选择区域中的日期 - 90天前和现在的日期.

Question

SELECT gameratingstblx245v.gameid,avg( gameratingstblx245v.rating ) as avgrating, count(gameratingstblx245v.rating) as count,gamedata.name ,gamedata.gameinfo
FROM gameratingstblx245v 
LEFT JOIN gamedata ON gamedata.id = gameratingstblx245v.game_id 
WHERE gameratingstblx245v.game_id=gameratingstblx245v.game_id 
GROUP BY gameid 
ORDER BY avg( gameratingstblx245v.rating ) DESC LIMIT 0,8

表gameratingstblx245v - gameid,评级
Rable gamedata - id,gameinfo,name,releasedate

这是我目前用于从两个表gamedata和gameratingstblx245v中提取数据的查询....我在这里做的是采取平均值.表gameratingstblx245v中所有评分的平均降序排列 我还从表游戏数据中提取与所选游戏人物相对应的相关信息...

现在我要提取的是最高平均值.来自game_ratingstblx245v的评分,但是对于那些从桌面游戏数据发布的游戏是在过去90天内...帮助将不胜感激..谢谢

Answer 1

这是我设计该查询的方式:

SELECT d.id, d.name, d.gameinfo,
  AVG(r.rating) AS avgrating, COUNT(r.rating) AS count
FROM gamedata d
LEFT JOIN gameratingstblx245v r ON (d.id = r.game_id)
WHERE d.releasedate BETWEEN NOW() - INTERVAL 90 DAY AND NOW()
GROUP BY d.id 
ORDER BY avgrating DESC LIMIT 0,8;