Question

“仅显示前2栋建筑及其相应的平均值会话的价格。“

非常感谢任何帮助，我已经尝试了很长时间这是我迄今为止最好的尝试（我认为）......

SELECT BUILDINGNO.BUILDING, AVG(SESSIONPRICE.CONFERENCESESSION)
FROM BUILDING
JOIN CONFERENCESESSION ON BUILDINGNO.BUILDING = BUILDINGNO.CONFERENCESESSION
WHERE BUILDINGNO.BUILDING = SESSIONPRICE.CONFERENCESESSION
GROUP BY BUILDINGNO.BUILDING
Order by BUILDINGNO.BUILDING DESC;

干杯

Answer 1

你可以使用它。

SELECT TOP 2  BUILDINGNO.BUILDING, AVG(SESSIONPRICE.CONFERENCESESSION)
FROM BUILDING
JOIN CONFERENCESESSION ON BUILDINGNO.BUILDING = BUILDINGNO.CONFERENCESESSION
WHERE BUILDINGNO.BUILDING = SESSIONPRICE.CONFERENCESESSION
GROUP BY BUILDINGNO.BUILDING
Order by AVG(SESSIONPRICE.CONFERENCESESSION) DESC;

Answer 2

使用行号/限制..取决于您使用的数据库

SELECT BUILDINGNO.BUILDING, AVG(SESSIONPRICE.CONFERENCESESSION)
FROM BUILDING
JOIN CONFERENCESESSION ON BUILDINGNO.BUILDING = 
        BUILDINGNO.CONFERENCESESSION
WHERE BUILDINGNO.BUILDING = SESSIONPRICE.CONFERENCESESSION
   AND rownum <3 /*if you are using oracle */
GROUP BY BUILDINGNO.BUILDING
Order by BUILDINGNO.BUILDING DESC;

Answer 3

要解决此问题，我必须知道您使用的是哪个数据库。因为如果你使用 oracle 而不是Rownum而Rowcount应该解决问题。如果你使用 sql server ，你应该使用{{ 1}}。

我将为Mysql提供解决方案：

TOP

对于oracle和sql server，你可以按照@Valli＆amp;的答案。 @sarslan分别。

希望它有所帮助！

查找多行AVG并显示最高AVG

3 个答案: