MySQL自定義選擇訂單

我在MySQL查詢中遇到自定義訂單問題。任何幫助，將不勝感激。MySQL自定義選擇訂單

所以我有一個表具有以下數據集：

id SCADA Date  Hour  Minute Second PlantNo Key Value 
1 2924 2014-01-02 00:00:12 00:00:00 00:00:32 25 1300  0 
2 2924 2014-01-02 00:00:15 00:00:41 00:00:33  7 1300 500 
3 2924 2014-01-02 00:00:16 00:00:03 00:00:12 25 1300 500 
4 2924 2014-01-02 00:00:21 00:00:53 00:00:05 25 1300 1000 
5 2924 2014-01-02 00:00:21 00:00:53 00:00:05 26 1300 2060 
6 2924 2014-01-02 00:00:21 00:00:53 00:00:09  7 1300 1000 
7 2924 2014-01-03 00:00:07 00:00:42 00:00:06 25 1300 2060 
8 2924 2014-01-03 00:00:07 00:00:42 00:00:07  7 1300 2060 
9 2924 2014-01-03 00:00:12 00:00:00 00:00:03  5 1300 20 
10 2924 2014-01-03 00:00:12 00:00:00 00:00:07  5 2501 18 
11 2924 2014-01-04 00:00:11 00:00:52 00:00:56 16 1031  0 
12 2924 2014-01-04 00:00:12 00:00:00 00:00:07  5 2501 8.5 
13 2924 2014-01-04 00:00:13 00:00:51 00:00:05  4 1030  0 
14 2924 2014-01-04 00:00:18 00:00:23 00:00:11  4 1030  1 
15 2924 2014-01-06 00:00:16 00:00:08 00:00:36 26 1300 1500 
16 2924 2014-01-07 00:00:17 00:00:11 00:00:00  5 1300 50 
17 2924 2014-01-07 00:00:19 00:00:31 00:00:38  5 1030  0 
18 2924 2014-01-07 00:00:21 00:00:00 00:00:53  5 1300 200 
19 2924 2014-01-07 00:00:21 00:00:59 00:00:17  5 1300 500 
20 2924 2014-01-08 00:00:08 00:00:28 00:00:53  5 1300 1000 
21 2924 2014-01-08 00:00:08 00:00:56 00:00:33 26 1300 500 
22 2924 2014-01-08 00:00:11 00:00:41 00:00:06 26 1300 1000 
23 2924 2014-01-08 00:00:11 00:00:41 00:00:41  5 1300 1500

，我需要選擇只從每個鍵和每個PlantNo的最後一個值。最後的意思是最早的日期和時間。

這是我到目前爲止的代碼：

SELECT * FROM (
    SELECT SCADA, PlantNo, tblpartemp.Date, MAKETIME(tblpartemp.Hour,tblpartemp.Minute,tblpartemp.Second) AS Time, tblpartemp.Key, tblparameter.description, tblpartemp.Value 
    FROM tblpartemp 
    LEFT JOIN tblparameter ON tblpartemp.Key = tblparameter.id 
    ORDER BY Date, Time DESC 
) AS T1 
GROUP BY T1.Key

來源

2014-05-09 devwebapp

爲＃1代碼幫助，而不是讓人們爲你寫代碼。向我們展示迄今爲止所做的工作，它出了什麼問題，我們可以幫助糾正它。 – Adam

對不起，我忘了把我的代碼！我編輯了我的問題 – devwebapp

雖然這不是一個解決方案，但您可以根據[這個想法]編寫子查詢（http://stackoverflow.com/questions/8125996/how-to-select-oldest-date-from-mysql ），所以選擇每個唯一的密鑰，然後找到每個的最舊記錄。 – Adam

您可以使用下面的查詢來實現的結果。

SELECT * FROM (SELECT * FROM plants ORDER BY Date DESC,Time DESC) AS t GROUP BY Plant

我假定表名爲'植物'，表中沒有其他字段。

編輯

來源

2014-05-09 14:47:01 bakee

謝謝，但它並不像我需要的那樣工作。請參閱我上傳的圖片。謝謝。 – devwebapp

我更新了我的答案。如果這次它不起作用，那麼我恐怕，我不明白你的問題，或者你想要的輸出與你的描述不一致。 – bakee

請注意，這個解決方案僅僅依靠MySQL的無證行爲。但是，這就是說，它*應該*工作。 – Strawberry

一旦你整理出你的設計缺陷，考慮以下因素：

DROP TABLE IF EXISTS my_table; 

CREATE TABLE my_table 
(id INT NOT NULL AUTO_INCREMENT PRIMARY KEY 
,SCADA INT NOT NULL 
,Date DATETIME NOT NULL 
,PlantNo INT NOT NULL 
,my_key INT NOT NULL 
,my_value INT NOT NULL 
); 

INSERT INTO my_table VALUES 
(1 ,2924 ,'2014-01-02 12:00:32' ,25 ,1300 , 0), 
(2 ,2924 ,'2014-01-02 15:41:33' , 7 ,1300 , 500), 
(3 ,2924 ,'2014-01-02 16:03:12' ,25 ,1300 , 500), 
(4 ,2924 ,'2014-01-02 21:53:05' ,25 ,1300 , 1000), 
(5 ,2924 ,'2014-01-02 21:53:05' ,26 ,1300 , 2060), 
(6 ,2924 ,'2014-01-02 21:53:09' , 7 ,1300 , 1000), 
(7 ,2924 ,'2014-01-03 07:42:06' ,25 ,1300 , 2060), 
(8 ,2924 ,'2014-01-03 07:42:07' , 7 ,1300 , 2060), 
(9 ,2924 ,'2014-01-03 12:00:03' , 5 ,1300 , 20), 
(10 ,2924 ,'2014-01-03 12:00:07' , 5 ,2501 , 18), 
(11 ,2924 ,'2014-01-04 11:52:56' ,16 ,1031 , 0), 
(12 ,2924 ,'2014-01-04 12:00:07' , 5 ,2501 , 8.5), 
(13 ,2924 ,'2014-01-04 13:51:05' , 4 ,1030 , 0), 
(14 ,2924 ,'2014-01-04 18:23:11' , 4 ,1030 , 1), 
(15 ,2924 ,'2014-01-06 16:08:36' ,26 ,1300 , 1500), 
(16 ,2924 ,'2014-01-07 17:11:00' , 5 ,1300 , 50), 
(17 ,2924 ,'2014-01-07 19:31:38' , 5 ,1030 , 0), 
(18 ,2924 ,'2014-01-07 21:00:53' , 5 ,1300 , 200), 
(19 ,2924 ,'2014-01-07 21:59:17' , 5 ,1300 , 500), 
(20 ,2924 ,'2014-01-08 08:28:53' , 5 ,1300 , 1000), 
(21 ,2924 ,'2014-01-08 08:56:33' ,26 ,1300 , 500), 
(22 ,2924 ,'2014-01-08 11:41:06' ,26 ,1300 , 1000), 
(23 ,2924 ,'2014-01-08 11:41:41' , 5 ,1300 , 1500); 

SELECT x.* 
    FROM my_table x 
    JOIN 
    (SELECT plantno,my_key 
      , MAX(date) max_date 
     FROM my_table 
     GROUP 
      BY plantno 
      , my_key 
    ) y 
    ON y.plantno = x.plantno 
    AND y.my_key = x.my_key 
    AND y.max_date = x.date 
ORDER 
    BY plantno,my_key; 
+----+-------+---------------------+---------+--------+----------+ 
| id | SCADA | Date    | PlantNo | my_key | my_value | 
+----+-------+---------------------+---------+--------+----------+ 
| 14 | 2924 | 2014-01-04 18:23:11 |  4 | 1030 |  1 | 
| 17 | 2924 | 2014-01-07 19:31:38 |  5 | 1030 |  0 | 
| 23 | 2924 | 2014-01-08 11:41:41 |  5 | 1300 |  1500 | 
| 12 | 2924 | 2014-01-04 12:00:07 |  5 | 2501 |  9 | 
| 8 | 2924 | 2014-01-03 07:42:07 |  7 | 1300 |  2060 | 
| 11 | 2924 | 2014-01-04 11:52:56 |  16 | 1031 |  0 | 
| 7 | 2924 | 2014-01-03 07:42:06 |  25 | 1300 |  2060 | 
| 22 | 2924 | 2014-01-08 11:41:06 |  26 | 1300 |  1000 | 
+----+-------+---------------------+---------+--------+----------+

相當於此查詢破解如下：

SELECT * 
    FROM 
    (SELECT * 
     FROM my_table 
     ORDER 
      BY Date DESC 
    ) t 
GROUP 
    BY plantno 
    , my_key;

來源

2014-05-09 15:22:51 Strawberry

MySQL自定義選擇訂單

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