Question

我需要你的支持，今天我得到了一个我需要这样做的场景。

该场景是包含用户名和ID的电子邮件，需要拆分并在两个不同的列中插入表中，即用户ID，用户名。

ex: ABC123@xyz.com, AAACC2356@mnc.com

ABC to be insert in table under username column and 123 to be insert in user id column

谢谢

纳伦德拉

Answer 1

你可以试试这个：

DECLARE @Email VARCHAR(100)= 'sample12312@test.com'

SET @Email = STUFF(@Email,CHARINDEX('@',@Email),LEN(@Email), '')
SELECT  SUBSTRING(@Email, 1, PATINDEX('%[0-9]%',@Email)-1) AS Username,
        SUBSTRING(@Email,PATINDEX('%[0-9]%',@Email), LEN(@Email)) AS UserID

OR

SELECT  SUBSTRING(@Email, 1, PATINDEX('%[0-9]%',@Email)-1) AS Username,,
        SUBSTRING(STUFF(@Email,CHARINDEX('@',@Email),LEN(@Email), ''), PATINDEX('%[0-9]%',@Email), LEN(@Email)) AS UserID

Answer 2

您可以使用少量功能来实现：

DECLARE @exp VARCHAR(1000) = 'AAACC2356@mnc.com'--'ABC123@xyz.com'

--finds where numbers starts
DECLARE @SplitPoint INT = (SELECT PATINDEX ( '%[0-9]%' , @exp ))
--finds where is @ sign
DECLARE @etaPoint INT = (SELECT PATINDEX ( '%@%' , @exp ))

--gets username
SELECT SUBSTRING(@exp, 0, @SplitPoint)
--gets number
SELECT SUBSTRING(@exp, @SplitPoint, @etaPoint-@SplitPoint)

Answer 3

Declare @X     varchar(100)
Select     @X=    'AAACC2356@mnc.com'

Select     @X=    SubString(@X,PATINDEX('%[0-9]%',@X),Len(@X))
Select     @X=    SubString(@X,0,PATINDEX('%[^0-9]%',@X))
Select     @X

Declare @Y    varchar(100)
Select     @Y =    'AAACC2356@mnc.com'

Select     @Y=    SubString(@Y,PATINDEX('%[A-Z]%',@Y),Len(@Y))
Select     @Y=    SubString(@Y,0,PATINDEX('%[^A-Z]%',@Y))
Select     @Y

使用Ms-SQL从电子邮件ID中提取用户名和用户ID

3 个答案: