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我的要求是显示国家名称、发票总数及其平均金额。此外，我只需要返回平均发票金额大于所有发票平均发票金额的国家。

查询 Oracle 数据库

SELECT cntry.NAME, 
       COUNT(inv.NUMBER), 
       AVG(inv.TOTAL_PRICE)
FROM   COUNTRY cntry JOIN
       CITY ct ON ct.COUNTRY_ID = cntry.ID JOIN
       CUSTOMER cst ON cst.CITY_ID = ct.ID JOIN
       INVOICE inv ON inv.CUSTOMER_ID = cst.ID
GROUP BY cntry.NAME,
         inv.NUMBER,
         inv.TOTAL_PRICE
HAVING AVG(inv.TOTAL_PRICE) > (SELECT AVG(TOTAL_PRICE)
                               FROM INVOICE);

结果: 奥地利 1 9500

预计：奥地利 2 4825

架构

国家

ID(INT)(PK) | NAME(VARCHAR)

城市

ID(INT)(PK) | NAME(VARCHAR) | POSTAL_CODE(VARCHAR) | COUNTRY_ID(INT)(FK)

顾客

ID(INT)(PK) | NAME(VARCHAR) | CITY_ID(INT)(FK) | ADDRS(VARCHAR) | POC(VARCHAR) | EMAIL(VARCHAR) | IS_ACTV(INT)(0/1)

发票

ID(INT)(PK) | NUMBER(VARCHAR) | CUSTOMER_ID(INT)(FK) | USER_ACC_ID(INT) | TOTAL_PRICE(INT)
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2 回答 2

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没有样本数据，我们无法真正判断这是否：

预计：奥地利 2 4825

是真是假。

无论如何：会将GROUP BY条款更改为

GROUP BY cntry.NAME

（即从中删除额外的两列）有什么好处吗？

于 2021-07-19T19:03:10.783 回答
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`SELECT C.COUNTRY_NAME,COUNT(I.INVOICE_NUMBER),AVG(I.TOTAL_PRICE) AS AVERAGE 
  FROM COUNTRY AS C JOIN CITY AS CS ON C.ID=CS.COUNTRY_ID
 JOIN CUSTOMER AS CUS ON CUS.CITY_ID=CS.ID
 JOIN INVOICE AS I ON I.CUSTOMER_ID=CUS.ID
 GROUP BY C.COUNTRY_NAME,C.ID
 HAVING AVERAGE>(SELECT AVG(TOTAL_PRICE) FROM INVOICE`
于 2021-09-17T18:23:35.353 回答