我一直堅持小查詢。有一個名爲Table_Activities的表。什麼可以是這個優化的查詢？

Table_Activities（Act_id，Dat E，Activity_name，description）。

我們需要生成一份報告。用戶將選擇月份和年份，他希望從下拉列表中生成報告。

我們需要按活動名稱顯示該月份和年份組的所有活動。

示例 - 用戶選擇June and 。

報告會是─

園藝

01/06/2012 - They have planted 100 trees. 
14/06/2012 - something 
27/06/2012 - something

培訓

02/06/2012 - Detail description 
15/06/2012 - something 
28/06/2012 - something

我的問題是什麼將是MySQL查詢以這種格式來提取數據？

來源

2012-08-13 user1545152

你可以發表你的表（S）的模式？ – 2012-08-13 06:33:41

是「他們種了100棵樹。」一個描述？ – hims056 2012-08-13 06:49:18

select `Date`,description from tm_activities 
where month(`Date`)='6' and year(`Date`)='2012' 
order by Activity_name,`date`

返回準確的格式在你的問題指出，請嘗試以下：如下

select concat(if(actdate='',activity_name,date_format(actdate,'%d/%m/%y')),if(description<>'',concat(' - ',description),'')) as labelm from 
(
(select ActDate,description,activity_name from tm_activities where month(ActDate)='6' and year(ActDate)='2012' 
) 
union all 
(Select distinct '','',activity_name from tm_activities where month(ActDate)='6' and year(ActDate)='2012') 
)m order by activity_name,actdate 
;

SQL FIDDLE HERE.

輸出：

Gardening 
01/06/12 - They have planted 100 trees. 
27/06/12 - Gar 2 
Training 
12/06/12 - Training 1 
28/06/12 - Traning 2 
30/06/12 - Traning 3

來源

2012-08-13 06:51:05 sel

偉大的解決方案。向你致敬。 – user1545152 2012-08-13 07:30:10

嗨，我們如何使用php在HTML頁面上顯示相同的內容。任何幫助將不勝感激。 – user1545152 2012-08-13 07:36:52

在html頁面上顯示相同的意思是什麼意思？ – sel 2012-08-13 07:46:34

要獲取ACTIVITY_NAME特定月份和年份組數據試試這個：

SELECT Activity_name 
,GROUP_CONCAT(DATE_FORMAT(Date,"%d/%m/%Y")) as `Date` 
,GROUP_CONCAT(Description) as `Description` 
FROM Table_Activities 
WHERE MONTH(Date) = MONTH('2012-06-01') 
AND YEAR(Date) = YEAR('2012-06-01') 
GROUP BY Activity_name

輸出

 
ACTIVITY_NAME DATE     DESCRIPTION 
----------------------------------------------------------------------------------- 
Gardening  01/06/2012,27/06/2012 They have planted 100 trees.,Description3 
Training   12/06/2012,28/06/2012 Description2,Description4

See this SQLFiddle

來源

2012-08-13 06:34:34 hims056

Select 
    DATE_FORMAT('date_column',"%D/%M/%Y") as `Date`, 
    other_column 
from Table_Activities 
where Month(Date) = Month('2012-06-01') 
AND Year(Date) = Year('2012-06-01')

來源

2012-08-13 06:36:28

什麼可以是這個優化的查詢？

回答

See this SQLFiddle

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