[英]MySQL subquery unknown column for complex query
我有以下sqlfiddle http://sqlfiddle.com/#!2/324628/1
我需要創建一個查詢,該查詢返回班級中每個學生的ID和位置(排名); 根據存儲在academic_averages
表中的學術平均值的值,該排名以降序排列。
(例如,第1類來自第一個,第二類來自第1,依此類推...第一個來自第2類,第二個來自第2類...)
這是查詢:
SELECT students.id,
(SELECT x.position
FROM (
SELECT t.student_id, t.value, @rownum := @rownum + 1 AS position
FROM (
SELECT aa.student_id, aa.value
FROM academic_averages AS aa
INNER JOIN students AS s ON s.id = aa.student_id
INNER JOIN classes_students AS cs ON cs.student_id = s.id
INNER JOIN classes_academic_years AS cas ON cas.id = cs.class_academic_year_id
INNER JOIN classes_academic_years as cas2 on cas2.class_id = cas.class_id
INNER JOIN classes_students as cs2 on cs2.class_academic_year_id = cas2.id
INNER JOIN students as s2 on s2.id = cs2.student_id
WHERE s2.id = 243
AND cas.academic_year_id = 4
AND aa.academic_year_id = 4
GROUP BY aa.student_id
ORDER BY abs(aa.value) DESC
) t
JOIN (SELECT @rownum := 0) r
) AS x WHERE x.student_id = students.id ) AS ranking_by_class
FROM students
但是,由於它包含一個子查詢,因此我無法將WHERE從最里面的查詢更改為s2.id = students.id
因為它會引發錯誤(未知列)。
我試過使用INNER JOIN代替子查詢,但是到目前為止還沒有運氣。
有沒有人有辦法解決嗎?
謝謝
LE:在性能方面,必須優化查詢
LE:這是表格的結構:
學術平均 :
id
student_id
value
academic_year_id
classes_academic_years :
id
class_id
name
grade
academic_year_id
classes_students :
id
class_academic_year_id
student_id
類 :
id
school_id
學生 :
id
所需的輸出應為student_id, position
。 sql小提琴似乎存在一些問題,同時這里是模式: http : //snippi.com/s/db8za8k
SELECT x.id
, x.position
, x.academic_average
FROM (SELECT
s.id
, @rownum := @rownum + 1 position
, av.value academic_average
FROM students s
JOIN classes_students cs ON s.id = cs.student_id
JOIN classes_academic_years cay ON cay.id = cs.class_academic_year_id
JOIN academic_averages av ON av.student_id = s.id
WHERE cay.academic_year_id = 4 -- change these two parameters in
AND av.academic_year_id = 4 -- the subquery for different years
ORDER BY av.value DESC) x,
(SELECT @rownum := 0) y
ORDER BY academic_average DESC
我認為上面的查詢應該為您工作。
我假設學術排名position
由學術平均水平降序確定。
我無權訪問您的數據集,因此我添加了三行,兩行選擇學生的學業平均值,一行根據學業成績以降序排列結果。 這應該可以幫助您驗證它是否按預期工作。 如果您運行查詢,並且該查詢有效,它將顯示position
從1開始並以1遞增的記錄。
在生產中,我將省略這些片段以獲取您指定的結果集:
1. , x.academic_average
2. , av.value academic_average
3.按ORDER BY academic_average DESC
該查詢應按班級為您提供學生的職位。 如果要擺脫某些字段,可以將SELECT
包裝在另一個SELECT
,或者在將數據集提取為另一種語言后忽略這些列。
SELECT
x.student_id
, x.cay_class_id
, x.academic_average
, if(@classid = x.cay_class_id, @rownum := @rownum + 1, @rownum := 1) position
, @classid := x.cay_class_id
FROM (SELECT
s.id student_id
, cay.class_id cay_class_id
, av.value academic_average
FROM students s
JOIN classes_students cs ON s.id = cs.student_id
JOIN classes_academic_years cay ON cay.id = cs.class_academic_year_id
JOIN academic_averages av ON av.student_id = s.id
WHERE cay.academic_year_id = 4 -- change these two parameters in
AND av.academic_year_id = 4 -- the subquery for different years
ORDER BY cay.class_id, av.value DESC) x,
(SELECT @classid := 0, @rownum := 0) y
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