Question

输入：得到o / p的逻辑：在开括号之前'（'将是fname，inside（）和@之后将是alias1 andalias2，具体取决于顺序顺序。

Full Name
------------------
ABC (PQR)
(ABC) PQR (XYZ)
ABC @ PQR
ABC @ PQR (XYZ)
ABC (PQR) @123

期望的输出：

FNAME   ALIAS1   ALIAS2
-----------------------
ABC     PQR
PQR     ABC      XYZ
ABC     PQR
ABC     PQR      XYZ
ABC     PQR      123

Answer 1

请使用类似的东西 -

;WITH CTE1 AS
(
    SELECT REPLACE(REPLACE(REPLACE([Full Name],'(',''),')',''),'@','') [Full Name] FROM Names
)
,CTE AS
(
    SELECT [Full Name],
    CAST(('<r><n>' + REPLACE([Full Name],' ', '</n><n>') + '</n></r>') AS XML) X
   FROM CTE1
)
SELECT 
FNAME, CASE WHEN ALIAS1 = '' THEN ALIAS2 ELSE ALIAS1 END ALIAS1,
CASE WHEN ALIAS1 = '' THEN '' ELSE ALIAS2 END ALIAS2 
FROM 
(
    SELECT
     i.value('n[1]','VARCHAR(50)') AS FNAME,
     ISNULL(i.value('n[2]','VARCHAR(50)'),'') AS ALIAS1,
     ISNULL(i.value('n[3]','VARCHAR(50)'),'') AS ALIAS2
    FROM CTE c
    CROSS APPLY c.X.nodes('/r') x(i)
)z

<强>输出

/*------------------------
OUTPUT
------------------------*/
FNAME                        ALIAS1    ALIAS2
---------------------------- --------- --------------------------------------------------
ABC                          PQR       
ABC                          PQR       XYZ
ABC                          PQR       
ABC                          PQR       
ABC                          PQR       123

(5 row(s) affected)

字符串隔离的SQL查询

1 个答案: