使用mysql每日選擇數據

Question

我有一個表tbl_worklog與列id（int），project_id（int），start_time（datetime），total（時間），user_id（id）。

我還有另一個表tbl_projects列id（int），project_name（文本），organization_id（int）。第一個表中的project_id對第二個表中的id有一個外鍵。

我也有organization表的列created_by（int）。組織表中的id對第二個表中的organization_id具有外鍵。

因此，這裏的問題，我想所有的項目成員，是CREATED_BY = 1，而在每一個成員，我需要每天得到他們的第一臺工作的組織下。例如，

Members | Sun | Mon | Tue | Wed | Thu | Fri | Sat | Weekly Total 
--------------- 
John | 0:30:00 | 0:30:00 | 0:30:00 | 0:30:00 | 0:30:00 | 0:30:00 | 1:00:00 | 4:00:00 

我在第一個表中使用了start_time來計算他們的日常工作。

這裏是我的查詢不工作，

SELECT (SELECT SEC_TO_TIME(SUM(TIME_TO_SEC(total))) AS total_time from tbl_worklog WHERE start_time < date_add('2016-09-04', INTERVAL 7 DAY) AND start_time >= '2016-09-04' AND user_id=tbl_worklog.user_id) as weekly_total, 
(SELECT SEC_TO_TIME(SUM(TIME_TO_SEC(total))) AS total_time from tbl_worklog WHERE start_time < date_add('2016-09-04', INTERVAL 1 DAY) AND start_time >= '2016-09-04' AND user_id=tbl_worklog.user_id) as sun, 
(SELECT SEC_TO_TIME(SUM(TIME_TO_SEC(total))) AS total_time from tbl_worklog WHERE start_time < date_add('2016-09-04', INTERVAL 2 DAY) AND start_time >= date_add('2016-09-04', INTERVAL 1 DAY) AND user_id=tbl_worklog.user_id) as mon, 
(SELECT SEC_TO_TIME(SUM(TIME_TO_SEC(total))) AS total_time from tbl_worklog WHERE start_time < date_add('2016-09-04', INTERVAL 3 DAY) AND start_time >= date_add('2016-09-04', INTERVAL 2 DAY) AND user_id=tbl_worklog.user_id) as tue, 
(SELECT SEC_TO_TIME(SUM(TIME_TO_SEC(total))) AS total_time from tbl_worklog WHERE start_time < date_add('2016-09-04', INTERVAL 4 DAY) AND start_time >= date_add('2016-09-04', INTERVAL 3 DAY) AND user_id=tbl_worklog.user_id) as wed, 
(SELECT SEC_TO_TIME(SUM(TIME_TO_SEC(total))) AS total_time from tbl_worklog WHERE start_time < date_add('2016-09-04', INTERVAL 5 DAY) AND start_time >= date_add('2016-09-04', INTERVAL 4 DAY) AND user_id=tbl_worklog.user_id) as thu, 
(SELECT SEC_TO_TIME(SUM(TIME_TO_SEC(total))) AS total_time from tbl_worklog WHERE start_time < date_add('2016-09-04', INTERVAL 6 DAY) AND start_time >= date_add('2016-09-04', INTERVAL 5 DAY) AND user_id=tbl_worklog.user_id) as fri, 
(SELECT SEC_TO_TIME(SUM(TIME_TO_SEC(total))) AS total_time from tbl_worklog WHERE start_time < date_add('2016-09-04', INTERVAL 7 DAY) AND start_time >= date_add('2016-09-04', INTERVAL 6 DAY) AND user_id=tbl_worklog.user_id) as sat, 
(SELECT name FROM tbl_accounts WHERE id=tbl_worklog.user_id) as name FROM tbl_worklog WHERE project_id IN (SELECT id FROM tbl_projects WHERE organization_id=1) 

有也任何其他方式使之成爲簡單更好的做法？ 如何使用PDO來做到這一點？

Answer 1

您沒有使用group by這是最重要的在這裏得到預期的結果。嘗試以下查詢以獲得每日結果。您可以修改以獲取PHP中的每週結果視圖。

select project_name, user_id, sum(time_to_sec(time)) as user_time, date(start_time) as 
daily_time , tw.project_id pID 
inner join tbl_projects tp on tp.id = tw.project_id 
inner join organization og on og.id = tp.organization_id 
where og.created_by = 1 group by user_id, pID, daily_time