SQL模糊联接-MSSQL

Question

我有两组数据。现有客户和潜在客户。

我的主要目标是弄清楚是否有任何潜在客户已经是现有客户。但是，跨数据集的客户命名约定不一致。

现有客户

Customer /  ID
Ed's Barbershop /   1002
GroceryTown /   1003
Candy Place /   1004
Handy Man / 1005

潜在客户

Customer
Eds Barbershop
Grocery Town
Candy Place
Handee Man
Beauty Salon
The Apple Farm
Igloo Ice Cream
Ride-a-Long Bikes

我想写一些如下所示的选择语句来达到我的目标：

SELECT a.Customer, b.ID
FROM PotentialCustomers a LEFT JOIN
     ExistingCustomers B
     ON a.Customer = b.Customer

结果如下所示：

Customer /  ID
Eds Barbershop  / 1002
Grocery Town    / 1003
Candy Place / 1004
Handee Man  / 1005
Beauty Salon /  NULL
The Apple Farm /    NULL
Igloo Ice Cream / NULL
Ride-a-Long Bikes / NULL

我对Levenshtein距离和Double Metaphone的概念含糊其词，但我不确定如何在此处应用它。

理想情况下，我希望SELECT语句的JOIN部分读取类似以下内容：LEFT JOIN ExistingCustomers as B WHERE a.Customer LIKE b.Customer但我知道语法不正确。

任何建议都欢迎。谢谢！

Answer 1

这是使用Levenshtein Distance可以完成的方法：

创建此函数：（首先执行此操作）

CREATE FUNCTION ufn_levenshtein(@s1 nvarchar(3999), @s2 nvarchar(3999))
RETURNS int
AS
BEGIN
 DECLARE @s1_len int, @s2_len int
 DECLARE @i int, @j int, @s1_char nchar, @c int, @c_temp int
 DECLARE @cv0 varbinary(8000), @cv1 varbinary(8000)

 SELECT
  @s1_len = LEN(@s1),
  @s2_len = LEN(@s2),
  @cv1 = 0x0000,
  @j = 1, @i = 1, @c = 0

 WHILE @j <= @s2_len
  SELECT @cv1 = @cv1 + CAST(@j AS binary(2)), @j = @j + 1

 WHILE @i <= @s1_len
 BEGIN
  SELECT
   @s1_char = SUBSTRING(@s1, @i, 1),
   @c = @i,
   @cv0 = CAST(@i AS binary(2)),
   @j = 1

  WHILE @j <= @s2_len
  BEGIN
   SET @c = @c + 1
   SET @c_temp = CAST(SUBSTRING(@cv1, @j+@j-1, 2) AS int) +
    CASE WHEN @s1_char = SUBSTRING(@s2, @j, 1) THEN 0 ELSE 1 END
   IF @c > @c_temp SET @c = @c_temp
   SET @c_temp = CAST(SUBSTRING(@cv1, @j+@j+1, 2) AS int)+1
   IF @c > @c_temp SET @c = @c_temp
   SELECT @cv0 = @cv0 + CAST(@c AS binary(2)), @j = @j + 1
 END

 SELECT @cv1 = @cv0, @i = @i + 1
 END

 RETURN @c
END

（功能由Joseph Gama开发）

然后只需使用此查询即可获取匹配项

SELECT A.Customer,
       b.ID,
       b.Customer
FROM #POTENTIALCUSTOMERS a
     LEFT JOIN #ExistingCustomers b ON dbo.ufn_levenshtein(REPLACE(A.Customer, ' ', ''), REPLACE(B.Customer, ' ', '')) < 5;

创建该函数后，请完成脚本：

IF OBJECT_ID('tempdb..#ExistingCustomers') IS NOT NULL
    DROP TABLE #ExistingCustomers;

CREATE TABLE #ExistingCustomers
(Customer VARCHAR(255),
 ID       INT
);

INSERT INTO #ExistingCustomers
VALUES
('Ed''s Barbershop',
 1002
);

INSERT INTO #ExistingCustomers
VALUES
('GroceryTown',
 1003
);

INSERT INTO #ExistingCustomers
VALUES
('Candy Place',
 1004
);

INSERT INTO #ExistingCustomers
VALUES
('Handy Man',
 1005
);

IF OBJECT_ID('tempdb..#POTENTIALCUSTOMERS') IS NOT NULL
    DROP TABLE #POTENTIALCUSTOMERS;

CREATE TABLE #POTENTIALCUSTOMERS(Customer VARCHAR(255));

INSERT INTO #POTENTIALCUSTOMERS
VALUES('Eds Barbershop');

INSERT INTO #POTENTIALCUSTOMERS
VALUES('Grocery Town');

INSERT INTO #POTENTIALCUSTOMERS
VALUES('Candy Place');

INSERT INTO #POTENTIALCUSTOMERS
VALUES('Handee Man');

INSERT INTO #POTENTIALCUSTOMERS
VALUES('Beauty Salon');

INSERT INTO #POTENTIALCUSTOMERS
VALUES('The Apple Farm');

INSERT INTO #POTENTIALCUSTOMERS
VALUES('Igloo Ice Cream');

INSERT INTO #POTENTIALCUSTOMERS
VALUES('Ride-a-Long Bikes');

SELECT A.Customer,
       b.ID,
       b.Customer
FROM #POTENTIALCUSTOMERS a
     LEFT JOIN #ExistingCustomers b ON dbo.ufn_levenshtein(REPLACE(A.Customer, ' ', ''), REPLACE(B.Customer, ' ', '')) < 5;

在这里，您可以在以下网址找到T-
SQL示例：http：//www.kodyaz.com/articles/fuzzy-string-matching-using-levenshtein-
distance-sql-server.aspx