Question

我的数据库中有两个表，每个表都有名称列。如何比较这些表列以及如何找到具有完全匹配的名称和这些名称，这些名称在table1和table2中类似？

例如：

表1

column1
---------------------
Tom Hawky
Antony Hendric Formen
Cathy Cassy
Bill Gates
Mary Diore

表2

column1
----------------------
Christopher Fridricson
Ken Lovely
Tom Hawky
Anthony Foreman
Chati Cassei

结果应该是这样的：

table1 - table2
Tom Hawky - Tom Hawky
Antony Hendric Formen - Anthony Henrich Foreman
Cathy Cassy - Chati Cassei

Answer 1

一种解决方案是执行以下操作：

1）将名称拆分为一些临时表，以便更容易进行比较

2）构建一个尝试根据距离

查找匹配项的查询

1）分词

<强>设置

-- drop table Table1
create table Table1 
(
    Id1 INT NOT NULL CONSTRAINT PK_Table1 PRIMARY KEY IDENTITY (1, 1),
    Column1 NVARCHAR(200) NOT NULL
)
GO

insert into Table1 (Column1)
VALUES ('Tom Hawky'), ('Antony Hendric Formen'), ('Cathy Cassy'), ('Bill Gates'), ('Mary Diore')
GO

-- drop table Table2
create table Table2
(
    Id2 INT NOT NULL CONSTRAINT PK_Table2 PRIMARY KEY IDENTITY (1, 1),
    Column2 NVARCHAR(200) NOT NULL
)
GO

insert into Table2 (Column2)
VALUES ('Christopher Fridricson'), ('Ken Lovely'), ('Tom Hawky'), ('Anthony Foreman'), ('Chati Cassei'), ('Tom X')
GO

select * from Table1
GO

select * from Table2
GO

拆分功能

可以使用多个split functions，我选择了XML方式：

CREATE FUNCTION dbo.SplitStrings_XML
(
   @List       NVARCHAR(MAX),
   @Delimiter  NVARCHAR(255)
)
RETURNS TABLE
WITH SCHEMABINDING
AS
   RETURN 
   (  
      SELECT Item = y.i.value('(./text())[1]', 'nvarchar(4000)')
      FROM 
      ( 
        SELECT x = CONVERT(XML, '<i>' 
          + REPLACE(@List, @Delimiter, '</i><i>') 
          + '</i>').query('.')
      ) AS a CROSS APPLY x.nodes('i') AS y(i)
   );
GO

拆分

drop table #T1Words
Go

create table #T1Words (Id1 INT NOT NULL, Word NVARCHAR(100))
GO

insert into #T1Words 
select T1.Id1, X1.Item
from Table1 T1
    cross apply dbo.SplitStrings_XML (T1.Column1, N' ') X1
GO

drop table #T2Words
GO 

create table #T2Words (Id2 INT NOT NULL, Word NVARCHAR(200))
GO

insert into #T2Words 
select T2.Id2, X2.Item
from Table2 T2
    cross apply dbo.SplitStrings_XML (T2.Column2, N' ') X2
GO

选择使用一段距离

可以使用的一个距离是Levenshtein。为了在SQL中使用它，它必须在CLR中实现，否则它非常慢。像这样：

[Microsoft.SqlServer.Server.SqlFunction(IsDeterministic = true, IsPrecise = false)]
public static int Levenshtein(SqlString S1, SqlString S2)
{
    if (S1.IsNull)
        S1 = new SqlString("");

    if (S2.IsNull)
        S2 = new SqlString("");

    int maxLen = 4096;

    // keeping only the first part of the string (performance reasons)
    String SC1 = S1.Value.ToUpper();
    String SC2 = S2.Value.ToUpper();

    if (SC1.Length > maxLen)
        SC1 = SC1.Remove(maxLen);
    if (SC2.Length > maxLen)
        SC2 = SC2.Remove(maxLen);

    int n = SC1.Length;
    int m = SC2.Length;

    short[,] d = new short[n + 1, m + 1];
    int cost = 0;

    if (n + m == 0)
    {
        return 0;
    }
    else if (n == 0)
    {
        return 0;
    }
    else if (m == 0)
    {
        return 0;
    }

    for (short i = 0; i <= n; i++)
        d[i, 0] = i;

    for (short j = 0; j <= m; j++)
        d[0, j] = j;

    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            if (SC1[i - 1] == SC2[j - 1])
                cost = 0;
            else
                cost = 1;

            d[i, j] = (short) System.Math.Min(System.Math.Min(d[i - 1, j] + 1, d[i, j - 1] + 1), d[i - 1, j - 1] + cost);
        }
    }

    // double percentage = System.Math.Round((1.0 - ((double)d[n, m] / (double)System.Math.Max(n, m))) * 100.0, 2);
    // return percentage;
    return d[n, m];
}

-- SQL to actually create scalar function that calls CLR code
ALTER FUNCTION dbo.Levenshtein(@S1 nvarchar(max), @S2 nvarchar(max))
    RETURNS INT as EXTERNAL NAME ClrUtils.StoredFunctions.Levenshtein
GO

-- CLR must be enabled for database
sp_configure 'clr enabled', 1
GO
RECONFIGURE
GO

2）查询

select T1.Id1, T1.Column1, T2.Id2, T2.Column2
from Table1 T1
    cross join Table2 T2 
where EXISTS (
    SELECT 1
    FROM #T1Words T1W
        JOIN #T2Words T2W ON [dbo].[Levenshtein](T2W.Word, T1W.Word) < 3   -- T2W.Word = T1W.Word
    WHERE T1W.Id1 = T1.Id1
        AND T2W.Id2 = T2.Id2
)

当然，您必须根据可接受的差异设置阈值。

Answer 2

select table1.column1 + '-' + table2.column1
from table1 join table2 on table1.column1 = table2.column1

Answer 3

您还没有明确定义用于确定相似性的参数。但有趣的方法可能是使用DIFFERENCE函数。这将比较两个字符串的语音表示（使用SOUNDEX函数）并返回0到4之间的值，其中4是最强匹配。所以你可以尝试这样的事情：

SELECT t1.column1 + ' - ' + t2.column1 AS 'table1 - table2'
FROM table1 t1
INNER JOIN table2 t2
ON DIFFERENCE(t1.column1,t2.column1)>= 3

我已将最低比较级别设置为3，但您可以将其调整为最适合您的数据。

问：如何在SQL中的不同表中查找相同和相似的字符串值

3 个答案: