解讀我對mysql語句的匹配

Question

它在我不想包含分數的情況下起作用。但是，這對於搜索結果的排序當然很重要。這不起作用：

SELECT * from 
      (SELECT distinct ts.users_user_id from subjects s 
      JOIN teachers_subjects ts ON s.subject_ID = ts.subject_id, MATCH (s.subject_en_name) AGAINST ('math') AS score 
      WHERE MATCH (s.subject_en_name) AGAINST ("math")) qualified 
      JOIN users u ON qualified.users_user_id = u.user_id 
      JOIN users_teachers ut ON qualified.users_user_id = ut.user_id 
      JOIN teachers_subjects ts2 ON qualified.users_user_id = ts2.users_user_id 
      JOIN subjects s2 ON ts2.subject_id = s2.subject_id 

這工作，但不包括各行的得分：

SELECT * from 
      (SELECT distinct ts.users_user_id from subjects s 
      JOIN teachers_subjects ts ON s.subject_ID = ts.subject_id 
      WHERE MATCH (s.subject_en_name) AGAINST ("math")) qualified 
      JOIN users u ON qualified.users_user_id = u.user_id 
      JOIN users_teachers ut ON qualified.users_user_id = ut.user_id 
      JOIN teachers_subjects ts2 ON qualified.users_user_id = ts2.users_user_id 
      JOIN subjects s2 ON ts2.subject_id = s2.subject_id 

Answer 1

的問題是，score應該是SELECT不FROM條款的一部分。試試吧

SELECT * 
    FROM 
(
    SELECT DISTINCT ts.users_user_id, 
     MATCH (s.subject_en_name) AGAINST ('math') AS score 
    FROM subjects s JOIN teachers_subjects ts 
     ON s.subject_ID = ts.subject_id 
    WHERE MATCH (s.subject_en_name) AGAINST ('math') 
) q JOIN users u 
    ON q.users_user_id = u.user_id JOIN users_teachers ut 
    ON q.users_user_id = ut.user_id JOIN teachers_subjects ts2 
    ON q.users_user_id = ts2.users_user_id JOIN subjects s2 
    ON ts2.subject_id = s2.subject_id 
ORDER BY q.score DESC