Question

我在mysql中有一个时间日期和军事格式的表（2014-01-22 16:10:02）

caller_data

|ID | caller_id   |       call_time      
| 1 | 123         | 2014-01-22 16:10:02          
| 2 | 123         | 2014-01-22 16:30:02          
| 3 | 123         | 2014-01-22 17:10:02          
| 4 | 123         | 2014-01-22 18:05:45

他们的时间范围如16:10和16:30。我使用HOUR（）对它们进行分组，结果为16.我将计算调用者在该小时范围内调用的次数。所以这将是2。我的问题：

SELECT caller_id,
HOUR(str_to_date(date_stamp, "%Y-%m-%d %H:%i:%s")) as hourly,
count(*) as total
from caller_data
group by hourly
order by hourly;

结果应该是：

caller_id    |  hourly |   total  
 123         |   16    |    2   
 123         |   17    |    1 
 123         |   18    |    1

我想在0-23上显示这3行。我一直试图用PHP显示它，但是当我创建一个$ i＆lt; = 23时它会显示3次。

for($i=0;i<=23;$i++)
{
   foreach ($rec as $row):
       echo $row['hourly'];
       echo $row['caller_id'];
       echo $row['total'];
   endforeach;
}

我不知道是否可以在mysql上显示这个...有人可以用php或mysql显示如何做到这一点吗？

| hours | caller_id   | total 
| 0     | 123         |  0
| 1     | 123         |  0        
| 2     | 123         |  0       
| 3     | 123         |  0     
| 4     | 123         |  0
| 5     | 123         |  0
| 6     | 123         |  0        
| 7     | 123         |  0       
| 8     | 123         |  0     
| 9     | 123         |  0
| 10    | 123         |  0
| 11    | 123         |  0
| 12    | 123         |  0        
| 13    | 123         |  0       
| 14    | 123         |  0     
| 15    | 123         |  0
| 16    | 123         |  2        
| 17    | 123         |  1       
| 18    | 123         |  1     
| 19    | 123         |  0
| 20    | 123         |  0
| 21    | 123         |  0     
| 22    | 123         |  0
| 23    | 123         |  0

Answer 1

您可以使用left outer join在SQL中执行此操作。你只需输入所有时间：

select caller_id, n.n as hourly, coalesce(total, 0) as total
from (select 1 as n union all select 2 union all select 3 union all select 4 union all
      select 5 union all select 6 union all select 7 union all select 8 union all
      select 9 union all select 10 union all select 11 union all select 12 union all
      select 13 union all select 14 union all select 15 union all select 16 union all
      select 17 union all select 18 union all select 19 union all select 20 union all
      select 21 union all select 22 union all select 23 union all select 24
     ) n left outer join
     (select caller_id, HOUR(str_to_date(date_stamp, "%Y-%m-%d %H:%i:%s")) as hourly,
             count(*) as total
      from caller_data
      group by hourly
     ) rep
     on n.n = rep.hourly
order by hourly;

使用php或mysql显示0到23小时的表

1 个答案: