Question

我需要计算特定日期的所有FirstExtracted，并且我需要计算同一日期的所有LastExtracted。所以，对于今天，我需要所有FirstExtracted和LastExtracted等于2012-10-24。

到目前为止，这是我所拥有的，但它没有提出LastExtracted。它输出LastExtracted与FirstExtracted相同的计数：

(SELECT LastExtracted,FirstExtracted,
COUNT(FirstExtracted) AS FirstCount,
COUNT(LastExtracted) AS LastCount,
DATE_FORMAT(`LastExtracted`,'%Y-%m-%d') AS Lastdate,
DATE_FORMAT(`FirstExtracted`,'%Y-%m-%d') AS Firstdate 
FROM results
WHERE DATE_FORMAT(`FirstExtracted`,'%Y-%m-%d') = DATE_FORMAT(`LastExtracted`,'%Y-%m-%d'))
UNION ALL
(SELECT LastExtracted,FirstExtracted,
COUNT(FirstExtracted) AS FirstCount,
COUNT(LastExtracted) AS LastCount,
DATE_FORMAT(`LastExtracted`,'%Y-%m-%d') AS Lastdate,
DATE_FORMAT(`FirstExtracted`,'%Y-%m-%d') AS Firstdate 
FROM results
WHERE DATE_FORMAT(`FirstExtracted`,'%Y-%m-%d') = DATE_FORMAT(`LastExtracted`,'%Y-%m-%d') GROUP BY Firstdate) 
ORDER BY Firstdate DESC
LIMIT 20

也许我应该使用内连接？

更新：所以使用你的查询我做了一些更改，让它现在为我做其他事情。如果你看看这个页面我把查询放了

semesterold.com/code2.html

我想要计算所有头衔和GROUP BY艺术家。这将是一个数组。那么我希望子查询按照艺术家匹配的searchtype计算。所以如果db有akon，rihanna，chris brown。我想要计算每个艺术家有多少个头衔，比如akon有100个。然后我想显示标题的数量，然后计算这100个标题中有多少是google，bing，akon特定的网站等等。

Answer 1

我建议在您的查询中添加DISTINCT：

(SELECT LastExtracted,FirstExtracted,
COUNT(DISTINCT FirstExtracted) AS FirstCount,
COUNT(DISTINCT LastExtracted) AS LastCount,
DATE_FORMAT(`LastExtracted`,'%Y-%m-%d') AS Lastdate,
DATE_FORMAT(`FirstExtracted`,'%Y-%m-%d') AS Firstdate 
FROM results
WHERE DATE_FORMAT(`FirstExtracted`,'%Y-%m-%d') = DATE_FORMAT(`LastExtracted`,'%Y-%m-%d'))
UNION ALL
(SELECT LastExtracted,FirstExtracted,
COUNT(DISTINCT FirstExtracted) AS FirstCount,
COUNT(DISTINCT LastExtracted) AS LastCount,
DATE_FORMAT(`LastExtracted`,'%Y-%m-%d') AS Lastdate,
DATE_FORMAT(`FirstExtracted`,'%Y-%m-%d') AS Firstdate 
FROM results
WHERE DATE_FORMAT(`FirstExtracted`,'%Y-%m-%d') = DATE_FORMAT(`LastExtracted`,'%Y-%m-%d') GROUP BY Firstdate) 
ORDER BY Firstdate DESC
LIMIT 20

Answer 2

我会使用union分别获取两个结果（FirstExtracted和LastExtracted），然后使用子查询合并它们。

SELECT
    sub.date, sum(sub.FirstCount) AS FirstCount,
    sum(sub.LastCount) AS LastCount 
  FROM (
      SELECT
          DATE_FORMAT(`FirstExtracted`,'%Y-%m-%d') AS date,
          COUNT(FirstExtracted) AS FirstCount, 0 AS LastCount
        FROM results
        GROUP BY date
      UNION ALL
        SELECT
            DATE_FORMAT(`LastExtracted`,'%Y-%m-%d'),
            0, COUNT(LastExtracted)
          FROM results
          GROUP BY date
    ) AS sub 
    GROUP BY sub.date

假设您今天首次提取了220个条目，而今天最近提取的条目为292个。在其他日期的条目中，这将给你：

+------------+------------+-----------+
| date       | FirstCount | LastCount |
+------------+------------+-----------+
| 2012-10-24 |        220 |       292 |
+------------+------------+-----------+

更新仅UNION将为您提供以下结果。注意零占位符。

+------------+------------+-----------+
| date       | FirstCount | LastCount |
+------------+------------+-----------+
| 2012-10-24 |        220 |         0 |
+------------+------------+-----------+
| 2012-10-24 |          0 |       292 |
+------------+------------+-----------+

如何计算同一查询中的所有结果和计数 - MySql

2 个答案: