Question

是否有可能选择增量数字，例如：

我试过了：

set @i=999;
select @i:=@i+1 as num
limit 1000

还有：

set @i=999;
select @i:=@i+1 as num
where @i <1999;

Answer 1

我假设（可能是错误地）你试图根据自动递增的主键选择行，看到这是MySQL吗？

“BETWEEN”关键字是否可以更好地使用，而不是使用变量？

例如，如果您想从表'myTable'中选择行，并且'num'是主键：

SELECT num FROM myTable
WHERE num BETWEEN 999 AND 1999

Answer 2

你必须在1000行表上操作才能得到你想要的东西：

SELECT @i := @i+1 AS num
FROM (
  SELECT 1 AS x UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL 
  SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 AS x UNION ALL 
  SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
) AS t
CROSS JOIN (
  SELECT 1 AS x UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL 
  SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 AS x UNION ALL 
  SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
) AS s
CROSS JOIN (
  SELECT 1 AS x UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL 
  SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 AS x UNION ALL 
  SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
) AS u
CROSS JOIN (SELECT @i := 999) AS var

在10 x 10 x 10表组合上使用CROSS JOIN，我们得到一个1000行内联表。以此为基础，我们可以计算@i值所需的序列。

Demo here

Answer 3

这是一个简单的查询，以设置增量值打印数字1000 - 1999。

DECLARE @firstNumber INT
DECLARE @lastNumber INT
DECLARE @incrementBy INT

SET @firstNumber = 1000
SET @lastNumber = 1999
SET @incrementBy = 1

CREATE TABLE #TEMP ( number INT) 

WHILE(@firstNumber <= @lastNumber)
BEGIN
    INSERT INTO #TEMP VALUES (@firstNumber)
    SET @firstNumber = @firstNumber + @incrementBy
END

SELECT * FROM #TEMP

DROP TABLE #TEMP

MySQL选择1000个增量数字

3 个答案: