[英]Joining 4 tables in php mysql
我嘗試使用以下基於兩個選定變量的查詢來連接3個表(class_section,section_name,student_detail):
SELECT A.cs_id
, A.class
, B.sec_id
, B.cs_id
, B.cs_name
, C.stud_id
, C.stud_class
, C.stud_section
, C.stud_adm_no
, C.stud_full_name
FROM class_section AS A
JOIN section_name as B
ON A.cs_id = b.cs_id
JOIN student_detail as c
ON C.stud_class = B.cs_id
WHERE A.cs_id = $cat
AND B.sec_id = $subcat
現在,我如何加入4個表,如上所述的3個表,第4個表是exam_schedule (ex_schedule_class,ex_schedule_section,ex_stud_id,ex_schedule_name)(其中(ex_schedule_class(與class_section中的cs_id的關系),ex_schedule_section(與section_name中的sec_id的關系)來自exam_schedule表的ex_stud_id(與student_detail中的stud_id的關系)我想檢索ex_schedule_name值。請幫助我。
您可以嘗試以下方法
SELECT A.cs_id
, A.class
, B.sec_id
, B.cs_id
, B.cs_name
, C.stud_id
, C.stud_class
, C.stud_section
, C.stud_adm_no
, C.stud_full_name
, D.name
FROM class_section AS A
JOIN section_name AS B ON A.cs_id = b.cs_id
JOIN student_detail AS c ON C.stud_class = B.cs_id
JOIN exam_schedule AS D ON D._class = A.cs_id
WHERE A.cs_id = $cat AND B.sec_id = $subcat
通過將“exam_shedule”與“class_section”相結合,您可以獲得exam_schedule_name(在查詢中:D.name)
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