I want to do this:
我想做这个：
Declare @a int;
Declare @b int;
SET @a,@b = (SELECT StartNum,EndNum FROM Users Where UserId = '1223')
PRINT @a
PRINT @b
But this is invalid syntax. How do I set multiple scalar variables in one select statement? I can do:
但这是无效的语法。如何在一个select语句中设置多个标量变量？我可以：
Declare @a int;
Declare @b int;
SET @a = (SELECT StartNum FROM Users Where UserId = '1223')
SET @b = (SELECT EndNum FROM Users Where UserId = '1223')
PRINT @a
PRINT @b
But this will take twice as long. What is the fastest way?
但这需要两倍的时间。什么是最快的方式？
3 个解决方案
#1
76
DECLARE @a int;
DECLARE @b int;
SELECT @a = StartNum, @b = EndNum
FROM Users
WHERE UserId = '1223'
#2
10
Do it like this:
像这样做：
Declare @a int;
Declare @b int;
SELECT @a=StartNum,@b=EndNum FROM Users Where UserId = '1223'
PRINT @a
PRINT @b
#3
1
If you are doing this in a stored procedure and don't want the result of the select in an output resultset you will need to use the word INTO.
如果您在存储过程中执行此操作并且不希望在输出结果集中使用select的结果，则需要使用单词INTO。
Declare @a int;
Declare @b int;
SELECT StartNum, EndNum
FROM Users
Where UserId = '1223'
INTO @a, @b;
It also can be used like this:
它也可以像这样使用：
SELECT StartNum, EndNum
INTO @a, @b
FROM Users
Where UserId = '1223';
#1
76
DECLARE @a int;
DECLARE @b int;
SELECT @a = StartNum, @b = EndNum
FROM Users
WHERE UserId = '1223'
#2
10
Do it like this:
像这样做：
Declare @a int;
Declare @b int;
SELECT @a=StartNum,@b=EndNum FROM Users Where UserId = '1223'
PRINT @a
PRINT @b
#3
1
If you are doing this in a stored procedure and don't want the result of the select in an output resultset you will need to use the word INTO.
如果您在存储过程中执行此操作并且不希望在输出结果集中使用select的结果，则需要使用单词INTO。
Declare @a int;
Declare @b int;
SELECT StartNum, EndNum
FROM Users
Where UserId = '1223'
INTO @a, @b;
It also can be used like this:
它也可以像这样使用：
SELECT StartNum, EndNum
INTO @a, @b
FROM Users
Where UserId = '1223';