Question

我有一个SQL语句，我在其中执行此操作

... group by date having date between '2010-07-01' and '2010-07-10';

结果如下：

sum(test)        day
--------------------
20        2010-07-03
120       2010-07-07
33        2010-07-09
42        2010-07-10

所以我有这些结果，但是有可能，我可以写一个语句，让我每天在“介于”条件下返回这种结果行：

sum(test)        day
--------------------
0         2010-07-01
0         2010-07-02
20        2010-07-03
0         2010-07-04
0         2010-07-05
0         2010-07-06
120       2010-07-07
...       ...
42        2010-07-10

否则，如果无法做到这一点，我必须在程序逻辑中执行此操作。

提前多多感谢＆amp;最诚挚的问候。

更新：如果我向您展示完整的SQL语句，也许会更好：

select COALESCE(sum(DUR), 0) AS "r", 0 AS "opt", DATE_FORMAT(date, '%d.%m.%Y') AS "day" from (
    select a.id as ID, a.dur as DUR, DATE(FROM_UNIXTIME(REVTSTMP / 1000)) as date, 
        a_au.re as RE, a_au.stat as STAT from b_c
        join c on b_c.c_id = c.id
        join a on c.id = a.c_id
        join a_au on a.id = a_au.id
        join revi on a_au.re = revi.re
        join (
            select a.id as ID, DATE(FROM_UNIXTIME(REVTSTMP / 1000)) as date, 
            max(a_au.re) as MAX_RE from b_c
                join c on b_c.c_id = c.id
                join a on c.id = a.c_id
                join a_au on a.id = a_au.id
                join revi on a_au.re = revi.re
                where b_c.b_id = 30 group by ID, date) x on
                x.id = a.id and x.date = date and x.MAX_RE = a_au.rev
                where a_au.stat != 7
            group by ID, x.date)
         AS SubSelTable where date between '2010-07-01' and '2010-07-15' group by date;

更新：我的新SQL语句（ - ＆gt; Dave Rix）：

select coalesce(`theData`.`real`, 0) as 'real', 0 as 'opt', DATE_FORMAT(`DT`.`ddDate`, '%d.%m.%Y') as 'date'
    from `dimdates` as DT 
    left join (
        select coalesce(sum(DUR), 0) AS 'real', 0 AS 'opt', date 
            from (
                select a.id as ID, a.dur as DUR, DATE(FROM_UNIXTIME(REVTSTMP / 1000)) as date, a_au.RE as RE, a_au.stat as STAT 
                    from b_c 
                        join c on b_c.c_id = c.id 
                        join a on c.id = a.c_id 
                        join a_au on a.id = a_au.id 
                        join revi on a_au.RE = revi.RE 
                        join ( 
                            select a.id as ID, DATE(FROM_UNIXTIME(REVTSTMP / 1000)) as date, max(a_au.RE) as MAX_RE 
                                from b_c 
                                join c on b_c.c_id = c.id 
                                join a on c.id = a.c_id 
                                join a_au on a.id = a_au.id 
                                join revi on a_au.RE = revi.RE 
                            where b_c.b_id = 30 GROUP BY ID, date
                        ) x 
                    on x.id = a.id and x.date = date and x.MAX_RE = a_au.RE 
                    where a_au.stat != 20 
                    group by ID, x.date
            ) AS SubTable 
        where date between '2010-07-01' and '2010-07-10' group by date) AS theData 
    ON `DT`.`ddDate` = `theData`.`date` where `DT`.`ddDate` between '2010-07-01' and '2010-07-15';

Answer 1

将Between逻辑放在Where子句中

Select Sum(day), day
From Table
Where day Between date1 and date2
Group By day

编辑：只应用于过滤聚合中的数据......即

Having Sum(day) > 10

Answer 2

查看我对以下问题的回答;

Select all months within given date span, including the ones with 0 values

这可能正是您所寻找的：）

您可以按照以下方式修改上面的查询（您可以将其集成，但这种方式更简单！）;


SELECT COALESCE(`theData`.`opt`, 0), `DT`.`myDate`
FROM `dateTable` AS DT
LEFT JOIN (
        ... INSERT YOUR QUERY HERE ...
) AS theData
ON `DT`.`myDate` = `theData`.`date`

您还需要将查询中的DATE_FORMAT(date, '%d.%m.%Y') AS "day"更改为date E.g。

select COALESCE(sum(DUR), 0) AS "r", 0 AS "opt", `date` from

对于@OMG Ponies的回答，您需要预先填充dateTable的大量数据行！

有谁知道如何将此表的SQL转储发布为可以附加的文件？它很大，但可能很有用......

Answer 3

假设您的日期列是DATETIME列，您需要使用某些内容来更改时间值，以便进行正确的分组。 IE：

  SELECT SUM(t.test),
         DATE_FORMAT(t.date, '%Y-%m-%d') AS day
    FROM TABLE t
   WHERE t.date BETWEEN @start AND @end
GROUP BY DATE_FORMAT(t.date, '%Y-%m-%d')

但是如果给定日期没有记录，则日期不会出现在结果集中。换句话说，输出中不会出现零日期。

要解决这个问题，你需要LEFT JOIN到一个日期表，MySQL没有生成的能力。它甚至无法生成数字列表，因此您必须创建一个包含单列的表：

DROP TABLE IF EXISTS `example`.`numbers`;
CREATE TABLE  `example`.`numbers` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

...并填充它：

INSERT INTO numbers (id) VALUES (NULL)

...在您使用数字值生成DATE_ADD function之前的日期列表：

之前

  SELECT COALESCE(SUM(t.test), 0),
         x.the_date AS day
     FROM (SELECT DATE_FORMAT(DATE_ADD(NOW(), INTERVAL n.id-1 DAY), '%Y-%m-%d') AS the_date
             FROM NUMBERS n) x
LEFT JOIN your_table yt ON DATE_FORMAT(yt.date, '%Y-%m-%d') = x.the_date
    WHERE x.the_date BETWEEN @start AND @end
 GROUP BY x.the_date

MySQL：在结果的条件之间使用日期

3 个答案: