如何結合這兩個sql語句

Question

我對sql不是很有經驗，但通常可以通過構造基本查詢來獲得，但這對我來說太複雜了。

我有兩個查詢，每個查詢都提供了我需要的結果，但是我需要將它們組合起來才能得到我需要的最終結果。

在一個查詢中，它爲特定學生抓取所有「分數」並返回平均值。在另一個查詢中，我從表格中除去最高和最低的記錄。

我想結合這些查詢來首先去除最高和最低的分數，然後從剩餘的記錄中返回平均分數。請幫助指導我使用正確的語法。

我的疑問：

---

這會將所有的最高分和最低分：

SELECT * 
FROM `scores` 
WHERE `tot_score` = (
SELECT MAX(`tot_score`) 
FROM `scores`) 
OR `tot_score` = (
SELECT MIN(`tot_score`) 
FROM `scores`) 

---

這將返回平均得分：

SELECT `pres_id`, COUNT(`pres_id`) , ROUND(AVG(`tot_score`) , 2) , `username`, `name`, `pres_day` 
FROM `scores`, `users` 
WHERE `users.id_user` = `scores.pres_id` 
GROUP BY `pres_id` 

我怎麼能結合這些來獲得我需要的結果？

Answer 1

我會建議你做最小的計算和最大值在一個子查詢中。然後，你可以簡單地認爲你想使用一個where子句可以過濾行：

SELECT s.`pres_id`, COUNT(`pres_id`), ROUND(AVG(`tot_score`) , 2) , 
     `username`, `name`, `pres_day` 
FROM `scores` s join 
    `users` u 
    on u.id_user = s.pres_id join 
    (select s.pres_id, max(tot_score) as maxts, min(tot_score) as mints 
     from scores s 
     group by s.pres_id 
    ) ssum 
    on u.id_user = ssum.pres_id 
where s.tot_score > ssum.mints and s.tot_score < ssum.maxts 
GROUP BY s.`pres_id`; 

我猜測你想要去除最高和最低分數爲每個學生。如果你只是想省略最高和最低所有學生，然後使用以下命令：

SELECT s.`pres_id`, COUNT(`pres_id`), ROUND(AVG(`tot_score`) , 2) , 
     `username`, `name`, `pres_day` 
FROM `scores` s join 
    `users` u 
    on u.id_user = s.pres_id cross join 
    (select max(tot_score) as maxts, min(tot_score) as mints 
     from scores s 
    ) ssum 
where s.tot_score > ssum.mints and s.tot_score < ssum.maxts 
GROUP BY s.`pres_id`; 

編輯：

修改，以保持得分時，只有一個分數的學生：

SELECT s.`pres_id`, COUNT(`pres_id`), ROUND(AVG(`tot_score`) , 2) , 
     `username`, `name`, `pres_day` 
FROM `scores` s join 
    `users` u 
    on u.id_user = s.pres_id join 
    (select s.pres_id, max(tot_score) as maxts, min(tot_score) as mints, 
      count(*) as cnt 
     from scores s 
     group by s.pres_id 
    ) ssum 
    on u.id_user = ssum.pres_id 
where (s.tot_score > ssum.mints and s.tot_score < ssum.maxts) or (cnt = 1) 
GROUP BY s.`pres_id`; 

Answer 2

你應該能夠反轉高/低成績查詢，然後把它作爲現有平均查詢的結果集：

SELECT `pres_id`, COUNT(`pres_id`) , ROUND(AVG(`tot_score`) , 2) , `username`, `name`, `pres_day` 
FROM (
    SELECT `tot_score` 
    FROM `scores` 
    WHERE `tot_score` <> (
    SELECT MAX(`tot_score`) 
    FROM `scores`) 
    OR `tot_score` <> (
    SELECT MIN(`tot_score`) 
    FROM `scores`) 
) AS `scores`, `users` 
WHERE `users.id_user` = `scores.pres_id` 
GROUP BY `pres_id`