SQL 多关键字查询并根据匹配程度排序

--创建测试表
IF  EXISTS (SELECT * FROM sys.objects WHERE object_id = OBJECT_ID(N'[dbo].[Score]') AND type in (N'U'))
DROP TABLE [dbo].[Score]
GO
CREATE TABLE [dbo].[Score](
    [Id] [int] IDENTITY(1,1) NOT NULL,
    [UserName] [nvarchar](50) NULL,
    [Subject] [nvarchar](50) NULL,
    [Source] [numeric](18, 0) NULL
) ON [PRIMARY]
GO

--插入测试数据
INSERT INTO [Score] ([UserName],[Subject],[Source])
    SELECT N'张三',N'语文',60  UNION ALL
    SELECT N'李四',N'数学',70  UNION ALL
    SELECT N'王五',N'英语',80  UNION ALL
    SELECT N'王五',N'数学',75  UNION ALL
    SELECT N'王五',N'语文',57  UNION ALL
    SELECT N'李四',N'语文',80  UNION ALL
    SELECT N'张三',N'英语',100
GO

--入参：
DECLARE @UserName VARCHAR(100)
DECLARE @Subject VARCHAR(100)
DECLARE @Source VARCHAR(100)
SET @UserName='张'
SET @Subject='语文'
SET @Source='70'


SELECT id, sum((CASE WHEN charIndex(@UserName,[UserName])>0 THEN 1 ELSE 0 END)
              +(CASE WHEN charIndex(@Subject, [Subject])>0 THEN 1 ELSE 0 END)
              +(CASE WHEN [Source]>@Source THEN 1 ELSE 0 END)
              ) AS cnt
INTO #temp
FROM Score GROUP BY id


SELECT * FROM  Score a JOIN #temp b ON a.id=b.id ORDER BY b.cnt DESC,UserName
DROP TABLE #temp