Question

我有一个记录表，用于记录各种数据。有时，每个时间戳会创建多个行，有时可能需要几秒钟的时间，更多的日志条目才能到达表。我想做的是将一组记录分组在一起，彼此之间相隔2-3分钟，并计算这两项的秒数差异。并且，如果可能，请选择最受欢迎的“类别”。

我真的不确定该怎么做，但说我的数据是这样的：

datetime                message                        category
2019-04-03 12:30:00     etc/bar/x did something        b
2019-04-03 12:30:00     there was a failure            b
2019-04-03 12:30:01     xyz is corrupt                 a
2019-04-03 12:31:00     something different            a
2019-04-03 12:31:00     drive on mnt/x is full         c
2019-04-03 12:31:00     extra info                     b
2019-04-03 12:31:05     /etc/bar/foo did something     c
2019-04-03 12:40:01     foo output x                   a
2019-04-03 12:41:30     another message                a

我想要得到的是：

Total time     category
00:01:05       b         # Note this is 1 min, 5 secs, not 1:05am!
00:01:29       a         # Note this is 1 min 29 secs, not 1:29am!

第一个列是hh：mm：ss（不是时间戳而是时间的计数，所以1分5秒）。关于如何解决这个问题有什么想法吗？

或者，或者，我很高兴获得上下限时间输出，例如：

Time between              avg category
2019-04-03 12:30:00       b                      # Note this is the start timestamp, 12:30:00pm
2019-04-03 12:31:05       c                      # Note this is the end timestamp, 12:31:05pm
2019-04-03 12:40:01       a
2019-04-03 12:41:30       a

Answer 1

对于第一个输出，您可以尝试以下查询：

SELECT SEC_TO_TIME(TIME_TO_SEC(MAX(timestamp))-TIME_TO_SEC(MIN(timestamp))) AS "Total Time",
Category FROM your_table GROUP BY category;

据此TIME_TO_SEC将您的TIME值转换为总秒数。因此，每个类别的MAX(timestamp)减去MIN(timestamp)值（均转换为秒值）。然后，通过使用SEC_TO_TIME操作将其结果转换回时间。因此，您将获得hh：mm：ss格式的Total Time。

对于第二个输出，您可以尝试以下查询：

SELECT MIN(timestamp) AS "Timestamp",category 
FROM your_table GROUP BY category UNION 
SELECT MAX(timestamp) AS "Timestamp",category 
FROM your_table GROUP BY category ORDER BY category;

在联合查询顶部获得MIN(timestamp)值，在下面获得MAX(timestamp)值。以ORDER BY category结尾。

如果您有DATE列，并且想要按日期分隔每个类别，只需在选择和分组依据中添加该列即可。例如：

SELECT Date,
SEC_TO_TIME(TIME_TO_SEC(MAX(timestamp))-TIME_TO_SEC(MIN(timestamp))) AS "Total Time",
Category FROM your_table GROUP BY date,category;

AND

SELECT Date,MIN(timestamp) AS "Timestamp",category 
FROM your_table GROUP BY category UNION 
SELECT Date,MAX(timestamp) AS "Timestamp",category 
FROM your_table GROUP BY Date,category ORDER BY date,category;

编辑： 在下面尝试以下查询：

SELECT * FROM 
(SELECT * FROM your_table 
WHERE category="b" 
GROUP BY DATE(timestamp),UNIX_TIMESTAMP(timestamp) DIV 180) sub1 
LEFT JOIN
(SELECT * FROM your_table WHERE category="b") sub2 
ON sub1.category=sub2.category AND DATE(sub1.timestamp)=DATE(sub2.timestamp) 
AND sub1.timestamp<>sub2.timestamp
AND sub2.timestamp BETWEEN sub1.timestamp AND sub1.timestamp + INTERVAL 3 MINUTE;

按时间间隔将时间戳分组在一起并以秒为单位计算差异

1 个答案: