選擇每個月最高分的獲勝者

我需要選擇最高得分者作爲月度獲勝者。我也想展示以前的獲獎者。我目前的查詢僅選擇前一個月的獲勝者，但我如何選擇以前的所有月度獲勝者？選擇每個月最高分的獲勝者

我的查詢：

function month_winner_now() 
    { 
     $query = $this->db->query("SELECT winner.id,winner.score,winner.user_id,winner.date, user.id,user.username,user.email,user_profile.user_image,user_profile.hometown,user_profile.country FROM `winner` LEFT JOIN `user` ON user.id = winner.user_id LEFT JOIN `user_profile` ON user_profile.user_id = winner.user_id WHERE MONTH(CURDATE())= MONTH(winner.date) ORDER BY winner.score DESC 
LIMIT 1"); 


      return $query->result_array(); 
    } 


My current output : 

"monthly winners":[ 
    { 
     "id":"5", 
     "score":"1256", 
     "user_id":"5", 
     "date":"2014-03-05", 
     "username":"", 
     "email":"", 
     "user_image":"", 
     "hometown":"", 
     "country":"" 
    }

但我需要輸出像

"monthly winners":[ 
     { 
      "id":"4", 
      "score":"233", 
      "user_id":"4", 
      "date":"2014-03-02", 
      "username":"Mahmudul Hasan Swapon", 
      "email":"", 
      "user_image":"", 
      "hometown":"", 
      "country":"" 
     }, 
     { 
      "id":"7", 
      "score":"123", 
      "user_id":"7", 
      "date":"2014-03-04", 
      "username":"Prosanto Biswas", 
      "email":"", 
      "user_image":"", 
      "hometown":"", 
      "country":"" 
     } 
    ],

每月優勝者JSON數組顯示所有以前的一個月贏家，但每個月都應該有一個贏家。

數據庫表的樣子

id | name | userid | score |  date  | 
    ------------------------------------------------------------ 
    1 | john | 1  | 44 | 2013-03-2 
    2 | mary | 2  | 59 | 2013-03-5 
    3 | john | 12 | 38 | 2013-03-8 
    4 | elvis | 3  | 19 | 2013-02-10 
    5 | john | 11 | 1002 | 2013-01-11 
    6 | johnah | 10 | 200 | 2013-01-11

來源

2014-03-12 user3407569

我想告訴所有以前的月冠軍

你檢查一個月的平等，因此其他月份（以前或以後的版本）省略。根據具體情況，可將其與<=或>=進行比較。

如果您使用LIMIT 1如果在當月存在記錄，則有可能不會提取其他月份的詳細信息。

試試這個：

SELECT 
    winner.id -- etc fields 
FROM 
    `winner` 
    LEFT JOIN `user` 
    ON user.id = winner.user_id 
    LEFT JOIN `user_profile` 
    ON user_profile.user_id = winner.user_id 
WHERE 
    date_format(winner.date, '%Y%m') <= date_format(CURDATE(), '%Y%m') 
ORDER BY 
    winner.score DESC

來源

2014-03-12 07:22:53

對於從上月的冠軍：

$query = $this->db->query(" 
    SELECT winner.id,winner.score,winner.user_id,winner.date, user.id,user.username,user.email,user_profile.user_image,user_profile.hometown,user_profile.country 
    FROM `winner` 
    LEFT JOIN `user` 
    ON user.id = winner.user_id 
    LEFT JOIN `user_profile` 
    ON user_profile.user_id = winner.user_id 
    WHERE MONTH(DATE_SUB(CURDATE(), INTERVAL 1 MONTH)) = MONTH(winner.date) ORDER BY winner.score DESC LIMIT 1 
");

來源

2014-03-12 07:24:02 sunshinejr

查詢有錯誤對你有所幫助＃1064 - 你在你的SQL錯誤句法;檢查對應於您的MySQL服務器版本的手冊，以在'1 MONTH））= MONTH（winner.date）ORDER BY winner.score DESC LIMIT 1'附近使用正確的語法 – user3407569

Ye，編輯，忘記了INTERVAL。 – sunshinejr

嘗試是這樣的查詢：

SELECT * FROM user 
INNSER JOIN 
(SELECT user_id, MAX(score) AS s, MONTH(date) AS d 
FROM winner 
GROUP BY MONTH(date)) monthlyWinner ON (user.id = monthlyWinner.user_id)

來源

2014-03-12 07:34:55 user3332631

試試這個，

;with cte as 
(
select Datename(mm,[date]) as m 
     --,max(amount_paid) 
     ,Rank() over(PARTITION BY Datename(mm,[date]) order by [score] desc) as rr 
     ,[score] 
     ,id 
from myTbl 
--where DATEDIFF(YY,[date],'1/1/2013') = 0 
) 
select * from cte 
left join myTbl as r on r.id=cte.id 
where rr = 1

來源

2014-03-12 07:46:46 AK47

我重新創建SQL查詢，並增加了一個場"month_of_year"，現在我認爲它會根據您的要求

SELECT 
    winner.id,winner.score,winner.user_id,winner.date, 
    user.id,user.username,user.email,user_profile.user_image, 
    user_profile.hometown,user_profile.country, 
    date_format(winner.date, '%Y-%m') AS month_of_year 
FROM 
    `winner` 
    LEFT JOIN `user` 
    ON user.id = winner.user_id 
    LEFT JOIN `user_profile` 
    ON user_profile.user_id = winner.user_id 
GROUP BY month_of_year 
ORDER BY winner.score DESC

來源

2014-03-12 07:52:48

選擇每個月最高分的獲勝者

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