Question

示例数据：

╔════╦════════════╦════════════╦═══════╦═══════════╦════════╗
║ ID ║   START    ║    STOP    ║ USER  ║   FILE    ║  SIZE  ║
╠════╬════════════╬════════════╬═══════╬═══════════╬════════╣
║  1 ║ 1330133409 ║ 1330133410 ║ user1 ║ file1.zip ║ 300000 ║
║  2 ║ 1330133409 ║ 1330133410 ║ user1 ║ file2.zip ║ 300500 ║
║  3 ║ 1330133409 ║ 1330133410 ║ user2 ║ file1.zip ║ 300000 ║
║  4 ║ 1330133409 ║ 1330133410 ║ user2 ║ file2.zip ║ 300500 ║
║  5 ║ 1330133409 ║ 1330133410 ║ user1 ║ file3.zip ║ 500000 ║
║  6 ║ 1330133409 ║ 1330133310 ║ user6 ║ file3.zip ║ 500000 ║
╚════╩════════════╩════════════╩═══════╩═══════════╩════════╝

我需要创建一个MySQL查询来计算PER_USER_AVERAGE_BANDWIDTH PER_USER_AVERAGE_BANDWIDTH = SUM(SIZE) / (STOP - START)，然后按PER_USER_AVERAGE_BANDWIDTH排序以生成如下结果：

╔═══════╦════════════════════════════╗
║ USER  ║ PER_USER_AVERAGE_BANDWIDTH ║
╠═══════╬════════════════════════════╣
║ user3 ║ 110.37                     ║
║ user1 ║ 100.25                     ║
║ user2 ║ 75.70                      ║
╚═══════╩════════════════════════════╝

清除泥土;）任何人？

Answer 1

我认为您的平均值应该是总持续时间的总大小，按用户分组：

SELECT   USER,
         SUM(SIZE) / SUM(STOP - START) AS PER_USER_AVERAGE_BANDWIDTH
FROM     my_table
GROUP BY USER
ORDER BY PER_USER_AVERAGE_BANDWIDTH DESC

在sqlfiddle上查看。

Answer 2

直接进行平均

SELECT 
  `user`,
  AVG( size / ( stop - start ) ) per_user_average_bandwidth
FROM
  tab_dl
GROUP BY `user`
ORDER BY per_user_average_bandwidth DESC

SQL Fiddle DEMO

Answer 3

此查询应该这样做：

SELECT USER, (SUM(SIZE) / (STOP - START)) AS PER_USER_AVERAGE_BANDWIDTH
FROM table
GROUP BY USER, stop, start
ORDER BY PER_USER_AVERAGE_BANDWIDTH DESC

这将为您提供每个唯一时间范围内每位用户的平均带宽（即，如果用户在时间1和时间5之间下载文件1和文件2，则在时间1和时间10之间为文件3提供2行）

复杂的mySQL查询顺序依据

3 个答案: