하는 방법은 아래의 작업에 대한 쿼리를 공식화 할 수 그룹

Question

PER 마지막 행을 가져옵니다. 내가 얻을하려는 각 행에 대해 , 대화 첫 번째 행의 "날짜 시간" "메시지"그것을

CREATE TABLE messages (
    ID INT NOT NULL AUTO_INCREMENT PRIMARY KEY, 
    FromID INT NOT NULL, 
    ToID INT NOT NULL,  
    ConversationID INT NOT NULL,  
    Subject varchar(255), 
    Message varchar(255), 
    DateTime DATETIME          
    ) ENGINE=InnoDB; 


CREATE TABLE conversations (
    ID INT NOT NULL AUTO_INCREMENT PRIMARY KEY          
    ) ENGINE=InnoDB; 



INSERT INTO conversations (ID) VALUES (1), (2), (3); 
INSERT INTO messages (FromID, ToID, ConversationID, Subject, Message, DateTime) VALUES (1,2, 1, "Hi", "This is a test message", "2010-08-08 16:23:48");   
INSERT INTO messages (FromID, ToID, ConversationID, Subject, Message, DateTime) VALUES (1,2, 1, "", "Hey again you have not answered", "2010-08-08 16:23:52");                                    
INSERT INTO messages (FromID, ToID, ConversationID, Subject, Message, DateTime) VALUES (2,1, 1, "", "Hi this is my answer", "2010-08-08 16:23:59"); 


INSERT INTO messages (FromID, ToID, ConversationID, Subject, Message, DateTime) VALUES (2,1, 2, "2.Hi", "2.This is a test message", "2010-08-08 16:25:48");   
INSERT INTO messages (FromID, ToID, ConversationID, Subject, Message, DateTime) VALUES (1,2, 2, "", "2.Hi back", "2010-08-08 16:25:52");                                    
INSERT INTO messages (FromID, ToID, ConversationID, Subject, Message, DateTime) VALUES (2,1, 2, "", "2.Hi this is my answer", "2010-08-08 16:25:59"); 


INSERT INTO messages (FromID, ToID, ConversationID, Subject, Message, DateTime) VALUES (2,1, 3, "3.Hi", "3.This is a test message", "2010-08-08 16:27:48");   
INSERT INTO messages (FromID, ToID, ConversationID, Subject, Message, DateTime) VALUES (1,2, 3, "", "2.Hi back", "2010-08-08 16:27:52"); 
INSERT INTO messages (FromID, ToID, ConversationID, Subject, Message, DateTime) VALUES (1,2, 3, "", "2.Hello are you there?", "2010-08-08 16:27:59");                                    

Answer 1

SELECT M.ConversationID, 
MAX(CASE WHEN M.DateTime = X.FirstRow THEN M.Subject END) AS Subject, 
CAST(COALESCE(MAX(CASE WHEN M.DateTime = X.LastRowSentByOtherUser 
         THEN M.DateTime END),X.LastRow) AS DateTime)AS LastTime, 
MAX(CASE WHEN M.DateTime = X.LastRow THEN M.Message END) AS Message, 
MAX(CASE WHEN FromID = 1 THEN ToID ELSE FromID END) AS OtherParticipantId 
FROM messages M 
JOIN (
    SELECT ConversationID, MIN(DateTime) AS FirstRow, MAX(DateTime) AS LastRow, 
    MAX(CASE WHEN FromID<>1 THEN DateTime END) AS LastRowSentByOtherUser 
    FROM messages 
    WHERE FromID=1 OR ToID=1 
    GROUP BY ConversationID 
) X ON X.ConversationID = M.ConversationID 
AND (M.DateTime IN (X.FirstRow, X.LastRow, X.LastRowSentByOtherUser)) 
GROUP BY M.ConversationID 
HAVING MAX(CASE WHEN M.DateTime = X.LastRowSentByOtherUser 
        THEN M.DateTime END) IS NOT NULL 

Answer 2

를 썼다 상관없이이 대화의 마지막 메시지에서 첫 번째 행의 "제목" @ROW_NUMBER() in MySQL을보십시오 - 문제에 확실히 적용 할 수 있습니다.

Answer 3

당신이 뭔가를 시도해야합니다 : 단 2 사람 사이의 대화가

SELECT 
    m1.Subject, 
    m1.DateTime, 
    m2.Message 
FROM conversations c 
INNER JOIN 
    (SELECT MIN(ID) AS minID, 
      MAX(ID) AS maxID, 
      ConversationID 
    FROM messages 
    WHERE FromID = @userID OR ToID = @userID 
    GROUP BY ConversationID) AS cGrouped 
ON c.ConversationID = cGrouped.ConversationID 
INNER JOIN messages m1 ON m1.ID = cGrouped.minID 
INNER JOIN messages m2 ON m2.ID = cGrouped.maxID