Question

在这个例子中，来自不同员工的薪水和奖金混合在while循环中，这使得计算无效。

如何独立迭代每个ID并根据该ID进行某些计算？

while (select count(*) from db.employee) > 0
begin   
    select top 1 ID, SALARY, [month] 
    from db.employee  
    order by [month] asc

    if deserve = yes
    begin
        set SALARY = SALARY + (bonus * 2)   
    end
end

这不是我的剧本。我有复杂的脚本，需要时间来理解目标。但是，我模仿这个简单想法的逻辑。所以稍后我可以根据建议调整我的原始脚本。原始的是Iterate through tables and do calculation based on one column SQL server

Answer 1

你可以通过多种方式做到这一点。

首先，我只是好奇你为什么不进行正常的更新？

update emp
set salary = emp.salary + (bonus * 2)
from employee emp
join bonuses bon on bon.employee_id = emp.employee_id
where bon.deserved = 1

另一种方法是使用CURSOR，但由于性能问题，人们不太喜欢它们：https://docs.microsoft.com/en-us/sql/t-sql/language-elements/declare-cursor-transact-sql

另一种方法是创建一个容器来保存员工数量，然后循环通过，而当前的itteration不等于计数并在每个循环中获取所需的员工，例如

declare @employeecount int = (select count(*) from employee)
declare @current int = 1

while @current <= @employeecount
begin

    declare @employeeid uniqueidentifier = (select top 1 employee_id from (select top (@current) * from employee order by employee_id asc) tmp order by employee_id desc)

    -- Grab Employee and run logic

set @current = @current + 1;
end

底部差不多就是CURSORS所做的，但我注意到游标做得更好。

以下是我最近采取的行动。我发现它更容易跟随。

declare @ids table (id uniqueidentifier)
insert into @ids
select employee_id
from employee

while (select count(*) from @ids) > 0
begin

    declare @employeeid uniqueidentifier = (select top 1 employee_id from @ids)

    -- Grab Employee and run logic

delete from @ids where id = @employeeid
end

如何在while循环SQL Server中独立迭代列ID

1 个答案: