1

挿入ステートメントでカンマに関するエラーが発生し続けます。これがなぜなのかについての考え。

エラーメッセージは次のとおりです。

メッセージ 102、レベル 15、状態 1、行 3
「,」付近の構文が正しくありません。

とINSERT INTO...SELECT声明

insert into custflag (cust_no, flag)
  select 
      customer.cust_no 
  from 
      customer, dupaddr 
  where 
      customer.cust_no = dupaddr.cust_no, select cast(flag as int) 
                                           from flag 
                                           where flag_desc = 'Dup Customer'

これが私のクエリの完全なコードです。

SET IDENTITY_INSERT flag ON
insert into flag (flag,flag_desc,available)
values ((select Max(flag) from flag) + 1, 'Dup Customer', 1)

create view dupaddr as 
select distinct c1.cust_no, c1.firstname, c1.lastname, c1.company, c1.predir + ' ' + c1.streetno + ' ' + c1.streetnm + ' ' + c1.suffix + ' ' + c1.postdir as fff ,c1.address2
from customer c1,customer c2 
where c1.cust_no <> c2.cust_no
and c1.firstname = c2.firstname 
and c1.lastname = c2.lastname 
and c1.company = c2.company
and c1.predir + ' ' + c1.streetno + ' ' + c1.streetnm + ' ' + c1.suffix + ' ' +   c1.postdir = c2.predir + ' ' + c2.streetno + ' ' + c2.streetnm + ' ' + c2.suffix + ' ' + c2.postdir 
and c1.address2 = c2.address2



insert into custflag (cust_no,flag)
select customer.cust_no from customer, dupaddr where customer.cust_no = dupaddr.cust_no , select cast(flag as int) from flag where flag_desc = 'Dup Customer'

ビューにフラグを追加し、挿入ステートメントを簡素化することができました。助けてくれてありがとう！

SET IDENTITY_INSERT flag ON
insert into flag (flag,flag_desc,available)
values ((select Max(flag) from flag) + 1, 'Dup Customer', 1)

create view dupaddr as 
select distinct c1.cust_no, 
c1.firstname, 
c1.lastname, 
c1.company, 
c1.predir + ' ' + c1.streetno + ' ' + c1.streetnm + ' ' + c1.suffix + ' ' +   c1.postdir as fff ,
c1.address2, 
(SELECT cast(flag as int) FROM flag  WHERE flag_desc = 'Dup Customer') as flag
from customer c1,customer c2 
where c1.cust_no <> c2.cust_no
and c1.firstname = c2.firstname 
and c1.lastname = c2.lastname 
and c1.company = c2.company
and c1.predir + ' ' + c1.streetno + ' ' + c1.streetnm + ' ' + c1.suffix + ' ' +  c1.postdir = c2.predir + ' ' + c2.streetno + ' ' + c2.streetnm + ' ' + c2.suffix + ' ' + c2.postdir 
and c1.address2 = c2.address2



insert into custflag (cust_no,flag)
select dupaddr.cust_no, dupaddr.flag from dupaddr
4

3 に答える 3

0

クエリを機能させるために、同等のコードを次に示します。

INSERT  INTO custFlag(cust_no, flag)
SELECT  a.cust_no, c.flag
FROM    customer a
        INNER JOIN dupaddr b
            ON a.cust_no = b.cust_no
        INNER JOIN
        (
            SELECT  cust_no, cast(flag as int) flag
            FROM    flag 
            WHERE   flag_desc = 'Dup Customer'
        ) c ON a.cust_no = c.cust_no

しかし、私はあなたが望むものではないと思います. tableが と にどのように関連してCROSS JOINいるか教えてもらえますか?flagcustomerdupaddr

于 2013-04-09T15:46:52.037 に答える