Question

出于测试目的，我想通过将colums中的位设置为随机值来更新表。

update [Planned] 
set [IsPlannable] = 1 * rand(cast(cast(newid() as binary(8)) as int))
WHERE [ComputerID] > 100

它似乎确实应该，但不是我想要的方式。我想问题是结果大部分时间都会高于1。

我如何将随机位翻转为随机值？

Answer 1

1 *仍然会产生一个小数＆amp;假设cast(0.1 as bit)将产生1，cast(0.9 as bit)将更新全部设置为1。

你可以;

update Planned set IsPlannable = case when rand(cast(newid() as binary(8))) < 0.5 then 0 else 1 end

Answer 2

根据您必须使用的位字段数，您可以使用以下内容生成所有可能的设置：

with test as (
    select 0 as myId, cast(0 as bit) col1, cast(0 as bit) col2, cast(0 as bit) col3
    union all
    select myId + 1, 
        case when myId & 1 = 1 then cast(1 as bit) else cast(0 as bit) end,
        case when myId & 2 = 2 then cast(1 as bit) else cast(0 as bit) end,
        case when myId & 4 = 4 then cast(1 as bit) else cast(0 as bit) end
        from test
        where myId<100
)
select distinct col1, col2, col3 from test

Answer 3

怎么样

cast(round(rand(), 0) as bit)

如何在SQL中翻转随机位

3 个答案: