如何按周分组带有Unix时间戳的行？

Question

我目前使用以下查询来获取按天分组的数据：

SELECT DATE(from_unixtime(timestampcolumn)) as date, COUNT(*)
FROM db.table
WHERE timestampcolumn BETWEEN :startTime AND :endTime
GROUP BY DATE(from_unixtime(timestampcolumn))
ORDER BY timestampcolumn

DATE()函数返回字符串YYYY-MM-DD，因此上述查询很简单，适用于按天分组数据，但如何修改它以返回按每周分组的数据？

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回应乔纳森的回答：

我尝试在SQL Fiddle中创建一个示例，但DATE() SQL函数由于某种原因在SQL Fiddle中不起作用（它在SQL Fiddle中不起作用，但它在我的生活和wamp上都有效）服务器）。

所以这里有一个例子，你可以试试，如果你想：

SETUP：

CREATE TABLE example(id INT NOT NULL AUTO_INCREMENT, PRIMARY KEY(id), timestamp INT, data INT);

INSERT INTO example (timestamp, data) VALUES (1355400000, 1);
INSERT INTO example (timestamp, data) VALUES (1355659200, 1);
INSERT INTO example (timestamp, data) VALUES (1357694861, 1);
INSERT INTO example (timestamp, data) VALUES (1355918400, 1);
INSERT INTO example (timestamp, data) VALUES (1356955200, 1);
INSERT INTO example (timestamp, data) VALUES (1358510400, 1);
INSERT INTO example (timestamp, data) VALUES (1358769600, 1);
INSERT INTO example (timestamp, data) VALUES (1358769600, 1);
INSERT INTO example (timestamp, data) VALUES (1371824368, 1);
INSERT INTO example (timestamp, data) VALUES (1371833476, 1);
INSERT INTO example (timestamp, data) VALUES (1371840324, 1);
INSERT INTO example (timestamp, data) VALUES (1371850523, 1);
INSERT INTO example (timestamp, data) VALUES (1371863191, 1);
INSERT INTO example (timestamp, data) VALUES (1371865401, 1);
INSERT INTO example (timestamp, data) VALUES (1371872379, 1);
INSERT INTO example (timestamp, data) VALUES (1372006190, 1);
INSERT INTO example (timestamp, data) VALUES (1372051945, 1);
INSERT INTO example (timestamp, data) VALUES (1372189402, 1);
INSERT INTO example (timestamp, data) VALUES (1372207830, 1);
INSERT INTO example (timestamp, data) VALUES (1372229733, 1);
INSERT INTO example (timestamp, data) VALUES (1372350338, 1);
INSERT INTO example (timestamp, data) VALUES (1372358259, 1);

QUERY：

SELECT DATE(from_unixtime(timestamp)) as date, COUNT(*)
FROM example
WHERE timestamp BETWEEN 0 AND 9999999999999999999999999999999
GROUP BY DATE(from_unixtime(timestamp))
ORDER BY timestamp

输出：

2012-12-13  1
2012-12-16  1
2012-12-19  1
2012-12-31  1
2013-01-08  1
2013-01-18  1
2013-01-21  2
2013-06-21  7
2013-06-23  1
2013-06-24  1
2013-06-25  2
2013-06-26  1
2013-06-27  2

现在，将时间戳除以(7 * 24 * 3600)。

QUERY：

SELECT DATE(from_unixtime(timestamp / (7 * 24 * 3600))) as date, COUNT(*)
FROM example
WHERE timestamp BETWEEN 0 AND 9999999999999999999999999999999
GROUP BY DATE(from_unixtime(timestamp / (7 * 24 * 3600)))
ORDER BY timestamp

输出：

1969-12-31  22

Answer 1

一个基本想法应该是按时间戳的整数值除以一周中的秒数来分组：timestampcolumn / (7 * 24 * 3600)

然后您需要考虑边缘效应和时区：

• 1970-01-01是在一周的适当日期（这是星期四，所以可能不是）。
• 当您的特定数据确实开始一周时;是受冬季和夏季时间（标准和夏令时）的影响吗？

您可以通过在分割之前向时间戳列添加适当的值来处理这些问题。最后一个转折：一些系统可能会弥补闰秒。 POSIX没有。你必须决定这对你是否重要。

另一个需要考虑的选择是，是否有办法将日期格式化为周值（例如，2013年第23周的2013-W23等ISO 8601表示法）。然后，您可以按该字符串进行分组。

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这就是我的意思：

SELECT timestamp / (7 * 24 * 3600) AS weekno, COUNT(*)
  FROM example
 WHERE timestamp BETWEEN 0 AND 9999999999999999999999999999999
 GROUP BY timestamp / (7 * 24 * 3600)
 ORDER BY weekno

您可能可以使用GROUP BY weekno，或者您可能需要ORDER BY timestamp / (7 * 24 * 3600)，但您可以制作周数。如果您还需要生成日期，则使用：

SELECT timestamp / (7 * 24 * 3600) AS weekno,
       DATE(FROM_UNIXTIME((7 * 24 * 3600) * INT(timestamp / (7 * 24 * 3600))))) AS weekstart,
       COUNT(*)
  FROM example
 WHERE timestamp BETWEEN 0 AND 9999999999999999999999999999999
 GROUP BY timestamp / (7 * 24 * 3600)
 ORDER BY weekno

除非您认为一周的第一天并不总是在数据中表示，否则您可能还会MIN(DATE(FROM_UNIXTIME(timestamp)))使用weekstart。

Answer 2

您是否只能使用原始查询并使用WEEK函数而不是DAY（）？

GROUP BY WEEK(from_unixtime(timestampcolumn))

Answer 3

这是一个基于上述建议的经过测试的干净解决方案：

SELECT  WEEK(from_unixtime(timestamp)) AS week_no, 
        date(from_unixtime(timestamp)) as week_start, 
        COUNT(*) as row_count
FROM api_tracker
GROUP BY WEEK(from_unixtime(timestamp))
ORDER BY week_no DESC

SQL响应：

感谢Shane N和Jonathan Leffler。