计算匹配词的百分比

Question

CREATE TABLE tbl_pat
(
    id int,
    name varchar(100),
    [address] varchar(500)
);
INSERT INTO tbl_pat VALUES(1,'Jack','Lane 1, 90 Road Street, SL');
INSERT INTO tbl_pat VALUES(2,'Will','SA, Lane 10, Street road');
INSERT INTO tbl_pat VALUES(3,'White','Lane 1 ZIM');
INSERT INTO tbl_pat VALUES(4,'Shaw','Street Road');
INSERT INTO tbl_pat VALUES(5,'Steve','Road Street');
INSERT INTO tbl_pat VALUES(6,'Brown','Nz Road 10');

预期结果：

搜索字符串是：Street Road

Name    Address                     Percentage
---------------------------------------------
Shaw    Street Road                 100
Steve   Road Street                 100
Will    SA, Lane 10, Street road    20
Jack    Lane 1, 90 Road Street, SL  17

注意：百分比是在假设上提到的，但前两个应该是 100%，因为它完全匹配。

我正在使用 PATINDEX 搜索单词。

查询：搜索街道

SELECT [Name],[Address] 
FROM tbl_pat 
WHERE PATINDEX('%Street%',[Address])>=1 AND PATINDEX('%Road%',[Address])>=1 

如何计算单个select语句中匹配词的百分比？

Answer 1

我认为您提供的预期百分比计算存在问题 例如 Jack，有一个匹配 2 output 6 words 。所以我希望它的结果是 %33

请测试我使用String_Split function 将文本拆分为单词的以下 SQL 查询

declare @str nvarchar(max) = 'Road Street'

; with tbl as (
select *, count(*) over (partition by id) word_count
from tbl_pat t
cross apply STRING_SPLIT(replace(t.address,',',' '), ' ')
where trim([value]) <> ''  
)
select distinct id, [name], word_count, count(search.[value]) over (partition by id),
    convert( decimal(5,2), (100.0 * (count(search.[value]) over (partition by id)) / word_count))
from tbl 
left join (
    select * from STRING_SPLIT(@str, ' ')
) search
    on search.[value] = tbl.[value]
order by id

输出是

Answer 2

我在这里使用来自SQL Server 2017 的string_split()。您可以替换为任何可用的字符串拆分功能。随便搜一下

它并不完美，但它适用于您的示例。

select  p.id, p.name, p.address, count(k.value) * 100.0 / count(*) as pecentage
from    tbl_pat p
        cross apply string_split(replace([address], ',', ' '), ' ') w
        left join
        (
            select  value
            from    string_split ('Road Street', ' ') 
        ) k on  w.value     = k.value
group by p.id, p.name, p.address