Table variable error: Must declare the scalar variable "@temp"
Solution 1
A table alias cannot start with a @
. So, give @Temp
another alias (or leave out the two-part naming altogether):
SELECT *
FROM @TEMP t
WHERE t.ID = 1;
Also, a single equals sign is traditionally used in SQL for a comparison.
Solution 2
Either use an Allias in the table like T and use T.ID, or use just the column name.
declare @TEMP table (ID int, Name varchar(max))
insert into @temp SELECT ID, Name FROM Table
SELECT * FROM @TEMP
WHERE ID = 1
Solution 3
There is one another method of temp table
create table #TempTable (
ID int,
name varchar(max)
)
insert into #TempTable (ID,name)
Select ID,Name
from Table
SELECT *
FROM #TempTable
WHERE ID = 1
Make Sure You are selecting the right database.
Solution 4
If you bracket the @ you can use it directly
declare @TEMP table (ID int, Name varchar(max))
insert into @temp values (1,'one'), (2,'two')
SELECT * FROM @TEMP
WHERE [@TEMP].[ID] = 1
Solution 5
You should use hash (#) tables, That you actually looking for because variables value will remain till that execution only. e.g. -
declare @TEMP table (ID int, Name varchar(max))
insert into @temp SELECT ID, Name FROM Table
When above two and below two statements execute separately.
SELECT * FROM @TEMP
WHERE @TEMP.ID = 1
The error will show because the value of variable lost when you execute the batch of query second time. It definitely gives o/p when you run an entire block of code.
The hash table is the best possible option for storing and retrieving the temporary value. It last long till the parent session is alive.
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objectWithoutClass
Updated on July 09, 2022Comments
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objectWithoutClass almost 2 years
I am trying to achieve:
declare @TEMP table (ID int, Name varchar(max)) insert into @temp SELECT ID, Name FROM Table SELECT * FROM @TEMP WHERE @TEMP.ID = 1 <--- ERROR AT @TEMP.ID
But I'm getting the following error:
Must declare the scalar variable "@temp".
What am I doing wrong?
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GriGrim almost 11 years
SELECT * FROM @TEMP T WHERE T.ID = 1
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Denziloe over 5 yearsA table variable is inherently temporary -- there is no such thing as a "temporary table variable". What you are trying to do is declare a table variable. There is also a related but different concept of a temporary table.
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third_eye almost 9 yearsVariables are NOT case sensitive.