Question

我想在Select中执行以下操作：

获取col1，col2，col3。
比较col2和col3是否匹配，如果是，则显示1，否则显示0。
计算col1中每个项目的比较百分比，以便它只显示col1中每个唯一值的一个百分比。

到目前为止，我有以下内容，但在这种情况下不知道如何获得百分比以及如何正确分组。有人可以告诉我如何调整它以获得我需要的东西吗？

BEGIN
    SET NOCOUNT ON;
    SELECT      (CASE WHEN A.col3 = A.col2 THEN '1' ELSE '0' END) AS match,
                (
                    SELECT      B.col1,
                                B.col2,
                                B.col3
                    FROM        LogTable B
                    WHERE       B.Category LIKE '2014-04%'
                    FOR XML     PATH(''), ELEMENTS, TYPE
                )
    FROM        LogTable A
    WHERE       A.Category LIKE '2014-04%'
    GROUP BY    CASE WHEN A.col3 = A.col2 THEN '1' ELSE '0' END
    FOR XML PATH('comparison'), ELEMENTS, TYPE, ROOT('ranks')

END

非常感谢你提供任何帮助，蒂姆。

Answer 1

它＆＃39;有点不同，但我会尝试这样的事情：

SELECT a.col1, a.total_count, b.match_count,
  (100*b.match_count/a.total_count) AS match_percentage
FROM (
  SELECT col1, COUNT(*) AS total_count
  FROM LogTable
  WHERE Category LIKE '2014-04%'
  GROUP BY col1
) a
JOIN (
  SELECT col1, COUNT(*) AS match_count
  FROM LogTable
  WHERE Category LIKE '2014-04%' AND col2=col3
  GROUP BY col1
) b ON a.col1=b.col1

作为替代......这应该给出相同的结果。不确定哪个更有效：

SELECT col1, total_count,
  (SELECT COUNT(*)
   FROM LogTable
   WHERE Category LIKE '2014-04%' AND col1=a.col1 AND col2=col3
  ) AS match_count,
  (100*match_count/total_count) AS match_percentage
FROM (
  SELECT col1, COUNT(*) AS total_count
  FROM LogTable
  WHERE Category LIKE '2014-04%'
  GROUP BY col1
) a

但是......注意......我不确定所有引擎都能在用于构建match_percentage列的表达式中直接引用子选择列match_count。

SQL Server：比较Select和count匹配中的两列

1 个答案: