如何计算一组单词在t-sql中给定数据中出现的次数？

Question

我的目标是得到这个数据集中“the”、“when”和“wolf”的总数。

我将在返回查询中总共有 14 个计数。

编辑：“the”也被添加到单词“they”的计数中

CREATE TABLE #BaseTable([Text] VARCHAR (500))

INSERT INTO #BaseTable ([Text]) VALUES 
('When the villagers heard the cry, they came running up the hill to drive the wolf away.'),
('But, when they arrived, they saw no wolf.'),
('The boy was amused when seeing their angry faces.')

version of SQL Server:

Microsoft SQL Server 2014 - 12.0.2000.8 (Intel X86) 
    Feb 20 2014 19:20:46 
    Copyright (c) Microsoft Corporation
    Express Edition on Windows NT 6.3 <X64> (Build 9600: ) (WOW64) (Hypervisor)

Answer 1

你可以拆分字符串并计数：

select count(*)
from t cross apply
     string_split(replace(replace(t.text, '.', ''), ',', ''), ' ') s
where s.value in ('the', 'when', 'wolf');

SQL Server 对于此类字符串操作不是最佳的，但它是可能的。

Answer 2

一个非常简单的方法可能如下所示：

DECLARE @BaseTable TABLE ([Text] VARCHAR(255))

INSERT INTO @BaseTable ([Text]) VALUES 
('When the villagers heard the cry, they came running up the hill to drive the wolf away.'),
('But, when they arrived, they saw no wolf.'),
('The boy was amused when seeing their angry faces.')

DECLARE @Val1 VARCHAR(255) = 'when'
,       @Val2 VARCHAR(255) = 'but'
,       @Val3 VARCHAR(255) = 'wolf'

SELECT  (LEN(BT.Text) - LEN(REPLACE(BT.Text, @Val1, ''))) /LEN(@Val1) 
,       (LEN(BT.Text) - LEN(REPLACE(BT.Text, @Val2, ''))) /LEN(@Val2)
,       (LEN(BT.Text) - LEN(REPLACE(BT.Text, @Val3, ''))) /LEN(@Val3)
FROM    @BaseTable AS BT

这个想法是计算目标词被替换为空时的长度差异。但是，如果您有选择的话，您并不想在数据库端执行此操作。

编辑： 将所有值相加非常简单：

SELECT  SUM(val1) + SUM(val2) + SUM(val3)
FROM    (
            SELECT  (LEN(BT.Text) - LEN(REPLACE(BT.Text, @Val1, ''))) /LEN(@Val1) AS val1
            ,       (LEN(BT.Text) - LEN(REPLACE(BT.Text, @Val2, ''))) /LEN(@Val2) AS val2
            ,       (LEN(BT.Text) - LEN(REPLACE(BT.Text, @Val3, ''))) /LEN(@Val3) AS val3
            FROM    @BaseTable AS BT
        ) AS subQ

Answer 3

您可以使用字符串函数和横向连接来做到这一点：

select bt.*, x.*
from #basetable bt
cross apply (
    select sum(no_matches) no_matches
    from (
        select (len(bt.text) - len(replace(bt.text, w.word, ''))) / len(w.word) no_matches
        from (values ('the'), ('when'), ('wolf')) w(word)
    ) x
) x

如果您只想要结果集中的匹配总数，那么：

select sum(no_matches) total_matches
from #basetable bt
cross apply (
    select (len(bt.text) - len(replace(bt.text, w.word, ''))) / len(w.word) no_matches
    from (values ('the'), ('when'), ('wolf')) w(word)
) x

Demo on DB Fiddle

Answer 4

如果您的 SQL Server 版本早于 2016 年且 STRING_SPLIT 不可用，则可以使用序号拆分器函数（例如 dbo.DelimitedSplit8K）来分隔单词。

dbo.DelimitedSplit8K

CREATE FUNCTION dbo.DelimitedSplit8K
--===== Define I/O parameters
        (@pString VARCHAR(8000), @pDelimiter CHAR(1))
--WARNING!!! DO NOT USE MAX DATA-TYPES HERE!  IT WILL KILL PERFORMANCE!
RETURNS TABLE WITH SCHEMABINDING AS
 RETURN
--===== "Inline" CTE Driven "Tally Table" produces values from 1 up to 10,000...
     -- enough to cover VARCHAR(8000)
  WITH E1(N) AS (
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
                 SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
                ),                          --10E+1 or 10 rows
       E2(N) AS (SELECT 1 FROM E1 a, E1 b), --10E+2 or 100 rows
       E4(N) AS (SELECT 1 FROM E2 a, E2 b), --10E+4 or 10,000 rows max
 cteTally(N) AS (--==== This provides the "base" CTE and limits the number of rows right up front
                     -- for both a performance gain and prevention of accidental "overruns"
                 SELECT TOP (ISNULL(DATALENGTH(@pString),0)) ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM E4
                ),
cteStart(N1) AS (--==== This returns N+1 (starting position of each "element" just once for each delimiter)
                 SELECT 1 UNION ALL
                 SELECT t.N+1 FROM cteTally t WHERE SUBSTRING(@pString,t.N,1) = @pDelimiter
                ),
cteLen(N1,L1) AS(--==== Return start and length (for use in substring)
                 SELECT s.N1,
                        ISNULL(NULLIF(CHARINDEX(@pDelimiter,@pString,s.N1),0)-s.N1,8000)
                   FROM cteStart s
                )
--===== Do the actual split. The ISNULL/NULLIF combo handles the length for the final element when no delimiter is found.
 SELECT ItemNumber = ROW_NUMBER() OVER(ORDER BY l.N1),
        Item       = SUBSTRING(@pString, l.N1, l.L1)
   FROM cteLen l
;

查询

select count(*)
from #BaseTable bt
     cross apply dbo.DelimitedSplit8K(bt.[Text], ' ') ds
where ds.Item in ('the', 'when', 'wolf');

输出

9