group_by和order_by與codeigniter

我試圖找出這一點，沒有運氣。我不斷得到一些奇怪的結果！首先，我有一個MySQL數據庫的一些行，該結構是這樣的，group_by和order_by與codeigniter

ID MESSAGE_ID 消息 TIME 日期現在

，該MESSAGE_ID可以是相同的多次，這是我的問題開始的地方。我想通過ID選擇最後一行復制的message_ID的排序。

所以，讓我們說我有一個正常的GET請求，結果是這樣的，

Array 
(
    [0] => Array 
     (
      [id] => 3 
      [message_id] => 1 
      [from_user_id] => 3 
      [to_user_id] => 1 
      [message] => last 
      [time_sent] => 1331874924 
      [date_sent] => 16/03/2012 
      [opened] => 0 
      [ip] => ::1 
      [deleted] => 0 
      [reported] => 0 
     ) 

    [1] => Array 
     (
      [id] => 2 
      [message_id] => 1 
      [from_user_id] => 3 
      [to_user_id] => 1 
      [message] => middle 
      [time_sent] => 1331874920 
      [date_sent] => 16/03/2012 
      [opened] => 0 
      [ip] => ::1 
      [deleted] => 0 
      [reported] => 0 
     ) 

    [2] => Array 
     (
      [id] => 1 
      [message_id] => 1 
      [from_user_id] => 3 
      [to_user_id] => 1 
      [message] => first 
      [time_sent] => 1331874916 
      [date_sent] => 16/03/2012 
      [opened] => 0 
      [ip] => ::1 
      [deleted] => 0 
      [reported] => 0 
     ) 

    [3] => Array 
     (
      [id] => 4 
      [message_id] => 2 
      [from_user_id] => 3 
      [to_user_id] => 1 
      [message] => test 
      [time_sent] => 1331874916 
      [date_sent] => 16/03/2012 
      [opened] => 0 
      [ip] => ::1 
      [deleted] => 0 
      [reported] => 0 
     ) 

)

現在，我想要什麼，是去除dubplicated MESSAGE_ID的，只有得到這個（注意如何應該由ID）

Array 
(
    [0] => Array 
     (
      [id] => 3 
      [message_id] => 1 
      [from_user_id] => 3 
      [to_user_id] => 1 
      [message] => last 
      [time_sent] => 1331874924 
      [date_sent] => 16/03/2012 
      [opened] => 0 
      [ip] => ::1 
      [deleted] => 0 
      [reported] => 0 
     ) 

    [1] => Array 
     (
      [id] => 4 
      [message_id] => 2 
      [from_user_id] => 3 
      [to_user_id] => 1 
      [message] => test 
      [time_sent] => 1331874916 
      [date_sent] => 16/03/2012 
      [opened] => 0 
      [ip] => ::1 
      [deleted] => 0 
      [reported] => 0 
     ) 

)

下訂單我一直試圖做之前，

$this->db->order_by(‘id’, ‘DESC’); 
$this->db->group_by(‘message_id’);

和

$this->db->group_by(‘message_id’); 
$this->db->order_by(‘id’, ‘DESC’);

，但我沒有得到來自MESSAGE_ID的最新成果，只有第一個（看起來像它不刪除或某件事之前soring？）

任何幫助將是巨大的！

來源

2012-03-18 Ninja__1881

'SELECT * FROM table GROUP BY message_id ORDER BY id ASC;' – hjpotter92 2012-03-18 16:13:26

你想要結果id是按升序還是降序？ – 2012-03-18 16:18:44

我做了hjpotter92，$ this-> db-> group_by（'message_id'）AND $ this-> db-> order_by（'id'，'ASC'）;但是如果我使用ASC或DESC，無論如何我都會得到第一個ID（ID = 1），看起來它並不像我在第一篇文章中所說的那樣排序。 – 2012-03-18 16:20:48

我成功了，與此對我的表：

select * from 
    (select * from da_user_messages order by id desc) t 
where receiver=1 group by msg_session

你的情況msg_session

是MESSAGE_ID，接收器to_user_id和da_user_messages是你的表名。

來源

2012-03-18 16:12:07

這應該爲什麼tpaksu？ – 2012-03-18 16:32:44

我的意思是不同的message_id。你爲什麼認爲這不起作用？（現在更正我的答案。） – 2012-03-18 16:37:27

那麼我仍然會得到很多重複ID的？ – 2012-03-18 16:45:55

group_by和order_by與codeigniter

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