在case语句中搜索字符串是否存在特定字符

时间:2019-02-01 19:56:32

标签: sql sql-server-2014

我有以下SQL,包括CASE的{​​{1}}语句。我想修改D.TIN_TYPE语句(或使用替代方法)以添加到第二WHEN条件,以便CASEWHEN D.TIN_TYPE = 'S' AND字符串中有一个逗号(,),然后中断名称使用LEFT / CHARINDEX和RIGHT / CHARINDEX函数分成两列,否则按原样显示字符串(如果字符串中不包含逗号)。该列的格式不统一为 Lastname,Firstname (有些是 Firstname Lastname ),因此这就是为什么我需要检查字符串中是否存在逗号的原因。我该怎么做?

D.PAYEE_NAME1

编辑:

这是经过编辑的SQL,在CASE语句中使用额外的WHEN。它运行,但是对于姓氏列(RIGHT函数),它同时返回名字和姓氏。我如何才能返回姓氏(逗号前的字符)?我也尝试为每个WHEN使用END AS'First name'和END AS'Last Name',但是出现语法错误。有没有一种方法可以为列指定不同的名称?

    SELECT D.PAYER_TIN, '001', YEAR( (CONVERT(CHAR(10),A.PYMNT_DT,121))), D.PAYEE_TIN, 
    CASE WHEN  D.TIN_TYPE = 'F' THEN '001' 
         WHEN  D.TIN_TYPE = 'S' THEN '002' 
         END AS 'Taxpayer Identification Type', 
    A.VENDOR_ID,
    CASE WHEN D.TIN_TYPE = 'F' THEN LEFT(D.PAYEE_NAME1, 30)
         WHEN D.TIN_TYPE = 'S' AND --Additional logic for existence of comma--
         ELSE D.PAYEE_NAME1
         END AS 'PAYEE_NAME1', 
    D.PAYEE_NAME1,
    D.PAYEE_NAME2,  
    D.PAYEE_ADDRESS1, 
    LEFT(PAYEE_CITYSTZIP,CHARINDEX(',', PAYEE_CITYSTZIP)-1), 
    SUBSTRING(PAYEE_CITYSTZIP, CHARINDEX(',', PAYEE_CITYSTZIP)+2, 2), 
    RIGHT(PAYEE_CITYSTZIP, LEN(PAYEE_CITYSTZIP) - CHARINDEX(',', PAYEE_CITYSTZIP)-7), 
    'US', 
    D.WTHD_BOX1, 
    D.WTHD_BOX2, 
    D.WTHD_BOX3, 
   FROM PS_WTHD_TRXN_TBL A, PS_TMP_1099_COPYB D, PS_GHS_1099_PA_VW2 E 
   WHERE ( A.WTHD_TYPE = D.WTHD_TYPE 
         AND A.VENDOR_ID = D.VENDOR_ID 
         AND YEAR( A.PYMNT_DT) = '2018')
   GROUP BY  D.PAYER_TIN,  YEAR( (CONVERT(CHAR(10),A.PYMNT_DT,121))),  D.PAYEE_TIN,  CASE WHEN  D.TIN_TYPE = 'F' THEN '001' 
    WHEN  D.TIN_TYPE = 'S' THEN '002' 
    END,  A.VENDOR_ID,  D.PAYEE_NAME1,  D.PAYEE_NAME2,   D.PAYEE_ADDRESS1, 
    LEFT(PAYEE_CITYSTZIP,CHARINDEX(',', PAYEE_CITYSTZIP)-1),  
    SUBSTRING(PAYEE_CITYSTZIP, CHARINDEX(',', PAYEE_CITYSTZIP)+2, 2),  
    RIGHT(PAYEE_CITYSTZIP, LEN(PAYEE_CITYSTZIP) - CHARINDEX(',', PAYEE_CITYSTZIP)-7), 
    D.WTHD_BOX1,  D.WTHD_BOX2,  D.WTHD_BOX3,  D.WTHD_BOX6,  D.WTHD_BOX7,  D.WTHD_BOX14,  D.WTHD_BOX16,  D.WTHD_BOX18
    , CASE WHEN D.TIN_TYPE = 'F' THEN LEFT(D.PAYEE_NAME1, 30)
         ELSE D.PAYEE_NAME1
         END

1 个答案:

答案 0 :(得分:1)

您将需要为此创建多个case语句。我的建议是创建3个case语句:一个用于非拆分,一个用于第一个拆分,一个用于第二个拆分。

要检查逗号,可以使用类似以下的内容:

select 
    split1=case
        WHEN CHARINDEX(',','a,b') > 0 AND D.TIN_TYPE = 'S' THEN left('a,b', CHARINDEX(',','a,b')-1)
    END,
    split2=case
        WHEN CHARINDEX(',','a,b') > 0 AND D.TIN_TYPE = 'S' THEN right('a,b', len('a,b')-CHARINDEX(',','a,b'))
    END  

Results:
split1  split2
a        b

请参见CASE (Transact-SQL) Example B

编辑: 尝试从您编辑过的查询中提取此较短的查询。使它起作用,然后开始在GROUP BY中添加其余条件,我想这就是您的问题所在。

看看我对PAYEE_FULLNAME,PAYEE_FIRSTNAME,PAYEE_LASTNAME做了什么。

    SELECT D.PAYER_TIN, '001', 13679089, YEAR( (CONVERT(CHAR(10),A.PYMNT_DT,121))), D.PAYEE_TIN, 
CASE WHEN  D.TIN_TYPE = 'F' THEN '001' 
     WHEN  D.TIN_TYPE = 'S' THEN '002' 
     END AS 'Taxpayer Identification Type', 
A.VENDOR_ID,
PAYEE_FULLNAME=CASE 
        WHEN D.TIN_TYPE = 'F' THEN LEFT(D.PAYEE_NAME1, 30)
        ELSE D.PAYEE_NAME1
     END , 
PAYEE_FIRSTNAME=CASE 
        WHEN D.TIN_TYPE = 'S' AND CHARINDEX(',',D.PAYEE_NAME1) > 0 THEN LEFT(D.PAYEE_NAME1,CHARINDEX(',', D.PAYEE_NAME1)-1)  --END AS 'First Name'
       END ,
PAYEE_LASTNAME=CASE 
        WHEN D.TIN_TYPE = 'S' AND CHARINDEX(',',D.PAYEE_NAME1) > 0 THEN RIGHT(D.PAYEE_NAME1, LEN(PAYEE_CITYSTZIP) - CHARINDEX(',', D.PAYEE_NAME1)+1 ) --END AS 'Last Name'
     END
FROM PS_WTHD_TRXN_TBL A, PS_TMP_1099_COPYB D, PS_GHS_1099_PA_VW2 E 
WHERE ( A.WTHD_TYPE = D.WTHD_TYPE 
     AND A.VENDOR_ID = D.VENDOR_ID 
     AND YEAR( A.PYMNT_DT) = '2018'
     AND A.WTHD_CLASS IN ('01','02','07') 
     AND A.VENDOR_ID = E.VENDOR_ID 
     AND A.VNDR_LOC = E.VNDR_LOC 
     AND E.YEAR = '2018' 
     AND D.WTHD_CNTL_ID = 'TGC' 
     AND D.WTHD_BOX18 > 0.00
     AND A.VENDOR_ID <> '80577A' )