这是一个复杂的过程.但是我有一个表,其中包含一个DATETIME字段,以及一些其他的int和float字段,需要对其求和并求平均值.我们希望基于时间戳在此表上进行求和和平均,最终旨在开发3个在某种意义上可以相互补充的查询.
所以能力看起来像这样
TIMESTAMP |subj_diff| SCR2 | SCR3
2011-09-20 09:01:37 | 1 | 0.02 | 1.6
2011-09-20 09:04:18 | 3 | 0.09 | 1.8
2011-09-20 14:24:55 | 5 | 0.21 | 1.2
2011-09-21 18:50:47 | 8 | 0.08 | 0.9
2011-09-21 18:54:21 | 9 | 0.12 | 2.1
我们要生成的三个查询是：
1.对以前的数据(包括当前选择的记录)中的所有前面的项进行汇总.还应该有一个总计列的另一列.例如,如果我们希望返回20号和21号之间的结果,则返回的表应如下所示：
TIMESTAMP |subj_diff| SCR2 | SCR3 | COUNT
2011-09-20 09:01:37 | 1 | 0.02 | ... | 1
2011-09-20 09:04:18 | 4 | 0.11 | | 2
2011-09-20 14:24:55 | 9 | 0.32 | | 3
2011-09-21 18:50:47 | 17 | ...
2011-09-21 18:54:21 | 26 |
2.以5分钟的时间间隔汇总结果-与上述类似,但是查询将返回3行作为第1行和第2行. 2和第4& 5行将以与上述相同的方式相加.在此查询中,如果每隔5分钟没有返回0,则返回0,计数为0.
TIMESTAMP |subj_diff| SCR2 | SCR3 | COUNT
2011-09-20 09:05:00 | 4 | 0.11 | 3.4 | 2
2011-09-20 14:25:00 | 5 | 0.21 | 1.2 | 1
2011-09-21 18:55:00 | 17 | 0.20 | 3.0 | 2
3.在一天中的每5分钟间隔(即从00:05:00到24:00:00),对第二个查询的结果集在第1个查询中执行相同的操作.
这是一个相当棘手的问题,我不知道如何开始.谁能编写SQL来解决这个问题？
这是一些使用游标和存储的proc的基本代码,但实际上并没有用.
DROP PROCEDURE curdemo;
DELIMITER $$
CREATE PROCEDURE curdemo()
BEGIN
DECLARE done INT DEFAULT 0;
DECLARE a datetime;
DECLARE b,c FLOAT;
DECLARE cur1 CURSOR FOR
SELECT msgDate, subj_diff FROM classifier_results
WHERE DATE(msgDate) >= DATE('2011-09-25');
DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1;
CREATE TEMPORARY TABLE IF NOT EXISTS temp_scores (d datetime, acc float);
OPEN cur1;
read_loop: LOOP
FETCH cur1 INTO a, b;
IF done THEN
LEAVE read_loop;
END IF;
INSERT temp_scores(d,acc)
SELECT a, SUM(subj_diff) FROM classifier_results
WHERE DATE(msgDate) >= DATE('2011-09-25')
AND msgDate <= a;
END LOOP;
CLOSE cur1;
SELECT * FROM temp_scores;
END;
干杯!
解决方法:
试试这个代码-
创建并填充表：
CREATE TABLE classifier_results(
`TIMESTAMP` DATETIME NOT NULL,
subj_diff INT(11) DEFAULT NULL,
scr2 FLOAT(10, 5) DEFAULT NULL,
scr3 FLOAT(10, 5) DEFAULT NULL
);
INSERT INTO classifier_results VALUES
('2011-09-20 09:01:37', 1, 0.02000, 1.60000),
('2011-09-20 09:04:18', 3, 0.09000, 1.80000),
('2011-09-20 14:24:55', 5, 0.21000, 1.20000),
('2011-09-21 18:50:47', 8, 0.08000, 0.90000),
('2011-09-21 18:54:21', 9, 0.12000, 2.10000);
并执行以下查询：
-- 1 query
SET @subj_diff = 0;
SET @scr2 = 0;
SET @scr3 = 0;
SET @cnt = 0;
SELECT timestamp,
@subj_diff:=IF(@subj_diff IS NULL, subj_diff, @subj_diff + subj_diff) subj_diff,
@scr2:=IF(@scr2 IS NULL, scr2, @scr2 + scr2) scr2,
@scr3:=IF(@scr3 IS NULL, scr3, @scr3 + scr3) scr3,
@cnt:=@cnt+1 count
FROM classifier_results;
+---------------------+-----------+---------+---------+-------+
| timestamp | subj_diff | scr2 | scr3 | count |
+---------------------+-----------+---------+---------+-------+
| 2011-09-20 09:01:37 | 1 | 0.02000 | 1.60000 | 1 |
| 2011-09-20 09:04:18 | 4 | 0.11000 | 3.40000 | 2 |
| 2011-09-20 14:24:55 | 9 | 0.32000 | 4.60000 | 3 |
| 2011-09-21 18:50:47 | 17 | 0.40000 | 5.50000 | 4 |
| 2011-09-21 18:54:21 | 26 | 0.52000 | 7.60000 | 5 |
+---------------------+-----------+---------+---------+-------+
-- 2 query
SELECT
DATE(timestamp) + INTERVAL 5 * (12 * HOUR(timestamp) + FLOOR(MINUTE(timestamp) / 5)) MINUTE new_timestamp,
SUM(subj_diff) subj_diff,
SUM(scr2) scr2,
SUM(scr3) scr3,
COUNT(*) count
FROM classifier_results
GROUP BY new_timestamp;
+---------------------+-----------+---------+---------+-------+
| new_timestamp | subj_diff | scr2 | scr3 | count |
+---------------------+-----------+---------+---------+-------+
| 2011-09-20 09:00:00 | 4 | 0.11000 | 3.40000 | 2 |
| 2011-09-20 14:20:00 | 5 | 0.21000 | 1.20000 | 1 |
| 2011-09-21 18:50:00 | 17 | 0.20000 | 3.00000 | 2 |
+---------------------+-----------+---------+---------+-------+
-- 3 query
SET @subj_diff = 0;
SET @scr2 = 0;
SET @scr3 = 0;
SET @cnt = 0;
SELECT new_timestamp timestamp,
@subj_diff:=IF(@subj_diff IS NULL, subj_diff, @subj_diff + subj_diff) subj_diff,
@scr2:=IF(@scr2 IS NULL, scr2, @scr2 + scr2) scr2,
@scr3:=IF(@scr3 IS NULL, scr3, @scr3 + scr3) scr3,
@cnt:=@cnt+1 count
FROM (
SELECT
DATE(timestamp) + INTERVAL 5 * (12 * HOUR(timestamp) + FLOOR(MINUTE(timestamp) / 5)) MINUTE new_timestamp,
SUM(subj_diff) subj_diff,
SUM(scr2) scr2,
SUM(scr3) scr3,
COUNT(*) count
FROM classifier_results
GROUP BY new_timestamp
) t;
+---------------------+-----------+---------+---------+-------+
| timestamp | subj_diff | scr2 | scr3 | count |
+---------------------+-----------+---------+---------+-------+
| 2011-09-20 09:00:00 | 4 | 0.11000 | 3.40000 | 1 |
| 2011-09-20 14:20:00 | 9 | 0.32000 | 4.60000 | 2 |
| 2011-09-21 18:50:00 | 26 | 0.52000 | 7.60000 | 3 |
+---------------------+-----------+---------+---------+-------+
祝好运！
标签：aggregation,sql,mysql
来源： https://codeday.me/bug/20191207/2087353.html
