select uEmail=replace(uEmail,substring(uEmail,5,3),'***') from table

------解决方案--------------------
SQL code
DECLARE @T TABLE (uEmail NVARCHAR(50))
INSERT @T SELECT 'henyabc@hotmail.com'
INSERT @T SELECT 'wcchfae@163.com'
INSERT @T SELECT 'bono@hello.com'


SELECT 
CASE WHEN  CHARINDEX('@',uEmail)<5 THEN LEFT(uEmail,CHARINDEX('@',uEmail)-1) ELSE LEFT(uEmail,4) end+'***'+RIGHT(uEmail,CHARINDEX('@',REVERSE(uEmail))) AS uEmail

FROM @T

/*
heny***@hotmail.com
wcch***@163.com
bono***@hello.com
*/

------解决方案--------------------
探讨


引用:

SQL code
DECLARE @T TABLE (uEmail NVARCHAR(50))
INSERT @T SELECT 'henyabc@hotmail.com'
INSERT @T SELECT 'wcchfae@163.com'
INSERT @T SELECT 'bono@hello.com'


SELECT
LEFT……


------解决方案--------------------

探讨


引用:

SQL code
DECLARE @T TABLE (uEmail NVARCHAR(50))
INSERT @T SELECT 'henyabc@hotmail.com'
INSERT @T SELECT 'wcchfae@163.com'
INSERT @T SELECT 'bono@hello.com'


SELECT
LEFT……