2013-04-10 54 views
2

我試着從3列,每列,但IM得到一個錯誤得到最後的60項,平均:mysql的獲取平均數列與極限

#1248 - Every derived table must have its own alias 

這是查詢:

SELECT AVG(Sensor1) , AVG(Sensor2) , AVG(Sensor3) 
FROM (

    SELECT Sensor1, Sensor2, Sensor3 
    FROM temperatur 
    ORDER BY Zeit DESC 
    LIMIT 0 , 60 
) 

林相當失去了別名,我相信..

回答

4

的MySQL要求所有派生表和子查詢的別名,你是右括號後失蹤別名:

SELECT AVG(d.Sensor1) , AVG(d.Sensor2) , AVG(d.Sensor3) 
FROM 
(
    SELECT Sensor1, Sensor2, Sensor3 
    FROM temperatur 
    ORDER BY Zeit DESC 
    LIMIT 0 , 60 
) d --- add this 
+0

快:)我會,如果你給予好評礦井給予好評你們的;) – 2013-04-10 23:41:52

+0

謝謝,我想現在我明白了! :) – Jrc 2013-04-10 23:54:11

2

我猜測在語法時才但我相信你需要更多的東西是這樣的:

SELECT AVG(T.Sensor1),AVG(T.Sensor2),AVG(T.Sensor3) FROM (

SELECT Sensor1, Sensor2, Sensor3 
FROM temperatur 
ORDER BY Zeit DESC 
LIMIT 0 , 60 

)筆