Question

我最近开始尝试从一个我不太熟悉的旧数据库中获取一些数据。它具有的一个功能是每日可用性列，它可以表示资源在一天内可用。不幸的是，该字段是十六进制字段，如下所示：

0x000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF3F0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

我相信这代表上午6点到下午4点半，我不确定怎么样，这不是我的强项。似乎每15个字符代表一个小时，唯一令人困惑的部分是4:30结束时的3F。

所以问题是我如何在SQL中操作/解析这一点，以便在弄清楚数据的格式化后获取开始值和结束值？

以下是更多样本数据：

0x000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000E0FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF0700000000000000

这个代表下午4:29 - 晚上10:59

0x000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000C0FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF0F00000000000000

这是下午4:30 - 晚上11:00

0x000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000C0FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF000000000000000000000000000000

这个代表下午4:30 - 晚上10:00

Answer 1

你的数学分解如下：

一小时3600秒
15个字符代表一小时
3600/15 =每个角色240秒
十六进制F = 16. 240/16 =每个数字15秒。

原来，它一次需要2位数，这就是为什么你看到0C和F0的原因。

这需要更多的润色，但它一直都是我的。

declare @binaryS binary(180)
set @binaryS = 0x000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000E0FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF0700000000000000
declare @s varchar(max)
set @s = Convert(varchar(max), @binaryS, 2)
declare @seconds int
set @seconds = 0
declare @i int
set @i = 1
declare @t datetime
set @t = Convert(datetime, '01/01/2014', 101)
WHILE (@i <= 360)
BEGIN
    declare @segment varchar(max)
    set @segment = substring(@s, @i, 2)
    --print '0x' + @segment
    set @seconds = (convert(int, convert(varbinary, '0x' + @segment, 1))) * 15
    set @t = DATEADD(ss,480,@t)
    --print @seconds
    if(@seconds > 0)
      print @t
    set @i = @i + 2
END

快速解释。 @binaryS是您的db值。它转换为字符串，然后一次解析2位数。对于循环的每次迭代，将480秒（FF）添加到日期。当@seconds不为零时，它会打印出时间。因此，结果是每4分钟启用一次。您需要捕获开始和结束时间中的第一个和最后一个。以上工作在SSMS查询窗口中，但SQLFiddle无法处理它。

将十六进制字段转换为每日可用性（通常和在T-SQL中）

1 个答案: