在SQL Server选择查询中将文本从一列解析为三列

Question

我有一张桌子，我需要在其中解析一个包含句子的列（即关于问题的Q和A），我想将这一列解析为3列，请参见下面的示例。

在此先感谢:-)

关于，哈菲兹

Answer 1

根据SUBSTRING, PATHINDEX or CHARINDEX的简单方法

DECLARE @Text VARCHAR(MAX) = '<Date & Time Stamp>- User 1. What has caused the issue ? There was a script which mistakenly included and executed during deployment which caused
loutage 2. How was the issue resolved ? The Negative Impact was resolved by reverting back the
I changes which were deployed and getting back to the previous state.
3. What action are being taken to ensure similar negative impact will not occur again in future ?
Considering a humar error, scripts are being reviewed more diligently over the future release and
we have advised developments teams to keep a track of what is being pushed to make sure it does
'

DECLARE @RevisedText VARCHAR(MAX) = @Text
SET @RevisedText = REPLACE(@RevisedText, '<Date & Time Stamp>- User 1. What has caused the issue ?', '|{1}')
SET @RevisedText = REPLACE(@RevisedText, '2. How was the issue resolved ?', '|{2}')
SET @RevisedText = REPLACE(@RevisedText, '3. What action are being taken to ensure similar negative impact will not occur again in future ?', '|{3}')

SELECT
    Col_1 = (SELECT LTRIM(RTRIM(REPLACE(value, '{1}', ''))) FROM string_split(@RevisedText, '|') WHERE value LIKE '{1}%'),
    Col_2 = (SELECT LTRIM(RTRIM(REPLACE(value, '{2}', ''))) FROM string_split(@RevisedText, '|') WHERE value LIKE '{2}%'),
    Col_3 = (SELECT LTRIM(RTRIM(REPLACE(value, '{3}', ''))) FROM string_split(@RevisedText, '|') WHERE value LIKE '{3}%')

注意：如果您的SQL Server的兼容级别为130，则可以使用UDF代替本地split_string

CREATE FUNCTION string_split (
      @String       NVARCHAR(MAX)
    , @Delimiter    CHAR(1)
)
RETURNS @Table TABLE (StrCol    NVARCHAR(MAX))

AS      
BEGIN   
    DECLARE @Delimiter1 CHAR(3) = '%' + @Delimiter + '%'
    WHILE PATINDEX(@Delimiter1 , @String) <> 0  
    BEGIN
        INSERT INTO @Table VALUES ( LEFT(@String,PATINDEX(@Delimiter1 , @String)-1) )
        SET @String =  STUFF(@String, 1, PATINDEX(@Delimiter1 , @String), '')       
    END
    INSERT INTO @Table VALUES (@String )

    RETURN
END