Mysql 从visitlogs表中计算每小时的访问量

我有一张表格，记录了一些地点的访问信息以及时间，可能还有一些关于这个地方的评论，如：

+---------+---------+------------+-----------------------+---------------------+
| visitId | userId  | locationId | comments              | time                |
+---------+---------+------------+-----------------------+---------------------+
|       1 |    3    |     12     | It's a good day here! | 2012-12-12 20:50:12 |
+---------+---------+------------+-----------------------+---------------------+

我想做的是按小时计算访问组的数量，我想这样生成结果：

+------------+-----+-----+-----+-----+-----+-----+-------+------+
| locationId |  0  |  1  |  2  |  3  |  4  |  5  |  ...  |  23  |
+------------+-----+-----+-----+-----+-----+-----+-------+------+
|      12    |  15 |  12 |  34 |  67 |  78 |  89 |  ...  |  34  | 
+------------+-----+-----+-----+-----+-----+-----+-------+------+

我该怎么做？

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我想评估一整天的访问量差异。

这将为您提供每个地点ID每小时的访问量

SELECT locationId, HOUR(time), COUNT(*)
FROM table
GROUP BY locationId, HOUR(time)

---

这将为您提供每个locationid每小时的访问次数

SELECT locationId, HOUR(time), COUNT(*)
FROM table
GROUP BY locationId, HOUR(time)

---

如果不进行测试，我认为这将起作用：

select locationid, hour(time) as hour, count(distinct userid) as usercnt
from table_1
group by locationid, hour;

---

它不会像你建议的那样被转换，但是数据是相同的，没有测试，我认为这会起作用：

select locationid, hour(time) as hour, count(distinct userid) as usercnt
from table_1
group by locationid, hour;

---

它不会按照您的建议进行转换，但数据是相同的

尝试使用如下查询：

SELECT  location_id, sum(case when HOUR(time) = 0 then 1 else 0) as '0', 
        sum(case when HOUR(time) = 1 then 1 else 0) as '1',-- till 23
FROM table
GROUP BY location_id

---

尝试使用如下查询：

SELECT  location_id, sum(case when HOUR(time) = 0 then 1 else 0) as '0', 
        sum(case when HOUR(time) = 1 then 1 else 0) as '1',-- till 23
FROM table
GROUP BY location_id