在 MYSQL 中关闭百分比数学答案

【问题标题】：Closing percent MATH in MYSQL在 MYSQL 中关闭百分比数学
【发布时间】：2013-10-17 21:15:12
【问题描述】：

我想在查询中做数学运算，想知道用 PHP 还是 MYSQL 更好。另外，如果我选择 MYSQL，任何人都可以帮助我进行查询。

目前为止

  SELECT COUNT(*) as total, booker, appdate,
  SUM(CASE WHEN status='DNS' THEN 1 ELSE 0 END) book,
  SUM(CASE WHEN status!='DNS' THEN 1 ELSE 0 END) tot
  FROM appts WHERE WEEK(app_date)= WEEK(CURDATE()) GROUP BY booker

我想从这个查询中获得更多统计信息。我想做book / (book+tot) 但显然只有book!=0 或tot!=0，因为显然我不想将任何东西除以零。

有没有办法在 MYSQL 查询中做到这一点？

我希望我的输出是......

       book  | 14
       tot   | 25
       hold  | 35%

我还想按持有百分比从高到低排序。这可能吗？？？？

【问题讨论】：

标签： php mysql math average

【解决方案1】：

您可以使用子查询来实现您的要求，如下所示：

SELECT *, IF(book + tot, 100*book/(book + tot), NULL) AS hold
FROM (
  SELECT COUNT(*) as total, booker, appdate,
  SUM(status='DNS') book, SUM(status!='DNS') tot
  FROM appts WHERE WEEK(app_date)= WEEK(CURDATE()) GROUP BY booker
) AS subquery
ORDER BY hold DESC

请注意，我在几个地方使用了 MySQL 使用数字作为逻辑值的事实。所以不用CASE可以总结条件，不用<> 0校验也可以写IF的公式。

【讨论】：

【解决方案2】：

朴素的方法：

SELECT
    COUNT(*) as total,
    SUM(CASE WHEN status='DNS' THEN 1 ELSE 0 END) book,
    SUM(CASE WHEN status!='DNS' THEN 1 ELSE 0 END) tot,

    IF(  ( SUM(CASE WHEN status!='DNS' THEN 1 ELSE 0 END) +
           SUM(CASE WHEN status='DNS' THEN 1 ELSE 0 END) ) = 0,
          0,
          SUM(CASE WHEN status='DNS' THEN 1 ELSE 0 END) /
           ( SUM(CASE WHEN status!='DNS' THEN 1 ELSE 0 END) +
             SUM(CASE WHEN status='DNS' THEN 1 ELSE 0 END) )
    ) AS hold
FROM appts
WHERE WEEK(app_date) = WEEK(CURDATE()) GROUP BY booker
ORDER BY hold;

或者为了不重复你的别名，使用子查询：

SELECT *, IF (book + tot = 0, 0, book / (book + tot) * 100)
FROM (
    SELECT
        COUNT(*) as total,
        SUM(CASE WHEN status='DNS' THEN 1 ELSE 0 END) book,
        SUM(CASE WHEN status!='DNS' THEN 1 ELSE 0 END) tot,
    FROM appts
    WHERE WEEK(app_date) = WEEK(CURDATE()) GROUP BY booker
) AS subq
ORDER BY hold;

或者，更聪明:) (book + tot = total)

SELECT *, IF (total = 0, 0, book / total * 100)
FROM (
    SELECT
        COUNT(*) as total,
        SUM(CASE WHEN status='DNS' THEN 1 ELSE 0 END) book,
        SUM(CASE WHEN status!='DNS' THEN 1 ELSE 0 END) tot,
    FROM appts
    WHERE WEEK(app_date) = WEEK(CURDATE()) GROUP BY booker
) AS subq
ORDER BY hold;

而且，只是为了好玩，最紧凑的形式：

SELECT *, COALESCE(book / total * 100, 0) AS hold -- a division by 0 returns NULL
FROM (
    SELECT
        COUNT(*) total,
        SUM(status='DNS') book, -- boolean "true" is internally integer "1"
        SUM(status!='DNS') tot,
    FROM appts
    WHERE WEEK(app_date) = WEEK(CURDATE()) GROUP BY booker
) AS subq
ORDER BY hold;

【讨论】：

@KyleK 顺便说一句，total 永远不会是 0，因为如果COUNT(*) = 0，那么根本就没有行:)