Question

我希望查询我的销售表并获得不同收据编号的计数按小时计算一周中的每一天

每个记录都有一个重复的收据编号，一个只是日期数据类型的字段中的datein，以及时间数据类型中的sepearte时间字段

这是我正在尝试使用的查询

select dayname(datein),dayofweek(datein), 
if(hour(timein) between 9 and 11, count(distinct salenbr),0) as `9-11`,
if(hour(timein) between 11 and 13, count(distinct salenbr),0) as `11-13`,
if(hour(timein) between 13 and 15, count(distinct salenbr),0) as `13-15`,
if(hour(timein) between 15 and 17, count(distinct salenbr),0) as `15-17`,
if(hour(timein) between 17 and 19, count(distinct salenbr),0) as `17-19`,
if(hour(timein) between 19 and 21, count(distinct salenbr),0) as `19-21`
from carmanslive.allsales 
where loccode = 'SF' and exitcode in ('64','65')  
and deptname = 'Photofinishing' 
and datein >=  DATE_SUB(NOW(), INTERVAL 3 MONTH) 
group by dayname(datein),dayofweek(datein) 
order by dayofweek(datein)

但输出错误

<table caption="UnknownTable (7 rows)">
  <thead>
    <tr>
      <th class="col0">dayname(datein)</th>
      <th class="col1">dayofweek(datein)</th>
      <th class="col2">9-11</th>
      <th class="col3">11-13</th>
      <th class="col4">13-15</th>
      <th class="col5">15-17</th>
      <th class="col6">17-19</th>
      <th class="col7">19-21</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td class="col0">Sunday</td>
      <td class="col1">1</td>
      <td class="col2">0</td>
      <td class="col3">118</td>
      <td class="col4">0</td>
      <td class="col5">0</td>
      <td class="col6">0</td>
      <td class="col7">0</td>
    </tr>
    <tr>
      <td class="col0">Monday</td>
      <td class="col1">2</td>
      <td class="col2">0</td>
      <td class="col3">0</td>
      <td class="col4">360</td>
      <td class="col5">360</td>
      <td class="col6">0</td>
      <td class="col7">0</td>
    </tr>
    <tr>
      <td class="col0">Tuesday</td>
      <td class="col1">3</td>
      <td class="col2">0</td>
      <td class="col3">390</td>
      <td class="col4">0</td>
      <td class="col5">0</td>
      <td class="col6">0</td>
      <td class="col7">0</td>
    </tr>
    <tr>
      <td class="col0">Wednesday</td>
      <td class="col1">4</td>
      <td class="col2">480</td>
      <td class="col3">0</td>
      <td class="col4">0</td>
      <td class="col5">0</td>
      <td class="col6">0</td>
      <td class="col7">0</td>
    </tr>
    <tr>
      <td class="col0">Thursday</td>
      <td class="col1">5</td>
      <td class="col2">0</td>
      <td class="col3">0</td>
      <td class="col4">0</td>
      <td class="col5">330</td>
      <td class="col6">0</td>
      <td class="col7">0</td>
    </tr>
    <tr>
      <td class="col0">Friday</td>
      <td class="col1">6</td>
      <td class="col2">0</td>
      <td class="col3">380</td>
      <td class="col4">380</td>
      <td class="col5">0</td>
      <td class="col6">0</td>
      <td class="col7">0</td>
    </tr>
    <tr>
      <td class="col0">Saturday</td>
      <td class="col1">7</td>
      <td class="col2">0</td>
      <td class="col3">0</td>
      <td class="col4">368</td>
      <td class="col5">0</td>
      <td class="col6">0</td>
      <td class="col7">0</td>
    </tr>
  </tbody>
</table>

每列都应该有数据，我已经确认数据存在。

Answer 1

我认为你必须将聚合物放在你的if周围。类似的东西：

allowDeferredLocationUpdatesUntilTraveled()

使用Mysql时间数据类型的时间段

1 个答案: