MYSQL星期五分组报告

我们需要在周五到周五的基础上运行整个报表。我没有看到week()这个功能。MYSQL星期五分组报告

假设我们有一个表Summary，有两列 - count和creation_date。我需要根据星期五至星期五的creation_date汇总计数。这需要运行几年。

任何建议，将不胜感激。

来源

2013-03-26 kuriouscoder

为什么week（）不合适？ – Strawberry 2013-03-26 17:29:54

周以周一，周一为起点。 – kuriouscoder 2013-03-26 17:55:43

不能用'my_date-INTERVAL 2 DAY'解决吗？ – Strawberry 2013-03-26 21:34:52

下面是一个查询告诉你上周五和下周四从今天

SELECT DATE(week_beg),DATE(week_end) FROM 
(SELECT daywk_beg + INTERVAL 0 second week_beg, 
daywk_beg + INTERVAL 604799 second week_end 
FROM (SELECT (DATE(NOW()) - INTERVAL daysbacktoday DAY) daywk_beg 
FROM (SELECT SUBSTR('2345601',wkndx,1) daysbacktoday 
FROM (SELECT DAYOFWEEK(dt) wkndx FROM (SELECT DATE(NOW()) dt) 
AAAAA) AAAA) AAA) AA) A;

，这是今天的结果

mysql> SELECT DATE(week_beg),DATE(week_end) FROM 
    -> (SELECT daywk_beg + INTERVAL 0 second week_beg, 
    -> daywk_beg + INTERVAL 604799 second week_end 
    -> FROM (SELECT (DATE(NOW()) - INTERVAL daysbacktoday DAY) daywk_beg 
    -> FROM (SELECT SUBSTR('2345601',wkndx,1) daysbacktoday 
    -> FROM (SELECT DAYOFWEEK(dt) wkndx FROM (SELECT DATE(NOW()) dt) 
    -> AAAAA) AAAA) AAA) AA) A; 
+----------------+----------------+ 
| DATE(week_beg) | DATE(week_end) | 
+----------------+----------------+ 
| 2013-03-22  | 2013-03-28  | 
+----------------+----------------+ 
1 row in set (0.00 sec)

I wrote a query like this in the DBA StackExchange

下面是一个示例表

CREATE TABLE summary 
(
    id int not null auto_increment, 
    ... 
    `count` int not null default 0, 
    creation_date date, 
    primary key (id) 
);

为了让您的查询的基础上周五相符的结果，你需要这个

SELECT SUM(`count`) count_sum,friday FROM 
(SELECT `count`,DATE(week_beg) friday 
(SELECT daywk_beg + INTERVAL 0 second week_beg, 
daywk_beg + INTERVAL 604799 second week_end,`count` 
FROM (SELECT `count`,(DATE(NOW()) - INTERVAL daysbacktoday DAY) daywk_beg 
FROM (SELECT `count`,SUBSTR('2345601',wkndx,1) daysbacktoday 
FROM (SELECT `count`,DAYOFWEEK(dt) wkndx FROM 
(SELECT `count`,creation_date FROM dt) 
AAAAA) AAAA) AAA) AA) A) fri 
GROUP BY friday;

试试看！

来源

2013-03-26 19:14:04 RolandoMySQLDBA

Select 
count, creation_date, 
from 
(
    SELECT 
    count, creation_date 
    DATEDIFF(week, '2013-03-30', date) AS WeekNumber 
    FROM Summary 
) 
GROUP BY 
count, 
creation_date, 
WeekNumber

你会，你可以按

YEARWEEK(`date` - INTERVAL 5 DAY)

代替2013-03-30

来源

2013-03-26 17:24:31

我确实希望一次性生成报告。我不想有相同的工作。 – kuriouscoder 2013-03-26 17:55:23

@ kuriouscoder更新了我的答案 – 2013-03-26 18:26:28

输入开始日期，或者你可以用这个计算您start_week_date和end_week_date：

SELECT 
    `date`, 
    `date` - INTERVAL (DAYOFWEEK(`date`) + 1) % 7 DAY start_week_date, 
    `date` + INTERVAL 6 - (DAYOFWEEK(`date`) + 1) % 7 DAY end_week_date 
FROM 
    dates

请参阅此fiddle。

所以您的查询可能是这样的：

SELECT 
    creation_date - INTERVAL (DAYOFWEEK(creation_date) + 1) % 7 DAY start_week_date, 
    SUM(count) --- or your aggregate function 
FROM 
    Summary 
GROUP BY 
    creation_date - INTERVAL (DAYOFWEEK(creation_date) + 1) % 7 DAY start_week_date

来源

2013-03-26 18:34:01 fthiella

MYSQL星期五分组报告

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