我很幸运能在Stack Overflow上找到这段令人敬畏的代码，但是我想改变它所以它显示每半小时而不是每小时，但是搞乱它，只会导致我破坏查询哈哈
这是SQL：
SELECT CONCAT(HOUR(created_at), ':00-', HOUR(created_at)+1, ':00') as hours,
COUNT(*)
FROM urls
GROUP BY HOUR(created_at)
ORDER BY HOUR(created_at) ASC
我如何每半小时获得一次结果？ ：）
另一件事是，如果它有半小时没有结果，我希望它返回0而不是仅仅跳过那一步。它看起来有点奇怪的胜利我对查询做了统计，当它只跳过一个小时因为没有：P
答案 0 :(得分：5)
如果格式不太重要，则可以为间隔返回两列。您甚至可能只需要间隔的开始，可以通过以下方式确定：
date_format(created_at - interval minute(created_at)%30 minute, '%H:%i') as period_start
别名可以在GROUP BY和ORDER BY子句中使用。如果您还需要间隔结束，则需要进行一些小修改：
SELECT
date_format(created_at - interval minute(created_at)%30 minute, '%H:%i') as period_start,
date_format(created_at + interval 30-minute(created_at)%30 minute, '%H:%i') as period_end,
COUNT(*)
FROM urls
GROUP BY period_start
ORDER BY period_start ASC;
当然，您也可以连接值：
SELECT concat_ws('-',
date_format(created_at - interval minute(created_at)%30 minute, '%H:%i'),
date_format(created_at + interval 30-minute(created_at)%30 minute, '%H:%i')
) as period,
COUNT(*)
FROM urls
GROUP BY period
ORDER BY period ASC;
演示：http://rextester.com/RPN50688
另一件事是，如果它有半个小时没有结果，我 希望它返回0
如果以过程语言使用结果，则可以在循环中初始化所有48行，然后从结果中“注入”非零行。
然而 - 如果你需要在SQL中完成它，你需要一个表至LEFT JOIN，至少有48行。这可以用“巨大的”UNION ALL语句内联完成，但（恕我直言）这将是丑陋的。所以我更喜欢让序列表有一个整数列，这对于报告非常有用。要创建该表，我通常使用information_schema.COLUMNS，因为它可以在任何MySQL服务器上使用，并且至少有几百行。如果您需要更多行 - 只需将其与自身连接即可。
现在让我们创建该表：
drop table if exists helper_seq;
create table helper_seq (seq smallint auto_increment primary key)
select null
from information_schema.COLUMNS c1
, information_schema.COLUMNS c2
limit 100; -- adjust as needed
现在我们有一个从1到100的整数表（虽然现在你只需要48 - 但这是为了演示）。
使用该表我们现在可以创建所有48个时间间隔：
select time(0) + interval 30*(seq-1) minute as period_start,
time(0) + interval 30*(seq) minute as period_end
from helper_seq s
where s.seq <= 48;
我们将得到以下结果：
period_start | period_end
00:00:00 | 00:30:00
00:30:00 | 01:00:00
...
23:30:00 | 24:00:00
演示：http://rextester.com/ISQSU31450
现在我们可以将它用作派生表（FROM子句中的子查询）和LEFT JOIN urls表：
select p.period_start, p.period_end, count(u.created_at) as cnt
from (
select time(0) + interval 30*(seq-1) minute as period_start,
time(0) + interval 30*(seq) minute as period_end
from helper_seq s
where s.seq <= 48
) p
left join urls u
on time(u.created_at) >= p.period_start
and time(u.created_at) < p.period_end
group by p.period_start, p.period_end
order by p.period_start
演示：http://rextester.com/IQYQ32927
最后一步（如果真的需要）是格式化结果。我们可以在外部选择中使用CONCAT或CONCAT_WS和TIME_FORMAT。最终的查询是：
select concat_ws('-',
time_format(p.period_start, '%H:%i'),
time_format(p.period_end, '%H:%i')
) as period,
count(u.created_at) as cnt
from (
select time(0) + interval 30*(seq-1) minute as period_start,
time(0) + interval 30*(seq) minute as period_end
from helper_seq s
where s.seq <= 48
) p
left join urls u
on time(u.created_at) >= p.period_start
and time(u.created_at) < p.period_end
group by p.period_start, p.period_end
order by p.period_start
结果如下：
period | cnt
00:00-00:30 | 1
00:30-01:00 | 0
...
23:30-24:00 | 3
答案 1 :(得分：2)
嗯，这可能有点冗长，但它确实有效：
SELECT hours, SUM(count) as count FROM (
SELECT CONCAT(HOUR(created_at), ':', LPAD(30 * FLOOR(MINUTE(created_at)/30), 2, '0'), '-',
HOUR(DATE_ADD(created_at, INTERVAL 30 minute)), ':', LPAD(30 * FLOOR(MINUTE(DATE_ADD(created_at, INTERVAL 30 minute))/30), 2, '0')) as hours,
COUNT(*) as count
FROM urls
GROUP BY HOUR(created_at), FLOOR(MINUTE(created_at)/30)
UNION ALL
SELECT '00:00-00:30'as hours, 0 as count UNION ALL SELECT '00:30-01:00'as hours, 0 as count UNION ALL
SELECT '01:00-01:30'as hours, 0 as count UNION ALL SELECT '01:30-02:00'as hours, 0 as count UNION ALL
SELECT '02:00-02:30'as hours, 0 as count UNION ALL SELECT '02:30-03:00'as hours, 0 as count UNION ALL
SELECT '03:00-03:30'as hours, 0 as count UNION ALL SELECT '03:30-04:00'as hours, 0 as count UNION ALL
SELECT '04:00-04:30'as hours, 0 as count UNION ALL SELECT '04:30-05:00'as hours, 0 as count UNION ALL
SELECT '05:00-05:30'as hours, 0 as count UNION ALL SELECT '05:30-06:00'as hours, 0 as count UNION ALL
SELECT '06:00-06:30'as hours, 0 as count UNION ALL SELECT '06:30-07:00'as hours, 0 as count UNION ALL
SELECT '07:00-07:30'as hours, 0 as count UNION ALL SELECT '07:30-08:00'as hours, 0 as count UNION ALL
SELECT '08:00-08:30'as hours, 0 as count UNION ALL SELECT '08:30-09:00'as hours, 0 as count UNION ALL
SELECT '09:00-09:30'as hours, 0 as count UNION ALL SELECT '09:30-10:00'as hours, 0 as count UNION ALL
SELECT '10:00-10:30'as hours, 0 as count UNION ALL SELECT '10:30-11:00'as hours, 0 as count UNION ALL
SELECT '11:00-11:30'as hours, 0 as count UNION ALL SELECT '11:30-12:00'as hours, 0 as count UNION ALL
SELECT '12:00-12:30'as hours, 0 as count UNION ALL SELECT '12:30-13:00'as hours, 0 as count UNION ALL
SELECT '13:00-13:30'as hours, 0 as count UNION ALL SELECT '13:30-14:00'as hours, 0 as count UNION ALL
SELECT '14:00-14:30'as hours, 0 as count UNION ALL SELECT '14:30-15:00'as hours, 0 as count UNION ALL
SELECT '15:00-15:30'as hours, 0 as count UNION ALL SELECT '15:30-16:00'as hours, 0 as count UNION ALL
SELECT '16:00-16:30'as hours, 0 as count UNION ALL SELECT '16:30-17:00'as hours, 0 as count UNION ALL
SELECT '17:00-17:30'as hours, 0 as count UNION ALL SELECT '17:30-18:00'as hours, 0 as count UNION ALL
SELECT '18:00-18:30'as hours, 0 as count UNION ALL SELECT '18:30-19:00'as hours, 0 as count UNION ALL
SELECT '19:00-19:30'as hours, 0 as count UNION ALL SELECT '19:30-20:00'as hours, 0 as count UNION ALL
SELECT '20:00-20:30'as hours, 0 as count UNION ALL SELECT '20:30-21:00'as hours, 0 as count UNION ALL
SELECT '21:00-21:30'as hours, 0 as count UNION ALL SELECT '21:30-22:00'as hours, 0 as count UNION ALL
SELECT '22:00-22:30'as hours, 0 as count UNION ALL SELECT '22:30-23:00'as hours, 0 as count UNION ALL
SELECT '23:00-23:30'as hours, 0 as count UNION ALL SELECT '23:30-00:00'as hours, 0 as count
) AS T
GROUP BY hours ORDER BY hours;
查询中最困难的部分是输出没有任何命中的间隔的统计信息。 SQL就是查询和聚合现有数据;选择或汇总表中缺失的数据是非常不寻常的任务。这就是为什么像Wolph在评论中所说的那样，没有完美的解决方案来完成这项任务。
我通过明确选择当天的所有半间隔来解决这个问题。如果间隔数量有限，则可以使用此解决方案。但是，如果您在很长一段时间内汇总了不同的日期，这将无效。
我不是这个问题的粉丝，但我无法提出更好的建议。使用循环存储过程可以实现更优雅的解决方案，但似乎您希望使用原始SQL查询来解决它。
答案 2 :(得分：1)
30*60半小时）LEFT JOIN甚至可以获得缺少的时间单位。GROUP BY。（步骤3和4是可选的。问题是＆＃34;每个＆＃34;，所以我认为它们是必需的。）
步骤1和2体现在类似
的内容中FLOOR(UNIX_TIMESTAMP(created_at) / (30*60))
例如：
mysql> SELECT NOW(), FLOOR(UNIX_TIMESTAMP(NOW()) / (30*60));
+---------------------+----------------------------------------+
| NOW() | FLOOR(UNIX_TIMESTAMP(NOW()) / (30*60)) |
+---------------------+----------------------------------------+
| 2018-03-02 08:24:48 | 844448 |
+---------------------+----------------------------------------+
步骤3需要进行一次并保存在永久表中。或者，如果您有MariaDB，请使用＆＃34; seq＆＃34;假表;例如，`seq_844448_to_900000会动态地提供一个可以在未来很长的表格。
第6步示例：
mysql> SELECT DATE_FORMAT(FROM_UNIXTIME((844448) * 30*60), "%b %d %h:%i");
+-------------------------------------------------------------+
| DATE_FORMAT(FROM_UNIXTIME((844448) * 30*60), "%b %d %h:%i") |
+-------------------------------------------------------------+
| Mar 02 08:00 |
+-------------------------------------------------------------+
+---------------------------------------------------------------+
| DATE_FORMAT(FROM_UNIXTIME((844448+1) * 30*60), "%b %d %h:%i") |
+---------------------------------------------------------------+
| Mar 02 08:30 |
+---------------------------------------------------------------+
答案 3 :(得分：0)
您可以添加一些数学来计算48个区间而不是24个区间，并将其放入另一个您要进行分组和排序的区域。
SELECT HOUR(created_at)*2+FLOOR(MINUTE(created_at)/30) as interval48,
if(HOUR(created_at)*2+FLOOR(MINUTE(created_at)/30) % 2 =0,
CONCAT(HOUR(created_at), ':00-', HOUR(created_at), ':30'),
CONCAT(HOUR(created_at), ':30-', HOUR(created_at)+1, ':00')
) as hours,
count(*)
FROM urls
GROUP BY HOUR(created_at)*2+FLOOR(MINUTE(created_at)/30)
ORDER BY HOUR(created_at)*2+FLOOR(MINUTE(created_at)/30) ASC
结果示例：
0 0:00-0:30 2017
1 0:30-1:00 1959
2 1:30-2:00 1830
3 1:30-2:00 1715
4 2:30-3:00 1679
5 2:30-3:00 1688
Jazerix发布的原始查询结果是：
0:00-1:00 3976
1:00-2:00 3545
2:00-3:00 3367
答案 4 :(得分：0)
另一种方法，无需创建其他表格。可能看起来像黑客： - ）
第1步：动态生成时间表
假设：INFORMATION_SCHEMA DB是可用的，并且有一个表COLLATIONS，通常有超过100条记录。您可以使用任何至少有48条记录的表
查询：
id select_type table type possible_keys key key_len ref rows Extra
1 PRIMARY t1 ref dp_tech_licence_history_mod_Property dp_tech_licence_history_mod_Property 766 const 4411 Using index condition
1 PRIMARY t2 eq_ref PRIMARY PRIMARY 8 t1.HistoryId 1
1 PRIMARY t3 eq_ref PRIMARY PRIMARY 8 t2.CollectionId 1 Using where
2 DEPENDENT SUBQUERY t5 ref PRIMARY,dp_tech_licence_history_Licence,dp_tech_licence_history_Collection dp_tech_licence_history_Licence 8 t2.LicenceId 3 Using temporary; Using filesort
2 DEPENDENT SUBQUERY t4 ref dp_tech_licence_history_mod_History,dp_tech_licence_history_mod_Property dp_tech_licence_history_mod_History 8 t5.HistoryId 1 Using where
2 DEPENDENT SUBQUERY t6 eq_ref PRIMARY PRIMARY 8 t5.CollectionId 1 Using where
以上查询将提供一个时间和时间的表格，间隔为30分钟。
第2步：使用第一个查询生成所需结果加入网址表
查询：
SELECT @time fromTime, ADDTIME(@time, '00:29:00') toTime,
@time := ADDTIME(@time, '00:30:00')
FROM information_schema.COLLATIONS
JOIN (SELECT @time := TIME('00:00:00')) a
WHERE @time < '24:00:00'
答案 5 :(得分：0)
我希望这适用于，
SELECT
@sTime:= CONCAT(HOUR(created_at),":",
(CASE WHEN MINUTE(created_at) > 30 THEN 30 ELSE 0 END)) as intVar,
(CONCAT(
AddTime(@sTime, '00:00:00'),
' to ',
AddTime(@sTime, '00:30:00')
)) as timeInterval,
COUNT(*) FROM urls
GROUP BY
(CONCAT(HOUR(created_at),":",(CASE WHEN MINUTE(created_at) > 30 THEN 30 ELSE 0 END)))
ORDER BY HOUR(created_at) ASC