Question

HI，

我正在尝试在SQL Server 2005中创建以下函数。当我尝试对@FULLNAME2 & @FULLNAME1中携带的值进行RTrim时，它会给出异常。要求争论Rtrim的异常，但它已经通过那里。请注意我已经能够创建此功能，不包括Rtrim部分。

另一个问题是我需要在视图中调用此函数，我使用select names ('val1','val2')调用它。它声明这不是一个公认的功能。但是我可以在sysobjects表中看到这个函数有'TF'类型。

请建议。

CREATE FUNCTION dbo.names( @CUSTID varchar(20),@effdt varchar(20))
RETURNS @FinalResults1 TABLE (Name1 nvarchar(254), Name2 nvarchar(254)) 
AS
BEGIN   

DECLARE @FinalResults TABLE  (Name254 nvarchar(254), SRNO nvarchar(3))


INSERT INTO @FinalResults
SELECT (C.NAME1),ROW_NUMBER() OVER(ORDER BY A.SEQ_NBR) 
 FROM PS_ARB_CU_CLST_STN A , PS_ARB_CU_STATIONS C 
 WHERE A.EFF_STATUS = 'A' 
   AND A.EFFDT = ( 
 SELECT MAX(B.EFFDT) 
  FROM PS_ARB_CU_CLST_STN B 
 WHERE A.SETID = B.SETID 
  AND A.CUST_ID = B.CUST_ID 
   AND B.EFFDT <= @effdt) 
  AND A.SETID = C.SETID 
   AND A.ARB_STATION_ID =C.CUST_ID 
 AND A.CUST_ID = @CUSTID
 AND C.EFFDT = (SELECT MAX(D.EFFDT) FROM PS_ARB_CU_STATIONS D WHERE C.CUST_ID = D.CUST_ID
 AND D.SETID = C.SETID AND D.EFFDT <= @effdt)
 ORDER BY A.SEQ_NBR


DECLARE @Name nvarchar(254), @FULLNAME1 nvarchar(128), @FREEZENAME1 nvarchar(10), @append NVARCHAR (254)
DECLARE @FULLNAME254 nvarchar(254), @FULLNAME2 nvarchar(128), @FREEZENAME2 nvarchar(10), @COUNT INT, @i INT

SET @Name = ''
SET @FREEZENAME1 = 'FALSE'
SET @FREEZENAME2 = 'FALSE'
SET @FULLNAME1 = ''
SET @FULLNAME2 = ''
SET @FULLNAME254 = ''
SET @COUNT = 0
SET @i  = 0
SET @append = ''

SELECT @COUNT = COUNT(*)   FROM @FinalResults 

WHILE @i < @COUNT
BEGIN
IF  @FULLNAME1 = ''  
 BEGIN  
   IF(LEN((SELECT NAME254 FROM @FinalResults WHERE SRNO = @i )+ '/')<= 40 AND @FREEZENAME1 = 'FALSE' )
          SET @FULLNAME1 = (SELECT NAME254 FROM @FinalResults WHERE SRNO = @i  + '/');
          ELSE
          SET   @FREEZENAME1 = 'TRUE';
          END



ELSE 
  BEGIN
         IF (LEN(@FULLNAME1 +(SELECT NAME254 FROM @FinalResults WHERE SRNO = @i )+ '/') <=40 AND @FREEZENAME1 = 'FALSE' )
         SET    @FULLNAME1 = (@FULLNAME1 +(SELECT NAME254 FROM @FinalResults WHERE SRNO = @i )+ '/') ;
         ELSE
         BEGIN
         SET    @FREEZENAME1 = 'TRUE';
                IF @FULLNAME2 = ''   
                   BEGIN
                     IF (LEN((SELECT NAME254 FROM @FinalResults WHERE SRNO = @i )+ '/')<= 40 AND @FREEZENAME2 = 'FALSE' ) 
                      SET   @FULLNAME2 = ((SELECT NAME254 FROM @FinalResults WHERE SRNO = @i )+ '/') ;
                      ELSE
                      SET   @FREEZENAME2 = 'TRUE';
                   END


ELSE
       BEGIN
                IF (LEN(@FULLNAME2 +(SELECT NAME254 FROM @FinalResults WHERE SRNO = @i )+ '/') <=40 AND @FREEZENAME2 = 'FALSE') 
                SET @FULLNAME2 = (@FULLNAME2 +(SELECT NAME254 FROM @FinalResults WHERE SRNO = @i )+ '/') ;
                ELSE
                SET    @FREEZENAME2 = 'TRUE';
                END
       END
       END
       END

IF @append = '' 

           SET @append = (SELECT NAME254 FROM @FinalResults WHERE SRNO = @i );

         Else

           SET @append = @append + '/'+ (SELECT NAME254 FROM @FinalResults WHERE SRNO = @i );



SET @i = @i +1

 If (Len(@append) < 40) 
   SET   @FULLNAME1 = RTrim(@FULLNAME1, '/');
   End
   If ((Len(@append) > 40) And
         (Len(@append) < 80)) 

     SET @FULLNAME2 = RTrim(@FULLNAME2, '/');
   End


INSERT INTO @FinalResults1 VALUES (@FULLNAME1, @FULLNAME2)


RETURN  

END

Answer 1

如果它是一个表值函数（返回一个表），你应该像这样调用它

select <fields> from dbo.name(<parameters>)

rtrim函数也不能按你想要的方式工作，它需要一个参数然后删除字符串右边的所有空格

您在调用RTRIM的条件中也缺少几个BEGIN，或者您也可以删除结束。

If (Len(@append) < 40) 
BEGIN
   SET   @FULLNAME1 = RTrim(@FULLNAME1, '/');
End
If ((Len(@append) > 40) And (Len(@append) < 80)) 
BEGIN
     SET @FULLNAME2 = RTrim(@FULLNAME2, '/');
End

我看了你的函数，而while循环有几个奇怪的东西，有一些代码的部分永远不会被执行，我试着重写一个等效函数，我希望这可以帮助你

CREATE FUNCTION dbo.names(@CUSTID varchar(20),@effdt datetime)
RETURNS @results TABLE (Name1 nvarchar(254), Name2 nvarchar(254)) 
AS
BEGIN   
DECLARE rCursor CURSOR FOR
 SELECT (C.NAME1) AS name
 FROM PS_ARB_CU_CLST_STN AS A , PS_ARB_CU_STATIONS AS C 
 WHERE A.EFF_STATUS = 'A' AND 
  A.EFFDT = (
   SELECT MAX(B.EFFDT) 
   FROM PS_ARB_CU_CLST_STN AS B 
   WHERE A.SETID = B.SETID AND 
    A.CUST_ID = B.CUST_ID AND 
    B.EFFDT <= @effdt
  ) AND 
  A.SETID = C.SETID AND 
  A.ARB_STATION_ID = C.CUST_ID AND 
  A.CUST_ID = @CUSTID AND 
  C.EFFDT = ( 
   SELECT MAX(D.EFFDT) 
   FROM PS_ARB_CU_STATIONS D 
   WHERE C.CUST_ID = D.CUST_ID AND 
    D.SETID = C.SETID AND 
    D.EFFDT <= @effdt
  )
  ORDER BY A.SEQ_NBR

DECLARE @name nvarchar(254), 
        @fullname1 nvarchar(128),
        @fullname2 nvarchar(128),
        @append NVARCHAR (254); 
SET @fullname1 = '';
SET @fullname2 = '';
SET @append = ''

OPEN rCursor;
FETCH NEXT FROM rCursor INTO @name

WHILE @@FETCH_STATUS = 0 AND LEN(@fullname1 + @name) < 40
BEGIN
    SET @fullname1 = @fullname1 + '/' + @name
    FETCH NEXT FROM rCursor INTO @name
END

WHILE @@FETCH_STATUS = 0 AND LEN(@fullname2 + @name) < 40
BEGIN
    SET @fullname2 = @fullname2 + '/' + @name;
    FETCH NEXT FROM rCursor INTO @name;
END

--Append is not used 
WHILE @@FETCH_STATUS = 0
BEGIN
    SET @append = @append + '/' + @name;
    FETCH NEXT FROM rCursor INTO @name;
END

CLOSE rCursor;
DEALLOCATE rCursor;

INSERT INTO @results VALUES (@fullname1, @fullname2)
RETURN
END
GO

Answer 2

函数RTRIM用于从字符串的右侧删除空格;

DECLARE @string VARCHAR(20)
SET @string = '  text and text     '
SELECT RTRIM(@string)

这将返回值：' text and text'

表值函数需要被视为select语句中的数据源：

SELECT * from dbo.values('val1', 'val2')

无法在函数内传递变量值

2 个答案: