在解决我的编码挑战时，我已经求助于您的洞察力很长时间了。我终于被难住了，经过 12 小时的尝试，我必须向专家提出我自己的问题。如果您能提供任何指导，我将不胜感激。
我有两个 mysql 表； POLLS 表有一个民意调查列表，DATA 表是这些民意调查中的数据。我试图按照过去 30 天的平均值确定的顺序列出候选人。
表格投票
poll_id | poll_name | poll_date
_________|_____________|___________
6 | poll 6 | 2015-08-22
5 | poll 5 | 2015-08-14
4 | poll 4 | 2015-08-09
3 | poll 3 | 2015-08-02
2 | poll 2 | 2015-07-28
1 | poll 1 | 2015-07-20
表格数据
data_id | data_key | candidate | percent
________|__________|___________|________
1 | 1 | Joe | 18
2 | 1 | Sue | 19
3 | 1 | Joy | 15
4 | 1 | Tim | 16
5 | 2 | Joe | 20
6 | 2 | Sue | 19
7 | 2 | Joy | 19
8 | 2 | Tim | 22
9 | 3 | Joe | 14
10 | 3 | Sue | 16
11 | 3 | Joy | 21
12 | 3 | Tim | 15
13 | 4 | Joe | 14
14 | 4 | Sue | 15
15 | 4 | Joy | 20
16 | 4 | Tim | 15
17 | 5 | Joe | 17
18 | 5 | Sue | 14
19 | 5 | Joy | 15
20 | 5 | Tim | 16
21 | 6 | Joe | 17
22 | 6 | Sue | 19
23 | 6 | Joy | 16
24 | 6 | Tim | 18
这是我正在寻找的表格结果。它包含表 DATA 中的投票百分比，但候选人是根据过去 30 天的平均百分比按顺序列出的。
candidate | poll 6 | poll 5 | poll 4 | poll 3 | poll 2
__________|________|________|________|________|_______
Joy | 16 | 15 | 20 | 21 | 19
Tim | 18 | 16 | 15 | 15 | 22
Sue | 19 | 14 | 15 | 16 | 19
Joe | 17 | 17 | 14 | 14 | 20
我很困惑，我什至不知道从哪里开始了。我尝试了很多可能性，但只能让它们根据候选人姓名排序。
感谢您的时间和帮助。
最佳答案
例如……
DROP TABLE IF EXISTS polls;
CREATE TABLE polls
(poll_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,poll_name VARCHAR(12) NOT NULL
,poll_date DATE NOT NULL
);
INSERT INTO polls VALUES
(6 ,'poll 6','2015-08-22'),
(5 ,'poll 5','2015-08-14'),
(4 ,'poll 4','2015-08-09'),
(3 ,'poll 3','2015-08-02'),
(2 ,'poll 2','2015-07-28'),
(1 ,'poll 1','2015-07-20');
DROP TABLE IF EXISTS data;
CREATE TABLE data
(data_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,poll_id INT NOT NULL
,candidate VARCHAR(12) NOT NULL
,percent INT NOT NULL
);
INSERT INTO data VALUES
( 1,1,'Joe',18),
( 2,1,'Sue',19),
( 3,1,'Joy',15),
( 4,1,'Tim',16),
( 5,2,'Joe',20),
( 6,2,'Sue',19),
( 7,2,'Joy',19),
( 8,2,'Tim',22),
( 9,3,'Joe',14),
(10,3,'Sue',16),
(11,3,'Joy',21),
(12,3,'Tim',15),
(13,4,'Joe',14),
(14,4,'Sue',15),
(15,4,'Joy',20),
(16,4,'Tim',15),
(17,5,'Joe',17),
(18,5,'Sue',14),
(19,5,'Joy',15),
(20,5,'Tim',16),
(21,6,'Joe',17),
(22,6,'Sue',19),
(23,6,'Joy',16),
(24,6,'Tim',18);
SELECT d.candidate
, p.poll_name
, d.percent
, 30_day_avg
FROM polls p
JOIN data d
ON d.poll_id = p.poll_id
JOIN (SELECT candidate, AVG(percent) 30_day_avg FROM polls p JOIN data d ON d.poll_id = p.poll_id WHERE poll_date >= CURDATE() - INTERVAL 30 day GROUP BY candidate) x
ON x.candidate = d.candidate
WHERE p.poll_date >= CURDATE() - INTERVAL 30 day
ORDER
BY 30_day_avg DESC, candidate, p.poll_id DESC;
+-----------+-----------+---------+------------+
| candidate | poll_name | percent | 30_day_avg |
+-----------+-----------+---------+------------+
| Joy | poll 6 | 16 | 18.2000 |
| Joy | poll 5 | 15 | 18.2000 |
| Joy | poll 4 | 20 | 18.2000 |
| Joy | poll 3 | 21 | 18.2000 |
| Joy | poll 2 | 19 | 18.2000 |
| Tim | poll 6 | 18 | 17.2000 |
| Tim | poll 5 | 16 | 17.2000 |
| Tim | poll 4 | 15 | 17.2000 |
| Tim | poll 3 | 15 | 17.2000 |
| Tim | poll 2 | 22 | 17.2000 |
| Sue | poll 6 | 19 | 16.6000 |
| Sue | poll 5 | 14 | 16.6000 |
| Sue | poll 4 | 15 | 16.6000 |
| Sue | poll 3 | 16 | 16.6000 |
| Sue | poll 2 | 19 | 16.6000 |
| Joe | poll 6 | 17 | 16.4000 |
| Joe | poll 5 | 17 | 16.4000 |
| Joe | poll 4 | 14 | 16.4000 |
| Joe | poll 3 | 14 | 16.4000 |
| Joe | poll 2 | 20 | 16.4000 |
+-----------+-----------+---------+------------+
不幸的是，PHP 不是我的强项，但这里有一个示例，说明您可以如何在 PHP 中转换数组(从而“转换”结果)...
<?php
require('path/to/connection/stateme.nts');
$query = "
SELECT d.candidate
, p.poll_name
, d.percent
, 30_day_avg
FROM polls p
JOIN data d
ON d.poll_id = p.poll_id
JOIN (SELECT candidate, AVG(percent) 30_day_avg FROM polls p JOIN data d ON d.poll_id = p.poll_id WHERE poll_date >= CURDATE() - INTERVAL 30 day GROUP BY candidate) x
ON x.candidate = d.candidate
WHERE p.poll_date >= CURDATE() - INTERVAL 30 day
ORDER
BY 30_day_avg DESC, candidate, p.poll_id DESC;
";
$result = mysqli_query($conn,$query);
$old_array = array();
while($row = mysqli_fetch_assoc($result)){
$old_array[] = $row;
}
$new_array = array();
foreach( $old_array as $v )
{
if(!isset( $new_array[$v["candidate"]][$v["poll_name"]]))
{
$new_array[$v["candidate"]][($v["poll_name"])] = $v["percent"];
}
}
print_r($new_array);
?>
输出:
Array
(
[Joy] => Array
(
[poll 6] => 16
[poll 5] => 15
[poll 4] => 20
[poll 3] => 21
[poll 2] => 19
)
[Tim] => Array
(
[poll 6] => 18
[poll 5] => 16
[poll 4] => 15
[poll 3] => 15
[poll 2] => 22
)
[Sue] => Array
(
[poll 6] => 19
[poll 5] => 14
[poll 4] => 15
[poll 3] => 16
[poll 2] => 19
)
[Joe] => Array
(
[poll 6] => 17
[poll 5] => 17
[poll 4] => 14
[poll 3] => 14
[poll 2] => 20
)
)
希望从那里您可以弄清楚如何使用 html 和/或 css 构建漂亮的表格。
关于php - 根据过去 30 天的平均值对两个表进行排序，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/32156003/