tsql 从数据表中找出缺失的最大、最小ID值。

sample

declare @a table(Account int) 
insert into @a values(1 )  
insert into @a values(2 )  
insert into @a values(3 )  
insert into @a values(4 )  
insert into @a values(9 )  
insert into @a values(10 )  
insert into @a values(11 )  
insert into @a values(20 ) 
--缺失的最小ID号 
SELECT CASE WHEN NOT EXISTS(SELECT * FROM @a WHERE Account = 1) THEN 1 ELSE MIN(Account) + 1 END FROM @a AS A WHERE NOT EXISTS (SELECT * FROM @a AS B WHERE B.Account = A.Account + 1) 
--缺失的最大ID号
SELECT CASE WHEN NOT EXISTS(SELECT * FROM @a WHERE Account = 1) THEN 1 ELSE MAX(Account) - 1 END FROM @a AS A WHERE NOT EXISTS (SELECT * FROM @a AS B WHERE B.Account = A.Account - 1) 

 

返回结果： 缺失最小值：5 缺失最大值：19
思路：将原表数据值与值+1/-1对比，找出缺失的相邻的值，然后条件“如果不存在最小值1则赋于值为1，要不周赋于缺失值表的最小(+1)，最大(-1)