"Must declare the scalar variable" 在 SQL 中尝试运行 UPDATE 语句时

Question

尝试使用两个 table 变量（@NameZip、@NameZip2）更新两个 table（City 和 Location）。城市名称有 ZipCodes，ZipCodes 有 Names，反之亦然。更新正在更改错误输入的城市名称和邮政编码。但是出现错误：

Msg 137, Level 16, State 1, Line 28
Must declare the scalar variable "@NameZip2".

Msg 137, Level 16, State 1, Line 32
Must declare the scalar variable "@NameZip".

我写的查询：

--first table variable
DECLARE @NameZip TABLE 
                 (
                     Zip_Code NVARCHAR(100),
                     Name NVARCHAR(100),
                     id_city INT
                 )

--second table variable
DECLARE @NameZip2 TABLE
                  (
                      Zip_Code nvarchar(100),
                      Name NVARCHAR(100),
                      id_city INT
                  )

--inserting into first table variable from City and Location table
INSERT INTO @NameZip (Zip_code, Name, id_city)
    SELECT B.Zip_Code, A.Name, A.id_city
    FROM City A 
    INNER JOIN Location B ON A.id_city = B.id_city 
                          AND Name NOT LIKE '%[^0-9]%'

--inserting into second table variable from first table variable
INSERT INTO @NameZip2(Zip_code, Name, id_city)
    SELECT Name, Zip_Code, id_city
    FROM @NameZip

UPDATE City 
SET Name = (SELECT Name FROM @NameZip2)
WHERE City.id_city = @NameZip2.id_city -- I get error on this line

UPDATE Location
SET Zip_Code = (SELECT Zip_Code FROM @NameZip2)
WHERE Zip_Code = @NameZip.Zip_Code -- I get error on this line

如有任何意见，我们将不胜感激。

Answer 1

使用update from join语法

update c
set Name = z.Name 
from City C
inner join @NameZip2 z on c.id_city =z.id_city

对 location table 更新也做同样的事情。

Answer 2

您想通过联接来执行此操作...但是您的 table 变量并不是真正需要的。

update c
set c.Name = n.Name
from City c
inner join @NameZip2 n on n.id_city = c.id_city

update L
set L.Zip_Code = n.Zip_Code
from Location L
inner join
@NameZip2 n on n.Zip_Code = L.Zip_Code

可以写成...

update c
set c.Name = n.Name, c.Zip_Code = n.Zip_Code
from City c
inner join    
    (SELECT B.Zip_Code, A.Name, A.id_city
    FROM City A 
    INNER JOIN Location B ON A.id_city = B.id_city 
                          AND Name NOT LIKE '%[^0-9]%') n

Answer 3

虽然其他答案可以帮助您更改代码，但我认为解释代码的问题很有趣：

SET Name = (SELECT Name FROM @NameZip2)

这一行可能会给您带来错误。如果您使用 = 运算符，则必须确保表达式 return 只有一个值。即使你是对的，@NameZip2 只有一个记录，这也不是一个好方法，你可以这样做：

SET Name = (SELECT Top 1 Name FROM @NameZip2)

还有这一行：

WHERE Zip_Code = @NameZip.Zip_Code

将不起作用，因为@NameZip 是 table，您应该使用 SELECT 命令而不是 =，这样：

WHERE Zip_Code = (SELECT TOP 1 Zip_Code FROM @NameZip)

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