如何計算最接近100的平均數？

Question

我有一個MySQL表，其中我有號碼的列表，從遊戲收集，用戶必須點擊一個按鈕去接近100

我想我會用AVG()，但是這是不行的顯然，因爲當你有0和200時，它會看起來好像你取得了一個完美的100.

我需要的是每一輪（有10）計數離100有多遠，並計數那些。

Example: 
User A: 98 + 102 + 102 = 6 (something) Loser! 
User B: 95 + 100 + 100 = 5 (something) Winner! 

我有什麼，但顯然是有缺陷的：

SELECT user_id, AVG(score) AS Average, (100-AVG(score)) AS Difference, 
DATE(playtime) AS Playdate 
FROM playtable 
GROUP BY user_id, DATE(playtime) 
ORDER BY Playdate DESC, user_id 

Answer 1

這應該這樣做

SELECT user_id, SUM(ABS(100-score)) as cumulative_error, 
    DATE(playtime) AS Playdate 
FROM playtable 
GROUP BY user_id, DATE(playtime) 
ORDER BY Playdate DESC, user_id 

cumulative_error將包含累計差異值

Answer 2

您可能能夠解決您的查詢返回的絕對值看比分多遠從給定值（100你的情況）

SELECT user_id, AVG(score) AS Average, SUM(ABS(100 - score)) AS Difference, 
DATE(playtime) AS Playdate 
FROM playtable 
GROUP BY user_id, DATE(playtime) 
ORDER BY Playdate DESC, user_id