Question

我试图解决的任务是：在一个长方形1450米的正方形牢房内，大英博物馆的自行车站，Bloomsbury在4个小时的时间从早上6点开始。

我有两张桌子，站表和使用表。但是，在运行我的查询后，所有行都是空的！

SELECT ce.stationID,
       il.name,
       il.easting,
       il.northing,
       il.capacity,
       ce.t,
       ROUND(AVG(availableBikes) ,1) as Average,
       ROUND(AVG(availableBikes)/capacity*100,1) AS Percentage
FROM tflBikeUsagehour ce,
     tflStations il
where il.usageId=ce.stationID
  and datetime (t,'06:00:00')>6
  AND datetime (t,' 10:00:00')<10
  AND il.easting BETWEEN (SELECT easting-(1450*0.5) FROM tflstations
                          WHERE name='British Museum, Bloomsbury')
                     AND (SELECT easting+(1450*0.5) FROM tflstations
                          WHERE name='British Museum, Bloomsbury')
  AND il.northing BETWEEN (SELECT northing-(1450*0.5) FROM tflstations
                           WHERE name='British Museum, Bloomsbury')
                      AND (SELECT northing+(1450*0.5) FROM tflstations
                           WHERE name='British Museum, Bloomsbury')
  and ce.t between (select datetime(t,'06:00:00')<6 as tf from tflbikeusagehour)
               and (select datetime(t,'10:00:00')>10 from tflbikeusagehour)
GROUP BY ce.t
ORDER BY ce.t;

我的问题是日期时间格式！我使用它错了吗？

Answer 1

datetime function不能以这种方式使用。

要检查时间戳是否在某个范围内，请使用如下表达式：

time(t) BETWEEN '06:00:00' AND '09:59:59'

如果列t已使用格式hh:mm:ss，则可以将time(t)替换为t。

使用运营商的日期时间

1 个答案: