我創建了這個查詢來匹配多個表的數據,但我覺得它有一點複雜。在多個表上有多個JOIN的SQL查詢
我需要從下表中選擇:
leads.sequence = instructors_options.instructor_seq
leads.sequence = instructor_calendar.instructor_seq
和:
- 使用下面的關係導致
- instructors_options
- instructor_calendar
以下過濾器必須匹配:
- lead.lead_type ='輔導員
- instructors_options.option_name = 'pupil_cap' AND instructors_options.value> 0
- instructors_options.option_name = 'car_type' AND instructors_options.value ='變量」
- instructors_options.option_name = 'areas_covered' AND instructors_options.value = '變量'
- instructor_calendar.pupil_name = '可用' AND instructor_calendar.date> = '變量' AND instructor_calendar.date < = '變量'
instructors_options.option_name = 'diary_updates' AND instructors_options.value = '1'
SELECT i.sequence as instructor_seq, ic.date AS date, ic.start_time AS start_time, ic.end_time AS end_time FROM leads i LEFT JOIN instructors_options io ON i.sequence = io.instructor_seq AND (io.option_name = 'pupil_cap' AND io.value > '0') RIGHT JOIN instructors_options io2 ON i.sequence = io2.instructor_seq AND io2.option_name = 'car_type' AND io2.value = '".$bookingData["car_type"]."' RIGHT JOIN instructors_options io3 ON i.sequence = io3.instructor_seq AND io3.option_name = 'areas_covered' AND (io3.value LIKE '".substr($postcode, 0, 1)."' OR io3.value LIKE '".substr($postcode, 0, 2)."' OR io3.value LIKE '".substr($postcode, 0, 3)."' OR io3.value LIKE '".substr($postcode, 0, 4)."') RIGHT JOIN instructors_options io4 ON i.sequence = io4.instructor_seq AND io4.option_name = 'diary_updates' AND io4.value = '1' RIGHT JOIN instructor_calendar ic ON i.sequence = ic.instructor_seq AND ic.pupil_name = 'AVAILABLE' AND ic.date >= '".ChangeDateFormat($date, 'Y-m-d')."' AND ic.date <= '".date('Y-m-d', strtotime($date. ' + 7 days'))."' WHERE i.lead_type = 'Instructor' GROUP BY date ORDER BY date ASC, start_time ASC
有人可以幫我更新我的查詢,以確保我已經正確地做了
謝謝!
這是MySQL問題還是SQLServer問題?這些是兩種非常不同的產品,並且你們都標記了它們。爲什麼? – pmbAustin
對不起 - 我的錯誤,必須無意中點擊sql-server – charlie
我注意到的第一件事是你正在尋找「必須」匹配,但有左和右連接。你確定你不想要INNER JOIN嗎?你可以有額外的加入(io4),或者你錯過了需求清單:) –