Question

让我解释一下我想要的东西：假设我的表中有100条记录（列名：id，imei，纬度，经度，日期时间，状态）

然后，我放了50条记录，10000001 imei号码，现在我必须得到状态为0且日期差异超过5分钟的记录与同一个imei。

如何获取状态为0且持续时间超过5的记录？

在排序中，我需要比较1个日期和2个日期，然后是2个日期和3个日期，然后是3个日期和4个日期，然后是4个日期和5个日期，依此类推。

Answer 1

这将对imei进行分组并按日期时间对它们进行排序，然后按照升序对每个日期时间进行比较。既然你没有给出更好的方法来排序它们，我会假设这就是你要找的东西

; WITH T AS (SELECT *, ROW_NUMBER() OVER (PARTITION BY imei ORDER BY datetime) RN
    FROM tblName)
SELECT T1.imei, T1.status, T1.datetime [date1], T2.datetime [date2], DATEDIFF(mi, T1.datetime, T2.datetime) [difference]
FROM T T1 
JOIN (SELECT imei, RN, datetime FROM T WHERE RN > 1) T2 ON T2.RN-1 = T1.RN AND T1.imei = T2.imei
WHERE T1.datetime < DATEADD(mm, -5, T2.datetime)
AND T1.Status = 0

或者，如果您只想要日期差异（无论它们是否大于或小于5），您可以这样做：

; WITH T AS (SELECT *, ROW_NUMBER() OVER (PARTITION BY imei ORDER BY datetime) RN
    FROM tblName)
SELECT T1.imei, T1.status, T1.datetime [date1], T2.datetime [date2], DATEDIFF(mi, T1.datetime, T2.datetime) [difference], CASE WHEN DATEDIFF(mi, T1.datetime, T2.datetime) > 5 THEN 'Y' ELSE 'N' END -- 'Y' means greater than 5, 'N' less than or equal to 5
FROM T T1 
LEFT JOIN (SELECT imei, RN, datetime FROM T WHERE RN > 1) T2 ON T2.RN-1 = T1.RN AND T1.imei = T2.imei
WHERE T1.Status = 0

如何使用sql select query获取所有记录中两个日期之间的差异

1 个答案: