SQL Server,ISABOUT,加权术语

Question

我试图弄清楚加密术语在SQL SERVER中的ISABOUT查询中的确切工作方式.这是我目前所处的位置:

每个查询都返回以下行:

QUERY 1(重量1): 初始排名

SELECT * FROM CONTAINSTABLE(documentParts, title, 'ISABOUT ("e" weight (1) ) ') ORDER BY RANK DESC, [KEY]

KEY     RANK
306342  249
272619  156
221557  114

QUERY 2(权重0.8): 排名增加,保留初始订单

SELECT * FROM CONTAINSTABLE(documentParts, title, 'ISABOUT ("e" weight (0.8) ) ') ORDER BY RANK DESC, [KEY]

 KEY     RANK
 306342  321
 272619  201
 221557  146

QUERY 3(权重0.2): 排名增加,保留初始订单

SELECT * FROM CONTAINSTABLE(documentParts, title, 'ISABOUT ("e" weight (0.2) ) ') ORDER BY RANK DESC, [KEY]

 KEY    RANK
 306342 998
 272619 877
 221557 692

QUERY 4(权重0.17): 排名降低,最佳匹配现在是最后一次,这些术语的倒置行为从0.17开始

SELECT * FROM CONTAINSTABLE(documentParts, title, 'ISABOUT ("e" weight (0.17) ) ') ORDER BY RANK DESC, [KEY]

 KEY      RANK
 272619   960
 221557   958
 306342   802

QUERY 5(体重0.16): 排名上升,最佳匹配现在排名第二

SELECT * FROM CONTAINSTABLE(documentParts, title, 'ISABOUT ("e" weight (0.17) ) ') ORDER BY RANK DESC, [KEY]

 KEY      RANK
 272619   978
 306342   935
 221557   841

QUERY 6(重量0.01): 排名降低,最佳匹配再次持续

SELECT * FROM CONTAINSTABLE(documentParts, title, 'ISABOUT ("e" weight (0.01) ) ') ORDER BY RANK DESC, [KEY]

 KEY    RANK
 221557 105
 272619 77
 306342 50

权重1的最佳匹配具有249的等级,而权重下降到最佳匹配的0.2等级增加到998.从0.2到0.17排名减少并且从0.16结果反转(重现此行为的权重值取决于术语和可能在列搜索...)

似乎有一个重量意味着相反的点,就像"不包括这个术语".
你对这种行为有什么解释吗？
为什么在体重下降时排名上升？
为什么排名会在某个点之后下降,直到结果被反转为止,您如何预测这一点？

当用户搜索创建以下查询的内容时,我使用自定义"断字符":

CONTAINSTABLE(documentParts, title, 
      'ISABOUT (
          "wordA wordB wordC" weight (0.8), 
          "wordA*" NEAR "wordB*" NEAR "wordC*" weight (0.6), 
          "wordA*" weight (0.1), 
          "wordB*" weight (0.1), 
          "wordC*" weight (0.1), 
       ) ')

我期待0.1字的大排名吗？

以下查询是否与上述相同,我是否期望0.1排名有一些奇怪的行为？

CONTAINSTABLE(documentParts, title, '
      ISABOUT ( "wordA wordB wordC" weight (0.8) ), 
      OR ISABOUT ( "wordA*" NEAR "wordB*" NEAR "wordC*" weight (0.6) ), 
      OR ISABOUT ( "wordA*" weight (0.1) ), 
      OR ISABOUT ( "wordB*" weight (0.1) ), 
      OR ISABOUT ( "wordC*" weight (0.1) ), 
      ')

编辑:
我发现这个主题:http://msdn.microsoft.com/en-us/library/ms142524(v = sql.105).aspx ,它回答了我的一些问题,但创造了一些新的!

我在两个表中搜索"文档"和"documentParts",并使用union all来对行进行求和并得到我的结果.根据这篇文章,这是错误的,因为索引行计算到计算排名所以RANK将像添加苹果和胡萝卜...

我现在的解决方案是计算每个CONTAINSTABLE的百分比,如下所示:

SELECT * FROM CONTAINSTABLE(documentParts, title, 'ISABOUT ("e" weight (1) ) ') ORDER BY RANK DESC, [KEY]

KEY     RANK
306342  249
272619  156
221557  114

并总结一下......

Answer 1

根据我的经验，当权重加起来为 1 时，我得到了最好的结果。

CONTAINSTABLE(documentParts, content, 
          'ISABOUT (
              "wordA wordB wordC" weight (0.5), 
              "wordA*" NEAR "wordB*" NEAR "wordC*" weight (0.2), 
              "wordA*" weight (0.1), 
              "wordB*" weight (0.1), 
              "wordC*" weight (0.1) 
           ) ')

Answer 2

由于时间在滴答作响，我最终得到了这样的结果，取得了相当好的结果......：

SELECT [KEY], SUM([RANK]) AS [RANK] FROM (
    SELECT [KEY], ([RANK]*1)/(SUM([RANK]) OVER( PARTITION BY 1)/ CAST(COUNT([RANK]) OVER( PARTITION BY 1) AS FLOAT)) AS [RANK] 
        FROM CONTAINSTABLE(documentParts, content, 
              'ISABOUT (
                  "wordA wordB wordC" weight (0.8), 
                  "wordA*" NEAR "wordB*" NEAR "wordC*" weight (0.6), 
                  "wordA*" weight (0.4), 
                  "wordB*" weight (0.4), 
                  "wordC*" weight (0.4) 
               ) ') c
        WHERE c.RANK>0
        UNION ALL      
        SELECT [KEY], ([RANK]*2)/(SUM([RANK]) OVER( PARTITION BY 1)/ CAST(COUNT([RANK]) OVER( PARTITION BY 1) AS FLOAT)) AS [RANK] 
        FROM CONTAINSTABLE(documents, title, 
              'ISABOUT (
                  "wordA wordB wordC" weight (0.8), 
                  "wordA*" NEAR "wordB*" NEAR "wordC*" weight (0.6), 
                  "wordA*" weight (0.4), 
                  "wordB*" weight (0.4), 
                  "wordC*" weight (0.4) 
               ) ') c
         WHERE c.RANK>0
    ) t 
    GROUP BY [KEY]
ORDER BY [RANK] DESC

我会将其传递给测试团队，然后就到此为止......