Question

'数据库用于存储大量数据'

我需要在上面的字符串中找到以元音开头的单词数。任何人都可以给我这样做的想法

Answer 1

一种方法是使用常用的“拆分”用户定义函数，并使用它：

SELECT COUNT(*)
FROM dbo.fnSplit(‘Data base is used to store large amount of data’, ' ') x
WHERE x.Value LIKE '[aeiouAEIOU]%'

可以在此处找到拆分UDF的一些示例：T-SQL: Opposite to string concatenation - how to split string into multiple records

如果这是您正在运行的常见查询，我想我会考虑在将值插入表中并将数字存储在行中时计算出该计数。

Answer 2

declare @str2 varchar(max)='Data base is used to store large amount of data'

Declare @data varchar(max)
Declare @Cnt int
Set @Cnt = 0

While (Charindex(' ',@str2)>0)
Begin
Select @Data = ltrim(rtrim(Substring(@str2,1,Charindex(' ',@str2)-1)))

Set @str2 = Substring(@str2,Charindex(' ',@str2)+1,len(@str2))
if(SUBSTRING(@Data,1,1)='a' or SUBSTRING(@Data,1,1)='e' or SUBSTRING(@Data,1,1)='i' or  SUBSTRING(@Data,1,1)='o' or SUBSTRING(@Data,1,1)='u')
 begin
Set @Cnt = @Cnt + 1
End
End

Select @Data = ltrim(rtrim(@str2))
if(SUBSTRING(@Data,1,1)='a' or SUBSTRING(@Data,1,1)='e' or SUBSTRING(@Data,1,1)='i' or    SUBSTRING(@Data,1,1)='o' or SUBSTRING(@Data,1,1)='u')
 begin
 Set @Cnt = @Cnt + 1
 End

select @Cnt

Answer 3

declare @S varchar(100) = 'Data base is used to store large amount of data'

;with cte as
(
  select case when upper(left(@S, 1)) in ('A','E','I','O','U') then 1 else 0 end as vow,
         stuff(@S, 1, charindex(' ', @S+' '), '')+' ' as rest
  where len(@S) > 0
  union all
  select case when upper(left(rest, 1)) in ('A','E','I','O','U') then 1 else 0 end as vow,
         stuff(rest, 1, charindex(' ', rest), '') as rest
  from cte         
  where len(rest) > 0
)
select coalesce(sum(vow), 0)
from cte

Answer 4

OP提到在没有函数的情况下进行此操作，因此请尝试使用单个字符串（假设您有"Numbers"表）：

--sp GEN_FN_GEN_ListToTable
DECLARE @String  varchar(500)
       ,@SplitOn char(1)
SELECT @String='the quick red fox jumped over the lazy brown dog, again.'
      ,@SplitOn=' '


SELECT
    COUNT(*) AS VowelCount
    FROM (SELECT
              LTRIM(RTRIM(SUBSTRING(List2, number+1, CHARINDEX(@SplitOn, List2, number+1)-number - 1))) AS ListValue
              FROM (
                       SELECT @SplitOn + @String + @SplitOn AS List2
                   ) AS dt
                  INNER JOIN Numbers n ON n.Number < LEN(dt.List2)
              WHERE SUBSTRING(List2, number, 1) = @SplitOn
         ) dt2
    WHERE ListValue IS NOT NULL AND ListValue!='' AND LEFT(ListValue,1) IN ('a','e','i','o','u')

输出：

VowelCount
-----------
2

(1 row(s) affected)

这里有一个如何计算表中许多字符串的元音（无函数）的例子（前提是你有一个"Numbers"表）：

DECLARE @YourTable table (RowID int, RowValue  varchar(500))

INSERT INTO @YourTable VALUES (1,'the quick red fox jumped over the lazy brown dog, again.')
INSERT INTO @YourTable VALUES (2,'just another example of some words.')
INSERT INTO @YourTable VALUES (3,'a b c d e f g h i j k l m n o p q r s t u v w x y z')

SELECT
    RowID,COUNT(*) AS VowelCount
    FROM (SELECT
              RowID, LTRIM(RTRIM(SUBSTRING(List2, number+1, CHARINDEX(' ', List2, number+1)-number - 1))) AS ListValue
              FROM (SELECT
                       RowID, ' ' + RowValue + ' ' AS List2
                       FROM @YourTable
                   ) AS dt
                  INNER JOIN Numbers n ON n.Number < LEN(dt.List2)
              WHERE SUBSTRING(List2, number, 1) = ' '
         ) dt2
    WHERE ListValue IS NOT NULL AND ListValue!='' AND LEFT(ListValue,1) IN ('a','e','i','o','u')
    GROUP BY RowID

输出：

RowID       VowelCount    
----------- -----------
1           2
2           3
3           5

(3 row(s) affected)

你基本上需要获得一个split函数，然后在一个空格here is a similar answer上拆分，它解释了一个split函数及其用法。我不知道为什么OP不想使用函数。

Answer 5

SELECT * FROM tbl_name WHERE col_name REGEXP 'vowels'

找到字符串中以元音开头的单词数

5 个答案: