你可以通过一个共同的表表达式可以算出加权做到这一点。例如:
--** Set up the example tables and data
DECLARE @Name TABLE (id INT IDENTITY, name VARCHAR(50));
DECLARE @SearchWords TABLE (word VARCHAR(50));
INSERT INTO @Name
(name)
VALUES ('Jack Nicholson')
,('Henry Jack Blueberry')
,('Pontiac Riddleson Jack')
,('Fred Bloggs');
INSERT INTO @SearchWords
(word)
VALUES ('Jack')
,('Pontiac');
--** Example SELECT with @Name selected and ordered by words in @SearchWords
WITH Order_CTE (weighting, id)
AS (
SELECT COUNT(*) AS weighting
, id
FROM @Name AS n
JOIN @SearchWords AS sw
ON n.name LIKE '%' + sw.word + '%'
GROUP BY id
)
SELECT n.name
, cte.weighting
FROM @Name AS n
JOIN Order_CTE AS cte
ON n.id = cte.id
ORDER BY cte.weighting DESC;
使用这种技术,如果您愿意,也可以将值应用于每个搜索词。所以你可以让杰克比庞蒂克更有价值。这看起来像这样:
--** Set up the example tables and data
DECLARE @Name TABLE (id INT IDENTITY, name VARCHAR(50));
DECLARE @SearchWords TABLE (word VARCHAR(50), value INT);
INSERT INTO @Name
(name)
VALUES ('Jack Nicholson')
,('Henry Jack Blueberry')
,('Pontiac Riddleson Jack')
,('Fred Bloggs');
--** Set up search words with associated value
INSERT INTO @SearchWords
(word, value)
VALUES ('Jack',10)
,('Pontiac',20)
,('Bloggs',40);
--** Example SELECT with @Name selected and ordered by words and values in @SearchWords
WITH Order_CTE (weighting, id)
AS (
SELECT SUM(sw.value) AS weighting
, id
FROM @Name AS n
JOIN @SearchWords AS sw
ON n.name LIKE '%' + sw.word + '%'
GROUP BY id
)
SELECT n.name
, cte.weighting
FROM @Name AS n
JOIN Order_CTE AS cte
ON n.id = cte.id
ORDER BY cte.weighting DESC;
你有没有考虑使用SQL SErver全文搜索? – 2010-06-26 11:14:36
是的,我有 - 它对我们来说效果不好,而且很难根据我们的要求进行定制。 – 2010-06-26 11:16:41
+1全文搜索 - 不一定是SQL Server,但例如lucene.net。 – 2010-06-26 11:16:43