Question

我有一张像这样的表：

Id     Word
---    ----
1      this
2      is
3      a
4      cat.
5      that
6      is
7      a
8      dog.
9      and
10     so
11     on

并且需要添加一个新的列，用于句号基于点字符：

Id     Word     S#
---    ----     --
1      this     1
2      is       1
3      a        1
4      cat.     1
5      that     2
6      is       2
7      a        2
8      dog.     2
9      and      3
10     so       3
11     on       3

效果方面的最佳解决方案是什么？

Answer 1

select table.id, table.word, count(*) + 1 as serial_number
from table left join
( select id, word from table where word like '%.' ) Z 
on table.id > Z.id
group by table.id, table.word

Answer 2

你假设句子是由提升的身份号码形成的。这是一个非常糟糕的主意。

此查询应该为您提供有关句子中断的信息。（将“T”替换为真实的表名。）

SELECT
    Break1.Id as BreakId,
    COALESCE(MAX(Break2.Id), 0) as PreviousBreakId,
    COALESCE(COUNT(Break2.Id), 0) + 1 as BreakNumber
FROM
    (SELECT Id FROM T WHERE Word LIKE '%.') as Break1,
    (SELECT Id FROM T WHERE Word LIKE '%.') as Break2
WHERE Break2.Id < Break1.Id
GROUP BY Break1.Id

以下是您在UPDATE中使用它的方法。

UPDATE T
SET SentenceNum = (
    SELECT B.BreakNumber
    FROM
        (
        SELECT
            Break1.Id as BreakId,
            COALESCE(MAX(Break2.Id), 0) as PreviousBreakId,
            COALESCE(COUNT(Break2.Id), 0) + 1 as BreakNumber
        FROM
            (SELECT Id FROM T WHERE Word LIKE '%.') as Break1,
            (SELECT Id FROM T WHERE Word LIKE '%.') as Break2
        WHERE
            Break2.Id < Break1.Id
        GROUP BY Break1.Id
        ) as B
    WHERE T.Id >= B.PreviousBreak AND T.Id < B.BreakId
)

我提供了教育价值的查询，但我不能根据您的信息宽恕这种方法。

修改

我的原始版本第一句话有问题，因为基本上逻辑会查找不存在的前一个句子中断。 @ cravoori的解决方案通过左连接处理此问题。这是一个与我自己的答案相同精神的工作版本，它返回完整的单词列表而不是中断。除了交叉连接和虚拟零行之外，它的核心是相同的。

SELECT T.Id, MIN(T.Word) as Word, COUNT(Breaks.Id) as SentenceNumber FROM T, (SELECT 0 as Id UNION ALL SELECT Id FROM T WHERE Word LIKE '%.') as Breaks WHERE Breaks.Id < T.Id GROUP BY T.Id;

Answer 3

检查此查询以获得所需的输出：

DECLARE  @Sentence TABLE(idn int identity,word varchar(50))
INSERT INTO @Sentence
VALUES('this'),('is'),('a'),('cat.'),('that'),('is'),('a'),('dog.'),('and'),('so'),('on.'),('I'),('love'),('india.')
--SELECT * FROM @Sentence 
DECLARE @seq int=1
;WITH CTE(idn,word,seq) AS(
SELECT idn,word,CASE WHEN word not like '%.' then @seq END from @Sentence where idn=1
union all
SELECT s.idn,s.word,CASE WHEN s.word like '%.' then c.seq+1  else c.seq END from @Sentence s inner join CTE c on s.idn-1=c.idn
)
,CTE1(idn,word,seq) As
(SELECT idn,word,CASE WHEN word like '%.' then seq-1 else seq end as seq from CTE)

SELECT * FROM CTE1

如何获取SQL中单词表中每个单词的句号？

3 个答案: