子查询返回多个值错误

Question

我正在使用以下查询，但它引发了错误。它在某些情况下可以正常工作，具体取决于 id，但并非适用于所有情况。

子查询返回超过 1 个值。这是不允许的，当 子查询遵循 =、!=、、>= 或当子查询用作 一个表达式。

Set @NoOfRows = (Select COUNT(*)  from @TempT)

While @i <= @NoOfRows
Begin
  SET @Voucher_No=( select V_No from @TempT where RowID=@i)
  SET @Voucher_Type_No=( select voucher_type_no from @TempT where RowID=@i)
  SET @Voucher_Id=( select Voucher_Id from @TempT where RowID=@i)
  set @strPartyName=''
  set @strPartyName = (Select Party_Name from Cheque_Book where Voucher_No=   @Voucher_No   and Voucher_Type_No=   @Voucher_Type_No   and Company_No=  @Comp_No   and Bank_Account_No= @dbc_Account )

  if NULLIF(@strPartyName,'') IS NULL
  begin
    set @strPartyName =(Select a.account_name from Voucher v inner join account a on v.Account_No = a.account_No where v.Voucher_Id= @Voucher_Id)
    Update @TempT Set Party_Name =@strPartyName Where RowID =@i
    set @i=@i+1
  end
End
Select * from @TempT

Answer 1

不是使用子查询设置变量，而是如下所示设置变量。

SELECT @NoOfRows = COUNT(*) FROM @TempT

WHILE @i <= @NoOfRows
BEGIN
    SELECT @Voucher_No = V_No FROM @TempT WHERE RowID=@i
    SELECT @Voucher_Type_No = voucher_type_no FROM @TempT WHERE RowID=@i
    SELECT @Voucher_Id = Voucher_Id FROM @TempT where RowID=@i
    SET @strPartyName=''

    SELECT @strPartyName =  Party_Name FROM Cheque_Book WHERE Voucher_No = @Voucher_No AND Voucher_Type_No = @Voucher_Type_No   AND Company_No=  @Comp_No   AND Bank_Account_No= @dbc_Account
    IF NULLIF(@strPartyName,'') IS NULL
    BEGIN
        SELECT @strPartyName = a.account_name FROM Voucher v INNER JOIN account a ON v.Account_No = a.account_No WHERE v.Voucher_Id= @Voucher_Id
        UPDATE @TempT SET Party_Name = @strPartyName WHERE RowID = @i
        SET @i = @i + 1
    END
END 
SELECT * FROM @TempT

Answer 2

它表明您的子查询返回多个值。

您必须使用 TOP 关键字修改您的子查询，或提供更多过滤器（where 条件和 order by），只返回一个值。

While @i <= @NoOfRows
 Begin

   SET @Voucher_No=( select top 1 V_No from @TempT where RowID=@i)
   SET @Voucher_Type_No=( select top 1 voucher_type_no from @TempT where RowID=@i)
   SET @Voucher_Id=( select top1 Voucher_Id from @TempT where RowID=@i)
......
END

要解决此问题，您只需在 while 循环中运行实际的 select 查询，它会给出超过 2 个值的结果，例如

While @i <= @NoOfRows
Begin

   --SET @Voucher_No=( 
   select  V_No from @TempT where RowID=@i --)
   --SET @Voucher_Type_No=( 
   select  voucher_type_no from @TempT where RowID=@i --)
   --SET @Voucher_Id=( 
   select  Voucher_Id from @TempT where RowID=@i --)
    ......
END

更新 这是用于了解您发生了什么以及如何解决它的示例数据。

declare @department table (deptid int, name varchar(50))
declare @emp table (id int, name varchar(50), deptid int)

insert into @department values (1,'d1'), (2,'d2'),(3,'d3'),(4,'d4')
insert into @emp values(1,'ajay',1), (2,'ajay1',1),(3,'ajay3' ,2),(4,'ajay4' ,3),(5,'ajay5' ,4)


select * from @department where deptid = 1

/* suppose your sub-query like */
--1. this give result
select * from @department where deptid = (select deptid from @emp where id = 1)

--2. this give result, but may be get wrong result as it take top 1.
select * from @department where deptid = (select top 1 deptid from @emp )

--3. this give error same you have
select * from @department where deptid = (select deptid from @emp)

--to resolve the above error, just uncomment below sub-query.
--select deptid from @emp

Answer 3

我在这里看到以下查询因子查询返回多个值而失败的问题的答案：

SET @Voucher_No = (SELECT [V_No] FROM @TempT WHERE [RowID] = @i);
SET @Voucher_Type_No = (SELECT [voucher_type_no] FROM @TempT WHERE [RowID] = @i);
SET @Voucher_Id = (SELECT [Voucher_Id] FROM @TempT WHERE [RowID] = @i);

这些答案主要有两种建议的解决方案：

1. SET @Variable = (SELECT TOP (1) [Field] FROM [Table] WHERE ...);，或
2. SELECT @Variable = [Field] FROM [Table] WHERE ...);。

令我惊讶的是，第二种方法被认为是首选方法。但是，有一些重要的事情需要考虑。

首先：如果 SELECT 查询返回多个结果，@Variable 的值将成为查询结果集中最后检索到的值。指定 ORDER BY 子句时必须考虑这一点。

所以下面的查询将返回5 而不是1：

DECLARE @TempTable TABLE ([ID] INT);
INSERT INTO @TempTable VALUES (1), (3), (2), (5), (4);

DECLARE @Variable INT = 42;
SELECT @Variable = [ID] FROM @TempTable ORDER BY [ID];

SELECT @Variable AS [@Variable];

第二：如果@Variable已经包含一个值并且SELECT查询没有产生任何结果（0行），@Variable的值保持不变！

所以下面的查询将产生结果42 而不是NULL：

DECLARE @TempTable TABLE ([ID] INT);
INSERT INTO @TempTable VALUES (1), (3), (2), (5), (4);

DECLARE @Variable INT = 42;
SELECT @Variable = [ID] FROM @TempTable WHERE [ID] = 6;

SELECT @Variable AS [@Variable];

出于所有这些原因，我个人偏爱SET @Variable = (SELECT TOP (1) ...) 方法。