通過在MySQL

Question

使用組，我有以下表創建腳本：

CREATE TABLE assetcost(
assettype VARCHAR(20) PRIMARY KEY, 
rentamt INT 
); 

CREATE TABLE furnishitem(
itemid VARCHAR(4) PRIMARY KEY CHECK(itemid LIKE 'I%'), 
description VARCHAR(30), 
availablesets INT, 
assettype VARCHAR(25), 
specialCharge CHAR(1) CHECK(specialcharge IN ('Y','N')), 
FOREIGN KEY(assettype) REFERENCES assetcost(assettype) 
); 

CREATE TABLE custdetail(
custid VARCHAR(5) PRIMARY KEY CHECK(custid LIKE 'C%'), 
custname VARCHAR(30) 
); 


CREATE TABLE transaction(
transid INT UNIQUE, 
custid VARCHAR(5) , 
itemid VARCHAR(4), 
sets INT, 
days INT, 
amount INT, 
returned char(1) Check (returned in('Y','N')), 
FOREIGN KEY(custid)REFERENCES custdetail(custid), 
FOREIGN KEY(itemid)REFERENCES furnishitem(itemid) 
); 

編寫查詢時顯示的那些誰也/支付總金額的最低客戶的客戶ID和CUSTNAME我得到的錯誤，不管返回的項目。

我的查詢是：

select t.custid,c.custname 
    -> from transaction t inner join custdetail c 
    -> on t.custid=c.custid 
    -> group by t.custid 
    -> having sum(amount)=(select min(sum(amount) from transaction group by custid); 

Answer 1

您可以在一個更優雅的方式來解決這個問題，但快速解決您的問題是：

SELECT t.custid, c.custname 
FROM TRANSACTION t 
INNER JOIN custdetail c ON t.custid = c.custid 
GROUP BY t.custid 
HAVING sum(amount) = (
    SELECT sum(amount) 
    FROM TRANSACTION 
    GROUP BY custid 
    ORDER BY 1 ASC 
    LIMIT 1 
    ); 

你不能真正一次去做min(sum(amount))。這種方法將爲您提供最小值的custid的總和。