Question

我想只选择当天的记录。所以我每天都能看到有多少访客（＆＃34; bezoekers＆＃34;）。

这是我使用的代码：

Select FROM_UnixTime((sensordata1.time), '%Y-%c-%d-%H') "Datum (Descending)", 
COUNT(DISTINCT address) "Bezoekers"  
FROM sensordata1  
ORDER BY 1
FROM_UnixTime((sensordata1.time), '%Y-%c-%d-%H') DESC

我尝试了什么：

Select FROM_UnixTime((sensordata1.time), '%Y-%c-%d-%H') "Datum (Descending)",
COUNT(DISTINCT address) "Bezoekers"  
FROM sensordata1  
WHERE 1, DATE('Datum') = DATE(CURDATE() 
ORDER BY 1
FROM_UnixTime((sensordata1.time), '%Y-%c-%d-%H') DESC

我想要的是什么：

Datum         | Bezoekers
2017-4-20-15  | 31
2017-4-20-14  | 34
2017-4-20-13  | 20
2017-4-20-12  | 26

我不想要例如那里的2017-04-21。如果我第二天回来，那就需要从那天开始，所以你不能约会。

Answer 1

您错过了GROUP BY。如果您想要每个FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H')一个结果行，则按其分组。

Select
  FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H') AS "Datum (Descending)", 
  COUNT(DISTINCT address) AS "Bezoekers"  
FROM sensordata1  
GROUP BY FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H')
ORDER BY FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H') DESC;

仅适用于当天（添加了WHERE条款）：

Select
  FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H') AS "Datum (Descending)", 
  COUNT(DISTINCT address) AS "Bezoekers"  
FROM sensordata1  
WHERE FROM_UnixTime(sensordata1.time, '%Y-%m-%d') = curdate()
GROUP BY FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H')
ORDER BY FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H') DESC;

Answer 2

我想指出您可以使用别名来简化查询。日期逻辑有点不清楚，但你似乎想要昨天的约会。您可以使用where子句执行此操作：

Select from_unixtime(sd.time, '%Y-%c-%d-%H') as Datum, 
       count(distinct sd.address) as Bezoekers 
from sensordata1 sd
where sd.time >= unixtime_timestamp(curdate() - interval 1 day) and
      sd.time < unixtime_timestamp(curdate())
group by datum
order by datum desc

如何仅显示当前的记录

2 个答案: