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我一直在努力让我的脑海围绕SQL Server全文搜索如何排名我的结果。使用跨多列的FREETEXTTABLE SQL Server排名奇怪
考虑以下FREETEXTTABLE搜索：
DECLARE @SearchTerm varchar(55) = 'Peter Alex'
SELECT ftt.[RANK], v.*
FROM FREETEXTTABLE (vMembersFTS, (Surname, FirstName, MiddleName, MemberRef, Passport), @SearchTerm) ftt
INNER JOIN vMembersFTS v ON v.ID = ftt.[KEY]
ORDER BY ftt.[RANK] DESC;
这将返回以下结果和排名情况：
RANK ID MemberRef Passport FirstName MiddleName Surname Salutation
----- ---- ---------- ----------- ----------- ------------ ---------- ------------
18 2 AB-002 Pete Peters
18 9 AB-006 George Alex Mr Alex
18 13 AB-009 Peter David Alex Mr Alex
14 3 AB-003 Peter Alex Jones
正如你可以从上面贴的结果告诉，最后一排，虽然有，我认为，在'彼得'和'亚历克斯'，出现在只有14的排名，其中第一排的结果只有一个匹配'彼得'（不可否认的是'彼得斯'）。
这是一个人为的例子，但有些方式可以说明我的挫折和缺乏知识。
我花了相当多的时间研究，但我现在感觉有点超出我的深度。我确信我正在做一些愚蠢的事情，比如搜索多个专栏。
我欢迎您的帮助和支持。提前致谢。
感谢，
凯恩
（顺便说一句，我使用SQL Server 2012）
这里是SQL，您可以使用自己重复测试：
-- Create the Contacts table.
CREATE TABLE dbo.Contacts
(
ID int NOT NULL PRIMARY KEY,
FirstName varchar(55) NULL,
MiddleName varchar(55) NULL,
Surname varchar(55) NOT NULL,
Salutation varchar(55) NULL,
Passport varchar(55) NULL
);
GO
-- Create the Members table.
CREATE TABLE dbo.Members
(
ContactsID int NOT NULL PRIMARY KEY,
MemberRef varchar(55) NOT NULL
);
GO
-- Create the FTS view.
CREATE VIEW dbo.vMembersFTS WITH SCHEMABINDING AS
SELECT c.ID,
m.MemberRef,
ISNULL(c.Passport, '') AS Passport,
ISNULL(c.FirstName, '') AS FirstName,
ISNULL(c.MiddleName, '') AS MiddleName,
c.Surname,
ISNULL(c.Salutation, '') AS Salutation
FROM dbo.Contacts c
INNER JOIN dbo.Members AS m ON m.ContactsID = c.ID
GO
-- Create the view index for FTS.
CREATE UNIQUE CLUSTERED INDEX IX_vMembersFTS_ID ON dbo.vMembersFTS (ID);
GO
-- Create the FTS catalogue and stop-list.
CREATE FULLTEXT CATALOG ContactsFTSCatalog WITH ACCENT_SENSITIVITY = OFF;
CREATE FULLTEXT STOPLIST ContactsSL FROM SYSTEM STOPLIST;
GO
-- Create the member full-text index.
CREATE FULLTEXT INDEX ON dbo.vMembersFTS
(Surname, Firstname, MiddleName, Salutation, MemberRef, Passport)
KEY INDEX IX_vMembersFTS_ID
ON ContactsFTSCatalog
WITH STOPLIST = ContactsSL;
GO
-- Insert some data.
INSERT INTO Contacts VALUES (1, 'John', NULL, 'Smith', NULL, NULL);
INSERT INTO Contacts VALUES (2, 'Pete', NULL, 'Peters', NULL, NULL);
INSERT INTO Contacts VALUES (3, 'Peter', 'Alex', 'Jones', NULL, NULL);
INSERT INTO Contacts VALUES (4, 'Philip', NULL, 'Smith', NULL, NULL);
INSERT INTO Contacts VALUES (5, 'Harry', NULL, 'Dukes', NULL, NULL);
INSERT INTO Contacts VALUES (6, 'Joe', NULL, 'Jones', NULL, NULL);
INSERT INTO Contacts VALUES (7, 'Alex', NULL, 'Phillips', 'Mr Phillips', NULL);
INSERT INTO Contacts VALUES (8, 'Alexander', NULL, 'Paul', 'Alex', NULL);
INSERT INTO Contacts VALUES (9, 'George', NULL, 'Alex', 'Mr Alex', NULL);
INSERT INTO Contacts VALUES (10, 'James', NULL, 'Castle', NULL, NULL);
INSERT INTO Contacts VALUES (11, 'John', NULL, 'Alexander', NULL, NULL);
INSERT INTO Contacts VALUES (12, 'Robert', NULL, 'James', 'Mr James', NULL);
INSERT INTO Contacts VALUES (13, 'Peter', 'David', 'Alex', 'Mr Alex', NULL);
INSERT INTO Members VALUES (1, 'AB-001');
INSERT INTO Members VALUES (2, 'AB-002');
INSERT INTO Members VALUES (3, 'AB-003');
INSERT INTO Members VALUES (5, 'AB-004');
INSERT INTO Members VALUES (8, 'AB-005');
INSERT INTO Members VALUES (9, 'AB-006');
INSERT INTO Members VALUES (11, 'AB-007');
INSERT INTO Members VALUES (12, 'AB-008');
INSERT INTO Members VALUES (13, 'AB-009');
-- Run the FTS query.
DECLARE @SearchTerm varchar(55) = 'Peter Alex'
SELECT ftt.[RANK], v.*
FROM FREETEXTTABLE (vMembersFTS, (Surname, FirstName, MiddleName, MemberRef, Passport), @SearchTerm) ftt
INNER JOIN vMembersFTS v ON v.ID = ftt.[KEY]
ORDER BY ftt.[RANK] DESC;