MySQLでの結合との時間差を正確に取得する

Question

mysqlの結合を使用して時間差を得るのに問題があります。私は時差のために最小値と最大値で演奏しましたが、それを動作させることはできません。何か助けてくれてありがとう。

SELECT DISTINCT 
    ab.record AS `record`, 
    ab.timeIn AS `timeIN`, 
    de.timeOut AS `timeOut`, 
    TIMEDIFF(de.timeOut,ab.timeIn) AS `timeDifference`, 
    ab.description AS `description`, 

    FROM 
    TableA ab 
    INNER JOIN TableB de ON de.record = ab.record 

    GROUP BY `timeIn`, `description` 

これはこれは、MySQLの時間差

結果希望するどのような

+----+--------+---------------------+---------------------+----------------+-------------+ 
| id | record | timeIn    | timeOut    | timeDifference | description | 
+----+--------+---------------------+---------------------+----------------+-------------+ 
| 1 | 9  | 2017-03-25 12:41:59 | 2017-03-25 12:47:17 | 00:05:18  | productA | 
+----+--------+---------------------+---------------------+----------------+-------------+ 
| 2 | 11  | 2017-03-25 12:42:00 | 2017-03-25 15:10:10 | 02:28:10  | productB | 
+----+--------+---------------------+---------------------+----------------+-------------+ 
| 4 | 14  | 2017-03-25 12:42:00 | 2017-03-25 12:47:18 | 00:05:18  | productC | 
+----+--------+---------------------+---------------------+----------------+-------------+ 
| 5 | 11  | 2017-03-25 15:00:35 | 2017-03-25 15:10:10 | 00:09:35  | productB | 
+----+--------+---------------------+---------------------+----------------+-------------+ 

とTableBの

+----+--------+---------------------+-------------+ 
| id | record | timeOut    | description | 
+----+--------+---------------------+-------------+ 
| 1 | 11  | 2017-03-25 15:10:10 | productB | 
+----+--------+---------------------+-------------+ 
| 2 | 9  | 2017-03-25 12:47:17 | productA | 
+----+--------+---------------------+-------------+ 
| 4 | 11  | 2017-03-25 12:47:17 | productB | 
+----+--------+---------------------+-------------+ 
| 5 | 14  | 2017-03-25 12:47:18 | productC | 
+----+--------+---------------------+-------------+ 

複合表であるTableAの

+----+--------+---------------------+-------------+ 
| id | record | timeIn    | description | 
+----+--------+---------------------+-------------+ 
| 1 | 9  | 2017-03-25 12:41:59 | productA | 
+----+--------+---------------------+-------------+ 
| 2 | 11  | 2017-03-25 15:00:35 | productB | 
+----+--------+---------------------+-------------+ 
| 4 | 11  | 2017-03-25 12:42:00 | productB | 
+----+--------+---------------------+-------------+ 
| 5 | 14  | 2017-03-25 12:42:00 | productC | 
+----+--------+---------------------+-------------+ 

です：

あなたはfirtsと最後の値を取得するためにあなたが分MXなどの集計関数を使用する必要があります
レコードごとにつ以上の値を持っている場合

+----+--------+---------------------+---------------------+----------------+-------------+ 
| id | record | timeIn    | timeOut    | timeDifference | description | 
+----+--------+---------------------+---------------------+----------------+-------------+ 
| 1 | 9  | 2017-03-25 12:41:59 | 2017-03-25 12:47:17 | 00:05:18  | productA | 
+----+--------+---------------------+---------------------+----------------+-------------+ 
| 2 | 11  | 2017-03-25 12:42:00 | 2017-03-25 12:47:17 | 00:05:18  | productB | 
+----+--------+---------------------+---------------------+----------------+-------------+ 
| 4 | 14  | 2017-03-25 12:42:00 | 2017-03-25 12:47:18 | 00:05:18  | productC | 
+----+--------+---------------------+---------------------+----------------+-------------+ 
| 5 | 11  | 2017-03-25 15:00:35 | 2017-03-25 15:10:10 | 00:09:35  | productB | 
+----+--------+---------------------+---------------------+----------------+-------------+ 

Answer 1

そしてあなたべきグループ

SELECT 
    ab.record AS `record`, 
    min(ab.timeIn)AS `timeIN`, 
    max(de.timeOut) AS `timeOut`, 
    TIMEDIFF(max(de.timeOut),min(ab.timeIn)) AS `timeDifference`, 
    ab.description AS `description`, 

    FROM 
    TableA ab 
    INNER JOIN TableB de ON de.record = ab.record 

    GROUP BY record 

Answer 2

持つことの必要なしBUそこにある機能別のグループ。あなたはすべてのレコードが時間と時間を必要とするので、必要な結果を得るには最大の違いが必要です。

SELECT DISTINCT 
    ab.record AS `record`, 
    ab.timeIn AS `timeIN`, 
    de.timeOut AS `timeOut`, 
    TIMEDIFF(de.timeOut,ab.timeIn) AS `timeDifference`, 
    ab.description AS `description` 
FROM 
    TableA ab 
    INNER JOIN TableB de ON de.record = ab.record and de.timeOut between ab.timeIn and 
date_add(ab.timeIn,INTERVAL 2 HOUR); 

ここでは、最大で2時間の間を取っています。

もう1つのことができます。私たちはそのレコードと時間との間にrow_numberを取ることができます。時間とレコードによってデータが順序づけられるため、簡単に違いを得ることができます。

以下は同じコードです。

select distinct ti.record AS record, ti.timeIn AS timeIN,to.timeOut as timeOut, ti.description AS description, TIMEDIFF(to.timeOut,ti.timeIn) AS timeDifference, 
from (SELECT ab.record AS record, ab.timeIn AS timeIN,ab.description AS description, 
@row_num:=if(@record_rn=record,@row_num+1,1) rownum 
FROM TableA ab , 
(select @row_num:=1) x, 
(select @record_rn:='') y 
order by ab.record ace ab.timeOut desc) ti 
inner join 
(SELECT de.record AS record,de.timeOut AS timeOut,de.description, AS description, 
@row_num:=if(@record_rn=record,@row_num+1,1) rownum2 
FROM TableB de, 
(select @row_num:=1) x, 
(select @record_rn:='') y 
order by de.record ace de.timeOut desc) to on (ti.rownum=to.rownum and ti.record=to.record)