如何从列中提取值并在另一列中更新结果

Question

我有这张桌子：

CREATE TABLE MyTable (
    IdDate int,
    FullDate varchar(255)
);

insert into MyTable (IdDate,FullDate)
VALUES (0, 'Nº1 (26) - Friday 4, January 2014'),
       (0,'Nº2 (64) - Monday 10, February 2015')

我想从 FullDate 中提取如下内容：

1 2014 01 04
2 2015 02 10

1st number is extracted from Nº1
2nd number is extracted from Year
3rd number is extracted from Month (convert January to 01)
4th number is extracted from day (if day < 10, add 0 at the beginning: 01,02... )

并更新在名为 IdDate 的第一列中提取的新值

我的最终结果应该是这样的：

IdDate        FullDate
120140104     Nº1 (26) - Friday 4, January 2014
220150210     Nº2 (64) - Monday 10, February 2015

Answer 1

如果对辅助表值函数开放：

示例

Declare @YourTable table (IdDate int,FullDate varchar(max))
Insert Into @YourTable values
 (0,'Nº1 (26) - Friday 4, January 2014')
,(0,'Nº2 (64) - Monday 10, February 2015')

Update A
   set IdDate = substring(Pos1,3,10)
              + try_convert(varchar(10),try_convert(date,Pos6+' '+Pos5+' '+Pos7),112)
 From  @YourTable A
 Cross Apply [dbo].[tvf-Str-Parse-Row](FullDate,' ') B

退货

IDDate      FullDate
120140104   Nº1 (26) - Friday 4, January 2014
220150210   Nº2 (64) - Monday 10, February 2015

如果它有助于可视化，则 TVF 会返回

感兴趣的功能

CREATE FUNCTION [dbo].[tvf-Str-Parse-Row] (@String varchar(max),@Delimiter varchar(10))
Returns Table 
As
Return (
    Select Pos1 = ltrim(rtrim(xDim.value('/x[1]','varchar(max)')))
          ,Pos2 = ltrim(rtrim(xDim.value('/x[2]','varchar(max)')))
          ,Pos3 = ltrim(rtrim(xDim.value('/x[3]','varchar(max)')))
          ,Pos4 = ltrim(rtrim(xDim.value('/x[4]','varchar(max)')))
          ,Pos5 = ltrim(rtrim(xDim.value('/x[5]','varchar(max)')))
          ,Pos6 = ltrim(rtrim(xDim.value('/x[6]','varchar(max)')))
          ,Pos7 = ltrim(rtrim(xDim.value('/x[7]','varchar(max)')))
          ,Pos8 = ltrim(rtrim(xDim.value('/x[8]','varchar(max)')))
          ,Pos9 = ltrim(rtrim(xDim.value('/x[9]','varchar(max)')))
    From  (Select Cast('<x>' + replace((Select replace(@String,@Delimiter,'§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml) as xDim) as A 
)

或者没有函数

Update A
   set IdDate = substring(Pos1,3,10)
              + try_convert(varchar(10),try_convert(date,Pos6+' '+Pos5+' '+Pos7),112)
 From  @YourTable A
 Cross Apply (
                Select Pos1 = ltrim(rtrim(xDim.value('/x[1]','varchar(max)')))
                      ,Pos2 = ltrim(rtrim(xDim.value('/x[2]','varchar(max)')))
                      ,Pos3 = ltrim(rtrim(xDim.value('/x[3]','varchar(max)')))
                      ,Pos4 = ltrim(rtrim(xDim.value('/x[4]','varchar(max)')))
                      ,Pos5 = ltrim(rtrim(xDim.value('/x[5]','varchar(max)')))
                      ,Pos6 = ltrim(rtrim(xDim.value('/x[6]','varchar(max)')))
                      ,Pos7 = ltrim(rtrim(xDim.value('/x[7]','varchar(max)')))
                From  (Select Cast('<x>' + replace((Select replace(FullDate,' ','§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml) as xDim) as A 
             ) B

编辑

这是 Shawn cleaner 解决方案的扩展版本

Update @YourTable 
   set IdDate = substring(left(FullDate,charindex(' ',FullDate)-1),3,25)
               +try_convert(varchar(10),try_convert(date,replace(substring(FullDate, charindex(',', FullDate) - 2, 100), ',', '')),112)


Select * from @YourTable

Answer 2

这将提取一个日期值。它只是查找逗号，备份几个字符并获取日期，去掉逗号并将其视为军事日期。

select convert(date,  
   replace(substring(FullDate, charindex(',', FullDate) - 2, 100), ',', ''), 106)

使用 format() 或日期样式 112 以您需要的方式获得输出。第一个字符显然只是substring(FullDate, 3, 1)，所以只需将其附加到前面即可。

Answer 3

如果您必须在不使用 John 发布的函数（我会这样做）的情况下解决这个问题，您可以深入挖掘并开始一条噩梦般的字符串函数路径。挑战在于 t-sql 不擅长字符串操作。

这似乎适用于您提供的示例数据。请注意，如果您有无法转换为日期的数据，则会失败。

update MyTable
set IdDate = substring(FullDate, 3, CHARINDEX(' ', FullDate) - 3)
    + Right(FullDate, 4)
    + right('0' + convert(varchar(2), datepart(month, convert(date, replace(substring(substring(FullDate, charindex('-', FullDate) + 2, len(FullDate)), charindex(' ', substring(FullDate, charindex('-', FullDate) + 2, len(FullDate))) + 1, len(FullDate)), ',', '')))), 2)
    + right('0' + convert(varchar(2), datepart(day, convert(date, replace(substring(substring(FullDate, charindex('-', FullDate) + 2, len(FullDate)), charindex(' ', substring(FullDate, charindex('-', FullDate) + 2, len(FullDate))) + 1, len(FullDate)), ',', '')))), 2)

select * from MyTable

然后去问问决定像这样存储数据的人为什么这样做？这根本不是处理数据的方式。

Answer 4

1st number is extracted from Nº1
2nd number is extracted from Year
3rd number is extracted from Month (convert January to 01)
4th number is extracted from day (if day < 10, add 0 at the beginning: 01,02... )

使用SQL Server's string functions 执行以下操作

1. 获取Nº 与其后的第一个空格之间的字符。
2. 获取最后 4 个字符
3. 使用 CASE 表达式根据 LIKE 字符串的月份名称生成月份数
4. 获取逗号前的 2 个字符，并将其中的任何空白字符替换为 0。

并将这 4 个值连接在一起。