【发布时间】:2020-08-08 14:13:29
【问题描述】:
数据
CREATE TABLE #tbl_LinkedInvoices(
InvoiceNbr varchar(50)
, AssociatedInvoiceNbr varchar(50)
, RowNbr int
, AssociatedRowNbr int
)
INSERT INTO #tbl_LinkedInvoices(
InvoiceNbr, AssociatedInvoiceNbr, RowNbr, AssociatedRowNbr)
VALUES
('A0001', 'A1001', 1, 4),
('A0002', 'A2002', 2, 5),
('A0002', 'A3002', 3, 6),
('A1001', 'A0001', 4, 1),
('A2002', 'A0002', 5, 2),
('A3002', 'A0002', 6, 3)
SELECT * FROM #tbl_LinkedInvoices
挑战/目标
tbl_LinkedInvoices 用于识别链接到的AssociatedInvoiceNbrs b 和InvoiceNbr a。因此,从(a, b) = (b,a) 起,一个集合可以在表中出现多次。为了解决这些重复出现的问题,添加了 RowNbr 和 AssociatedRowNbr 字段以提供分组序列。
使用已识别的重复行,删除重复行,在表中保留一条唯一记录。当前脚本产生错误,期待可能有更好的方法来编写查询。
脚本
使用计数器检查重复行是否仍然存在,如果确实删除该行,直到FALSE。
DECLARE @RowCounter int
DECLARE @RemoveRow int
SET @RowCounter = 1
IF EXISTS (SELECT
RowNbr
FROM #tbl_LinkedInvoices WHERE RowNbr = (SELECT AssociatedRowNbr FROM #tbl_LinkedInvoices)
)
BEGIN
SET @RemoveRow = (SELECT RowNbr FROM #tbl_LinkedINvoices
WHERE RowNbr = (
SELECT AssociatedRowNbr FROM #tbl_LinkedInvoices WHERE RowNbr =@RowCounter ))
BEGIN
DELETE FROM #tbl_LinkedInvoices
WHERE
RowNbr = @RemoveRow
END
BEGIN
SET @RowCounter = @RowCounter + 1
END
END
错误
Msg 512, Level 16, State 1, Line 212
Subquery returned more than 1 value. This is not permitted when the subquery follows =, !=, <, <= , >, >= or when the subquery is used as an expression.
【问题讨论】:
-
错误不是很清楚吗?使用
in而不是=。 -
@GordonLinoff -
IN返回全部或不返回,因为值存在于两列中。
标签: sql sql-server tsql duplicates sql-delete