('-',@Data)) + @Data    

SELECT 
    SUBSTRING
    (
        (REPLICATE('0',6-CHARINDEX('-',@Data)) +@Data)
        ,0
        ,6
    )
    +'-'+'BA'+  CAST('<x>' + REPLACE(@Data,'-','</x><x>') + '</x>' AS XML).value('/x[2]','varchar(max)')
    +'-'+ 'IN'+ '-' + CAST('<x>' + REPLACE(@Data,'-','</x><x>') + '</x>' AS XML).value('/x[3]','varchar(max)')

Answer 5

另一种方法是使用PARSENAME()，您可以使用此查询：

 WITH t AS (
SELECT 
    PARSENAME(REPLACE(REPLACE(s, '.', '@@@'), '-', '.'), 3) AS p1,
    REPLACE(PARSENAME(REPLACE(REPLACE(s, '.', '@@@'), '-', '.'), 2), '@@@', '.') AS p2,
    PARSENAME(REPLACE(REPLACE(s, '.', '@@@'), '-', '.'), 1) AS p3
FROM yourTable)
SELECT RIGHT('00000' + p1, 5) + '-BA' + p2 + '-IN-' + p3
FROM t;

在第一次爆炸之前填充零并移除空格并添加BA和IN

5 个答案: