Question

我有一张表格，代表玩过Flash游戏的用户的得分和ID。我想返回普通用户尝试游戏的次数 - 每次完成“尝试”时，用户的得分和id都输入到mysql表中。结构是 -

id int(11)                      
score int(11)                               
fb_id int(25)

这为我提供了每个用户的尝试次数

SELECT count(score) AS counted FROM `ifa_scores` GROUP BY `fb_id`

但我想要这样的东西

我想要总体上玩的平均次数 - 即具有相同FB_ID的平均行数

Answer 1

SELECT AVG(t.counted)
FROM
(
    SELECT COUNT(id) AS counted
    FROM ifa_scores
    GROUP BY fb_id
) AS t

Answer 2

SELECT `fb_id`, avg(score) AS counted FROM `ifa_scores` GROUP BY `fb_id`

Answer 3

你需要计算一下已玩过多少游戏，并计算玩家数量，将他们加在一起，然后适当划分;

之类的东西

SELECT GAMES_PLAYED,
       PLAYERS,
       GAMES_PLAYED / PLAYERS AS AVERAGE_GAMES_PER_PLAYER
FROM
  (SELECT COUNT(*) AS GAMES_PLAYED FROM 'ifa_scores') g,
  (SELECT COUNT(*) AS PLAYERS FROM (SELECT DISTINCT 'fb_id' FROM 'ifa_scores')) p;

应该有效。或者你可以使用

SELECT AVG(GAMES_PLAYED) AS AVERAGE_GAMES_PER_PLAYER
FROM (SELECT FB_ID, COUNT(*) AS GAMES_PLAYED
        FROM 'ifa_scores'
        GROUP BY 'fb_id');

分享并享受。

Answer 4

以下应该有效： count（*）返回每个用户玩的游戏数。平衡这个应该给你你正在寻找的答案。请检查

SELECT fb_id，avg（count（*））AS从ifa_scores GROUP BY fb_id计算

返回列的平均数量

4 个答案: