Для простых имен это будет работать. Остерегайтесь особых случаев (например, «Рональд Макдональд» в приведенном ниже примере).
В SQL Server:
SELECT --step 2: combine broken parts into a final name
NAME_PARTS.FNAME_INITIAL + NAME_PARTS.REST_OF_FNAME AS FNAME
,NAME_PARTS.LNAME_INITIAL + NAME_PARTS.REST_OF_LNAME AS LNAME
FROM
( --step 1: break name into 1st letter and "everything else"
SELECT
UPPER(SUBSTRING(TEST.FNAME,1,1)) AS FNAME_INITIAL
,UPPER(SUBSTRING(TEST.LNAME,1,1)) AS LNAME_INITIAL
,LOWER(SUBSTRING(TEST.FNAME,2,LEN(TEST.FNAME))) AS REST_OF_FNAME
,LOWER(SUBSTRING(TEST.LNAME,2,LEN(TEST.LNAME))) AS REST_OF_LNAME
FROM
( --step 0: generate some test data
SELECT 'john' AS FNAME, 'doe' as LNAME
UNION SELECT 'SUZY', 'SMITH'
UNION SELECT 'bIlLy', 'BOb'
UNION SELECT 'RoNALD', 'McDonald'
UNION SELECT 'Edward', NULL
UNION SELECT NULL, 'Jones'
) TEST
) NAME_PARTS
В Oracle
SELECT --step 2: combine broken parts into a final name
NAME_PARTS.FNAME_INITIAL || NAME_PARTS.REST_OF_FNAME AS FNAME
,NAME_PARTS.LNAME_INITIAL || NAME_PARTS.REST_OF_LNAME AS LNAME
FROM
( --step 1: break name into 1st letter and "everything else"
SELECT
UPPER(SUBSTR(TEST.FNAME,1,1)) AS FNAME_INITIAL
,UPPER(SUBSTR(TEST.LNAME,1,1)) AS LNAME_INITIAL
,LOWER(SUBSTR(TEST.FNAME,2,LENGTH(TEST.FNAME))) AS REST_OF_FNAME
,LOWER(SUBSTR(TEST.LNAME,2,LENGTH(TEST.LNAME))) AS REST_OF_LNAME
FROM
( --step 0: generate some test data
SELECT 'john' AS FNAME, 'doe' as LNAME FROM DUAL
UNION SELECT 'SUZY', 'SMITH' FROM DUAL
UNION SELECT 'bIlLy', 'BOb' FROM DUAL
UNION SELECT 'RoNALD', 'McDonald' FROM DUAL
UNION SELECT 'Edward', NULL FROM DUAL
UNION SELECT NULL, 'Jones' FROM DUAL
) TEST
) NAME_PARTS