我有大量的描述,每个描述可以是5到20个句子。我试图将一个脚本放在一起,找到并删除包含带有数字的单词之前或之后的句子。
之前的例子:Hello world。今天的部门有345名员工。祝你有美好的一天。 例子:你好世界。祝你有个美好的一天。
我现在的主要问题是确定违规行为 这里“345名员工”是导致判决被删除的原因。但是,每个描述将具有不同的数字,并且可能是雇员一词的不同变体。 我想避免创建一个包含员工所有不同变体的表格。
JTB
答案 0 :(得分:3)
这将成为一个很好的SQL拼图。
免责声明:可能会有很多边缘案例会将其搞砸
这将采用一个字符串,将其拆分为一个表,每个句子都有一行,然后删除匹配条件的行,然后最后将它们全部连接回一个字符串。
CREATE FUNCTION dbo.fn_SplitRemoveJoin(@Val VARCHAR(2000), @FilterCond VARCHAR(100))
RETURNS VARCHAR(2000)
AS
BEGIN
DECLARE @tbl TABLE (rid INT IDENTITY(1,1), val VARCHAR(2000))
DECLARE @t VARCHAR(2000)
-- Split into table @tbl
WHILE CHARINDEX('.',@Val) > 0
BEGIN
SET @t = LEFT(@Val, CHARINDEX('.', @Val))
INSERT @tbl (val) VALUES (@t)
SET @Val = RIGHT(@Val, LEN(@Val) - LEN(@t))
END
IF (LEN(@Val) > 0)
INSERT @tbl VALUES (@Val)
-- Filter out condition
DELETE FROM @tbl WHERE val LIKE @FilterCond
-- Join back into 1 string
DECLARE @i INT, @rv VARCHAR(2000)
SET @i = 1
WHILE @i <= (SELECT MAX(rid) FROM @tbl)
BEGIN
SELECT @rv = IsNull(@rv,'') + IsNull(val,'') FROM @tbl WHERE rid = @i
SET @i = @i + 1
END
RETURN @rv
END
go
CREATE TABLE #TMP (rid INT IDENTITY(1,1), sentence VARCHAR(2000))
INSERT #tmp (sentence) VALUES ('Hello world. Todays department has 345 employees. Have a good day.')
INSERT #tmp (sentence) VALUES ('Hello world. Todays department has 15 emps. Have a good day. Oh and by the way there are 12 employees somewhere else')
SELECT
rid, sentence, dbo.fn_SplitRemoveJoin(sentence, '%[0-9] Emp%')
FROM #tmp t
返回
rid | sentence | |
1 | Hello world. Todays department has 345 employees. Have a good day. | Hello world. Have a good day.|
2 | Hello world. Todays department has 15 emps. Have a good day. Oh and by the way there are 12 employees somewhere else | Hello world. Have a good day. |
答案 1 :(得分:2)
我也使用了分割/移除/连接技术。
要点是:
.
或!
或?
这是SqlFiddle demo和代码:
-- Split descriptions into sentences (could use period, exclamation point, or question mark)
-- Delete any sentences that, without whitespace, are like '%[0-9]employ%'
-- Join sentences back into descriptions
;with Splitter as (
select ID
, ltrim(rtrim(Data)) as Data
, cast(null as varchar(max)) as Sentence
, 0 as SentenceNumber
from Descriptions -- Your table here
union all
select ID
, case when Data like '%[.!?]%' then right(Data, len(Data) - patindex('%[.!?]%', Data)) else null end
, case when Data like '%[.!?]%' then left(Data, patindex('%[.!?]%', Data)) else Data end
, SentenceNumber + 1
from Splitter
where Data is not null
), Joiner as (
select ID
, cast('' as varchar(max)) as Data
, 0 as SentenceNumber
from Splitter
group by ID
union all
select j.ID
, j.Data +
-- Don't want "digit+employ" sentences, remove whitespace to search
case when replace(replace(replace(replace(s.Sentence, char(9), ''), char(10), ''), char(13), ''), char(32), '') like '%[0-9]employ%' then '' else s.Sentence end
, s.SentenceNumber
from Joiner j
join Splitter s on j.ID = s.ID and s.SentenceNumber = j.SentenceNumber + 1
)
-- Final Select
select a.ID, a.Data
from Joiner a
join (
-- Only get max SentenceNumber
select ID, max(SentenceNumber) as SentenceNumber
from Joiner
group by ID
) b on a.ID = b.ID and a.SentenceNumber = b.SentenceNumber
order by a.ID, a.SentenceNumber
答案 2 :(得分:0)
一种方法。请注意,只有在所有句子中都有一个数字时才有效。
declare @d VARCHAR(1000) = 'Hello world. Todays department has 345 employees. Have a good day.'
declare @dr VARCHAR(1000)
set @dr = REVERSE(@d)
SELECT REVERSE(RIGHT(@dr,LEN(@dr) - CHARINDEX('.',@dr,PATINDEX('%[0-9]%',@dr))))
+ RIGHT(@d,LEN(@d) - CHARINDEX('.',@d,PATINDEX('%[0-9]%',@d)) + 1)