我有这张桌子,我需要根据日期选择访客数量。
CREATE TABLE `visitors` (
`id` BIGINT(20) UNSIGNED NOT NULL AUTO_INCREMENT,
`ip` VARCHAR(32) NOT NULL,
`browser` VARCHAR(500) NOT NULL,
`version` VARCHAR(500) NOT NULL,
`platform` ENUM('w','l','m') NOT NULL,
`date` DATE NOT NULL,
PRIMARY KEY (`id`),
UNIQUE KEY `person` (`ip`,`date`)
) ENGINE=INNODB AUTO_INCREMENT=1 DEFAULT CHARSET=utf8;
我试试这个:
SELECT DATE(`date`) AS `date`, COUNT(*) AS `visitor`
FROM `visitors` WHERE (`date` >= (NOW() - INTERVAL 7 DAY))
GROUP BY `date`;
返回此内容:
date visitor
2015-12-19 2
2015-12-22 5
但我真正想要的是:
date visitor
2015-12-17 0 // Make 0 for the day which is not present
2015-12-17 0 // Make 0 for the day which is not present
2015-12-18 0 // Make 0 for the day which is not present
2015-12-19 2
2015-12-20 0 // Make 0 for the day which is not present
2015-12-21 0 // Make 0 for the day which is not present
2015-12-22 5
如果我尝试使用工会并获得日期,这与示例1周没有问题。但如果我需要1年,我该怎么办?
答案 0 :(得分:1)
使用序列引擎。所以你可以在没有表格的情况下生成日期
https://mariadb.com/kb/en/mariadb/sequence/
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" href="style.css">
<script src="script.js"></script>
</head>
<body bgcolor="#F0F0c0" onload="blinker()">
<span style="font-weight:bold; font-size:30px;">Type of Alarm:</span>
<div id="blink">Fire</div>
<span style="font-weight:bold; font-size:30px;"># of Employees Clocked In:</span>
<div id="blink">5</div>
</body>
</html>
这将在没有其他引擎的情况下工作。您只需要一个包含更多记录为天的表。向后的天数是LIMIT。
MariaDB [mysql]> SELECT date(now() - INTERVAL seq DAY) FROM seq_0_to_6;
+--------------------------------+
| date(now() - INTERVAL seq DAY) |
+--------------------------------+
| 2015-12-22 |
| 2015-12-21 |
| 2015-12-20 |
| 2015-12-19 |
| 2015-12-18 |
| 2015-12-17 |
| 2015-12-16 |
+--------------------------------+
7 rows in set (0.00 sec)
请告诉你这件事。