Question

CREATE TABLE    CUSTOMER (
CUSID   VARCHAR(25) NOT NULL,
CNAME   VARCHAR(50),
CONSTRAINT CUSTOMER_PKEY PRIMARY KEY (CUSID),
);

CREATE TABLE    SHOP (
SHOPID  VARCHAR(10) NOT NULL,
ADDRESS VARCHAR(25),
CONSTRAINT SHOP_PKEY PRIMARY KEY (SHOPID),
);

CREATE TABLE    VISIT (
CUSID   VARCHAR(25) NOT NULL,
SHOPID  VARCHAR(10) NOT NULL,
VDATE   DATE        NOT NULL,
CONSTRAINT VISIT_PKEY PRIMARY KEY (CUSID, SHOPID, VDATE),
CONSTRAINT VISIT_FKEY1 FOREIGN KEY (CUSID) REFERENCES CUSTOMER(CUSID),
CONSTRAINT VISIT_FKEY2 FOREIGN KEY (SHOPID) REFERENCES SHOP(SHOPID)
);

如何查找客户至少被访问过2次的商店地址，名称为“john”？

从商店中选择地址自然联合访问哪里是CUSID（从客户中选择CUST = CEND ='john'GROUP BY CUSID HAVING COUNT（CUSID）＆gt; 2）;

我尝试了很多种连接，似乎在我将count和equal条件放在一起之后，我的结果将是0行。

Answer 1

    SELECT DISTINCT s.address
    FROM shop s 
    JOIN visit v ON s.shopid = v.shopid 
    JOIN customer c ON v.customerid = c.customerid 
    WHERE c.cname = 'John' 
    GROUP BY 
    s.address
    , c.customerid 
    HAVING COUNT(*) > 1

具有计数条件和相同条件的Oracle Sql 3表

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