Question

我有以下两个表：

表1：

CREATE TABLE tbl_str_match_1
(
    enumber int,
    ename varchar(100),
    eaddress varchar(500)
);

INSERT INTO tbl_str_match_1 VALUES(1,'John Mak','Hno 12 Street Road, USA');
INSERT INTO tbl_str_match_1 VALUES(2,'Shai Lee','UK');
INSERT INTO tbl_str_match_1 VALUES(3,'Smith Watson','Street X01 UAE');
INSERT INTO tbl_str_match_1 VALUES(4,'Ray Gibbs','SA 124');

表2：

CREATE TABLE tbl_str_match_4
(
    name varchar(100),
    [address] varchar(500)
);

INSERT INTO tbl_str_match_4 VALUES('Mak John','Street Road, Hno 12, USA');
INSERT INTO tbl_str_match_4 VALUES('Shai A Lee','UK');
INSERT INTO tbl_str_match_4 VALUES('A watson Smeeth ','UAE Street X01');
INSERT INTO tbl_str_match_1 VALUES('Henry Jay','RUS OP124');

我想用传递的数字从表tbl_str_match_1中搜索名称，并在下一次搜索时输入名称作为名称，并从另一个称为tbl_str_match_4的表中找到名称和地址。

注意：

名字可以按照任何顺序排列，例如名字中间名，名字中间名或名字中间名，任何可能性都是可能的。
我想从第二个表中查找名称和地址，该表具有一个额外的列，即字符串的百分比匹配。
将进行两次搜索，第一个在表tbl_str_match_1上获取名称，第二个在表tbl_str_match_4上获取名称和地址。
对于第一条记录John Mak，它应显示与Mak John的100％匹配。
对于第二条记录，Shai Lee应该显示Shai A Lee与A的90％匹配。
最后Ray Gibbs条记录将不会显示在结果集中，因为它与其他表值不匹配。

-查询：

WITH CTE1 AS
(   
    SELECT ename FROM tbl_str_match_1 WHERE enumber = 1
)
SELECT name,[address] FROM tbl_str_match_4 WHERE name LIKE '%'+(SELECT ename from CTE1)+'%'

预期结果：

方案1：如果我通过enumber = 1，那么结果应该是：

    Name        Address                     Matching Percentage
    ------------------------------------------------------------
    Mak John    Street Road, Hno 12, USA    100

方案2：如果我通过enumber = 2，那么结果应该是：

    Name        Address                     Matching Percentage
    ------------------------------------------------------------
    Shai A Lee  UK                          90

方案3：如果我通过enumber = 3，那么结果应该是：

    Name                Address             Matching Percentage
    ------------------------------------------------------------
    A watson Smeeth     UAE Street X01      70

方案4：如果我通过enumber = 4，那么结果应该是：

没有结果，因为我们没有任何相关比赛。

    Name        Address                     Matching Percentage
    ------------------------------------------------------------

Answer 1

希望以下帮助。

我首先通过

标记tbl_1和tbl_4名称中的名称

之后，我将tbl_1中的标记与tbl_4比较

有关匹配百分比的问题。在“ Shai A Lee”的示例中，您有3个匹配项（“ Shai”，“ Lee”）中有2个匹配项（“ Shai”，“ A”，“ Lee”），因此匹配百分比不应为66.67吗？ >

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Answer 2

您可以结合使用CTE和STRING SPLIT来完成工作

我在tbl_str_match_4中添加了一个Identity列，以简化此操作

 DECLARE @enumber INT = 2

;WITH c1 AS 
( 
  --To split the ename from first  table 

   SELECT s.value AS name
   FROM tbl_str_match_1 t
   CROSS APPLY STRING_SPLIT(t.ename, ' ') AS s
   WHERE enumber=@enumber
)
,c2 AS
( 
   --To split the matching names from second table of matched records

   SELECT t.id,s.value AS name 
   FROM tbl_str_match_4 t
   CROSS APPLY STRING_SPLIT(t.name, ' ') AS s
   WHERE EXISTS(SELECT 1 FROM c1 c WHERE t.name LIKE '%'+c.name+'%')
)
,c3 AS 
( 
   --To calculate the percentage of match

   SELECT id,
   CAST (COUNT(c1.name) AS FLOAT )/ CAST (COUNT(c2.name) AS FLOAT ) * 100 As Percentage
   FROM c2
   LEFT JOIN  c1 on c1.name =c2.name
   GROUP BY id
) 
--display the details
SELECT t.*,c3.Percentage FROM tbl_str_match_4 t
JOIN c3 ON t.Id=c3.Id

FOR DEMO

Answer 3

希望对您有所帮助。

with CTE1 as

(
Select enumber,Ltrim(SubString(ename,1,Isnull(Nullif(CHARINDEX(' ',ename),0),1000))) As Firstename,

Ltrim(SUBSTRING(ename,CharIndex(' ',ename),
CAse When (CHARINDEX(' ',ename,CHARINDEX(' ',ename)+1)-CHARINDEX(' ',ename))<=0 then 0 
else CHARINDEX(' ',ename,CHARINDEX(' ',ename)+1)-CHARINDEX(' ',ename) end )) as Middleename,

Ltrim(SUBSTRING(ename,Isnull(Nullif(CHARINDEX(' ',ename,Charindex(' ',ename)+1),0),CHARINDEX(' ',ename)),
Case when Charindex(' ',ename)=0 then 0 else LEN(ename) end)) as Lastename

From tbl_str_match_1
),

CTE2 as

(
Select *,Ltrim(SubString(name,1,Isnull(Nullif(CHARINDEX(' ',name),0),1000))) As FirstName,

Ltrim(SUBSTRING(name,CharIndex(' ',name),
CAse When (CHARINDEX(' ',name,CHARINDEX(' ',name)+1)-CHARINDEX(' ',name))<=0 then 0 
else CHARINDEX(' ',name,CHARINDEX(' ',name)+1)-CHARINDEX(' ',name) end )) as MiddleName,

Ltrim(SUBSTRING(name,Isnull(Nullif(CHARINDEX(' ',name,Charindex(' ',name)+1),0),CHARINDEX(' ',name)),
Case when Charindex(' ',name)=0 then 0 else LEN(name) end)) as LastName

From tbl_str_match_4
)

select CTE2.name,CTE2.address from CTE1 inner join CTE2 on  CTE1.Firstename = CTE2.FirstName and CTE1.Lastename = CTE2.LastName
where CTE1.enumber = 1

任意顺序的字符串搜索

3 个答案: