Oracle实现分组统计记录

旧街凉风 提交于 2019-12-23 02:26:19

   今天刚上班不久,QQ滴滴的响个不停,看了下信息是一个网友要我帮忙下一个SQL语句,大体意思是:统计heart_active字段为不同情况的记录数,然后按时间来分组。

   我想了下,心里想这好办,于是马上建了一个表,语句如下:

CREATE TABLE rfid_fixed_heart (input_date date,

heart_active   VARCHAR2(2));

    接下来往rfid_fixed_heart表中插入了数据,heart_active字段为0和1, input_date中插入YYYY-MM-DD格式的数据。

后来就写了下面两个SQL给她,语句一和语句二有点区别,语句一快一统计出heart_active字段为不同情况的记录数,而语句二则只统计heart_active字段为0和1情况的记录数,两个语句的输出格式也有不同。具体如下:

语句一:

SELECT a.input_date, a.heart_active, SUM(decode(a.heart_active, 1, 1, 0, 1))

      FROM rfid_fixed_heart a

 GROUP BY a.heart_active, a.input_date

 ORDER BY a.input_date DESC;

语句二:

SELECT a.input_date, SUM(decode(a.heart_active, '0', '1')) AS heart_active_0,

                   SUM(decode(a.heart_active, '1', '1')) AS heart_active_1

      FROM rfid_fixed_heart a

 GROUP BY a.input_date;

很快就反馈过结果来了,没有达到预期的效果,但从她的结果可以看出是由于input_date插入的是YYYY-MM-DD 24HH:MI:SS格式的数据导致无法按日期来分组。

既然插入的是YYYY-MM-DD 24HH:MI:SS格式得数据,要按日期来排序就需要对input_date使用trunc函数来截取日期值。

最终把原来的两个SQL改成如下语句:

语句三:

SELECT trunc(a.input_date, 'dd'), a.heart_active, SUM(decode(a.heart_active, 1, 1, 0, 1))

      FROM rfid_fixed_heart a

 GROUP BY a.heart_active, trunc(a.input_date, 'dd')

 ORDER BY trunc(a.input_date, 'dd') DESC;

  

    语句四:

SELECT trunc(a.input_date, 'dd'), SUM(decode(a.heart_active, '0', '1')) AS heart_active_0,

                   SUM(decode(a.heart_active, '1', '1')) AS heart_active_1

      FROM rfid_fixed_heart a

 GROUP BY trunc(a.input_date, 'dd');

把语句给那网友后,运行满足要求,OK。对于SQL语句的编写需要认真考虑数据特殊性和表结构,那样才能够实现SQL语句对不同环境的适用。

   附未使用decode函数的实现SQL:

SELECT op_date, heart_active, SUM(heart_active_0) AS heart_active_0,

                   SUM(heart_active_1) AS heart_active_1

      FROM (SELECT to_char(rfid_fixed_heart.input_date, 'yyyy-mm-dd') AS op_date,

                                                heart_active AS heart_active,

                                                CASE heart_active

                                                             WHEN '0' THEN

                                                                  COUNT(heart_active)

                                                             ELSE

                                                                  0

                                                 END AS heart_active_0,

                                                CASE heart_active

                                                             WHEN '1' THEN

                                                                  COUNT(heart_active)

                                                             ELSE

                                                                  0

                                                 END AS heart_active_1

                               FROM rfid_fixed_heart

                              GROUP BY input_date, heart_active) a

 GROUP BY op_date, heart_active

 ORDER BY op_date DESC

      结果如下:

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!