1

SQLに問題があります。構築している登録システムでクラスの生徒の出席を計算しようとしていますが、サブクエリで親列を使用できません。

SELECT A.student_id, TRUNCATE((100 - ((100/B.reg_num) * C.abs_num)), 0) AS attendance FROM
students A
JOIN (
SELECT COUNT(*) AS reg_num
FROM students
JOIN seminargroup_student ON seminargroup_student.student_id = students.student_id
JOIN seminar_group ON seminar_group.seminar_group_id = seminargroup_student.seminar_group_id
JOIN modules ON modules.module_id = seminar_group.seminar_group_module_id
JOIN register_seminar ON register_seminar.seminar_id = seminar_group.seminar_group_id
JOIN registers ON registers.register_id = register_seminar.register_id
WHERE modules.module_id =1 AND students.student_id = A.student_id
) B
JOIN (
SELECT COUNT(*) AS abs_num
FROM students
JOIN seminargroup_student ON seminargroup_student.student_id = students.student_id
JOIN seminar_group ON seminar_group.seminar_group_id = seminargroup_student.seminar_group_id
JOIN modules ON modules.module_id = seminar_group.seminar_group_module_id
JOIN absence ON students.student_id = absence.student_id
WHERE modules.module_id =1 AND students.student_id = A.student_id
) C

これは次を返します：＃1054-不明な列'A.student_id' in'whereclause'

助けてくれてありがとう！

4

2 に答える 2

1

サブクエリで親列を使用できません

一般的に言えば、FROM にあるサブクエリで親を参照する必要はありません。

代わりに、結合フィールドをサブクエリのSELECTand句に追加してから結合するだけですGROUP BY

例えば

SELECT students.student_id, 
       Truncate(( 100 - ( ( 100 / b.reg_num ) * c.abs_num ) ), 0) AS attendance 
FROM   students 
       JOIN (SELECT a.studentid, 
                    Count(*) AS reg_num 
             FROM   students A 
                    JOIN seminargroup_student 
                      ON seminargroup_student.student_id = A.student_id 
                    JOIN seminar_group 
                      ON seminar_group.seminar_group_id = 
                         seminargroup_student.seminar_group_id 
                    JOIN modules 
                      ON modules.module_id = 
                         seminar_group.seminar_group_module_id 
                    JOIN register_seminar 
                      ON register_seminar.seminar_id = 
                         seminar_group.seminar_group_id 
                    JOIN registers 
                      ON registers.register_id = register_seminar.register_id 
             GROUP  BY a.studentid) A 
         ON students.studentid = a.student.id 
       JOIN (SELECT a.studentid, 
                    Count(*) AS abs_num 
             FROM   students aA 
                    JOIN seminargroup_student 
                      ON seminargroup_student.student_id = a.student_id 
                    JOIN seminar_group 
                      ON seminar_group.seminar_group_id = 
                         seminargroup_student.seminar_group_id 
                    JOIN modules 
                      ON modules.module_id = 
                         seminar_group.seminar_group_module_id 
                    JOIN absence 
                      ON a.student_id = absence.student_id 
             GROUP  BY a.studentid) b 
         ON students.studentid = b.student.id

補足として、左結合を使用し、* の代わりに PK フィールドで DISTINCT COUNT を実行する場合、2 つのサブクエリを実行する必要はありません。

SELECT 
    A.student_id, 
    TRUNCATE((100 - ((100/counts.reg_num) * counts.abs_num)), 0) AS attendance
FROM
students A   
JOIN 
(SELECT 
    COUNT(DISTINCT absence.absence_id) AS abs_num , --OR whatever the PK is
    COUNT(DISTINCT registers.regeister_id) as reg_num,
    students.student_id


FROM   students 
       JOIN seminargroup_student 
         ON seminargroup_student.student_id = students.student_id 
       JOIN seminar_group 
         ON seminar_group.seminar_group_id = 
            seminargroup_student.seminar_group_id 
       JOIN modules 
         ON modules.module_id = seminar_group.seminar_group_module_id 

       LEFT JOIN register_seminar 
         ON register_seminar.seminar_id = seminar_group.seminar_group_id 
       LEFT JOIN registers 
         ON registers.register_id = register_seminar.register_id   


       LEFT JOIN absence 
         ON students.student_id = absence.student_id 
GROUP BY
    students.student_id) COUNTS
ON a.student_id = coutnts.student_ID
于 2012-10-23T18:19:16.473 に答える
0

クエリを次のように書き直してみてください。

SELECT students.student_id, TRUNCATE((100 - ((100/B.reg_num) * C.abs_num)), 0) AS attendance FROM 
students 
JOIN ( 
SELECT COUNT(*) AS reg_num 
FROM students A
JOIN seminargroup_student ON seminargroup_student.student_id = A.student_id 
JOIN seminar_group ON seminar_group.seminar_group_id = seminargroup_student.seminar_group_id 
JOIN modules ON modules.module_id = seminar_group.seminar_group_module_id 
JOIN register_seminar ON register_seminar.seminar_id = seminar_group.seminar_group_id 
JOIN registers ON registers.register_id = register_seminar.register_id 
HAVING modules.module_id =1 AND students.student_id = A.student_id 
) B 
JOIN ( 
SELECT COUNT(*) AS abs_num 
FROM students A
JOIN seminargroup_student ON seminargroup_student.student_id = A.student_id 
JOIN seminar_group ON seminar_group.seminar_group_id = seminargroup_student.seminar_group_id 
JOIN modules ON modules.module_id = seminar_group.seminar_group_module_id 
JOIN absence ON A.student_id = absence.student_id 
HAVING modules.module_id =1 AND students.student_id = A.student_id 
) C

UPD：WHERE句はHAVINGに置き換えられ、エイリアスはサブクエリ内に移動されます。SELECTの前にWHEREが計算されるため、エラーが発生する可能性があります。

また、これらの質問は問題の解決に役立つ場合があります
。1.Where句の不明な列
2.WHEREとHAVING

于 2012-10-23T10:22:47.100 に答える