提问者:小点点

在另一列不相同的情况下对另一列的值求和


Hi StackOverFlow成员

reports=表名。

数据库

CREATE TABLE `reports` (
  `id` int(11) NOT NULL auto_increment,
  `report_day_name` varchar(20) NOT NULL,
  `report_day` varchar(20) NOT NULL,
  `report_month` varchar(20) NOT NULL,
  `report_year` varchar(20) NOT NULL,
  `report_result_number` varchar(20) NOT NULL,
  `report_result_text` varchar(20) NOT NULL,
  `report_since` varchar(20) NOT NULL,
  `report_date` varchar(20) NOT NULL,
  `catid` int(11) NOT NULL,
  `subjectid` int(11) NOT NULL,
  `userid` int(11) NOT NULL,
  `groupid` int(11) NOT NULL,
  PRIMARY KEY  (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=utf8 AUTO_INCREMENT=78 ;

INSERT INTO `reports` VALUES (73, 'day', '14', '1', '1434 h', '5', 'rate', '1234567890', '1434-1-14', 1, 132, 33, 35);
INSERT INTO `reports` VALUES (74, 'day', '12', '2', '1435 h', '4', 'rate', '1234567890', '1434-2-12', 2, 136, 36, 35);
INSERT INTO `reports` VALUES (75, 'day', '14', '1', '1434 h', '2', 'rate', '1354488730', '1434-1-14', 1, 132, 33, 35);
INSERT INTO `reports` VALUES (76, 'day', '12', '2', '1435 h', '4', 'rate', '1354488730', '1434-2-12', 2, 137, 36, 35);
INSERT INTO `reports` VALUES (77, 'day', '12', '2', '1435 h', '1', 'rate', '1354488730', '1434-2-12', 2, 134, 33, 35);

这是数据库表:

id  report_result_number    subjectid   userid
73  5                       132         33
74  4                       136         36
75  2                       132         33
76  4                       137         36
77  1                       134         33

我想sum(reports.report_result_number)where(reports.subjectId)是distinct

当我运行这段代码时..

SELECT
  users.user_id, users.user_name, users.user_country, SUM(reports.report_result_number) AS AllTotal, COUNT(DISTINCT reports.subjectid) AS TotalSubjects
FROM
  users
  INNER JOIN reports ON users.user_id = reports.userid
GROUP BY
  users.user_id
  ORDER BY
  AllTotal DESC LIMIT 4

它返回AllTotal

user_id user_name   user_country    AllTotal    TotalSubjects
36       name         country        8 (correct)        2
33        name        country        8 (not correct)    2

共1个答案

匿名用户

这个问题有两种解释。

<罢工> 如果您想要的是 result_report_number仅当给定的subjectid和userid只有一行时才包含在和聚合中(如果同一subjectid有多行,则希望排除所有这些行的report_result_number...

那么类似这样的事情就会起作用:

SELECT u.user_id
     , u.user_name
     , u.user_country
     , SUM(s.report_result_number) AS AllTotal
     , COUNT(DISTINCT r.subjectid) AS TotalSubjects
  FROM users u
  JOIN reports r
    ON r.userid = u.user_id
  JOIN ( SELECT d.userid
              , d.subjectid
              , d.report_result_number
           FROM reports d
          GROUP
             BY d.userid
              , d.subjectid
         HAVING COUNT(1) = 1
       ) s
    ON s.userid = r.userid
 GROUP
    BY u.user_id
 ORDER
    BY AllTotal DESC
 LIMIT 4

这只是请求的结果集的一种(奇数)解释。示例数据和预期的结果集将有助于澄清规范。

对于添加到问题中的数据,该查询应该返回,例如。

36 fee fi   8  2 
33 foo bar  1  2 

有两行用户33的subjectid值为132,因此这些行的report_result_number被排除在总和之外。subjectid有两个不同的值(132和134),因此我们返回一个:distinct:count of 2。

<罢工> 如果您要求SUM仅在给定用户的subjectid没有重复值时才返回值...

SELECT u.user_id
     , u.user_name
     , u.user_country
     , IF(COUNT(DISTINCT r.subjectid) = COUNT(r.subjectid)
         ,SUM(r.report_result_number)
         ,NULL
       ) AS AllTotal
     , COUNT(DISTINCT r.subjectid) AS TotalSubjects
  FROM users u
  JOIN reports r
    ON r.userid = u.user_id
 GROUP
    BY u.user_id
 ORDER
    BY AllTotal DESC
 LIMIT 4

哈桑说...“如果[给定用户ID的subjectid]有重复值,请获取其中一个值”

只需从别名为s的内联视图中删除HAVING子句。这将返回一行的report_result_number值。(从哪个“匹配”行返回值是任意的:

SELECT u.user_id
     , u.user_name
     , u.user_country
     , SUM(r.report_result_number) AS AllTotal
     , COUNT(DISTINCT r.subjectid) AS TotalSubjects
  FROM users u
  JOIN ( SELECT d.userid
              , d.subjectid
              , d.report_result_number
           FROM reports d
          GROUP
             BY d.userid
              , d.subjectid
       ) r
    ON r.userid = u.user_id
 GROUP
    BY u.user_id
 ORDER
    BY AllTotal DESC
 LIMIT 4

为了使resultset可重复,为了始终获得最低或最高值,可以添加一个聚合函数来指定返回哪个值。

替换...

          , d.report_result_number

与...

          , MAX(d.report_result_number)  AS report_result_number

使用MAX()聚合,将返回:

36 fee fi   8  2
33 foo bar  6  2

(查询将获得subjectid=132 userid=33的值“5”,并将省略相同subjectid的值“2”。)如果没有MAX聚合,查询可以有效地(且任意地)返回一个“3”来代替“6”。(它可以包含“5”或“2”,并省略另一个。)

Q:如何在代码中使用(where report_month='number')?

A:在内联视图中,在GROUP BY子句之前的FROM子句之后添加WHERE子句。替换此:

       FROM reports d
      GROUP

与例如。

       FROM reports d
      WHERE d.report_month = 'number'
      GROUP

仅返回满足指定谓词的行。