SQL 查询 returns 只有 1 行,连接三个以上的表的正确方法是什么?

SQL query returns only 1 row and what is the correct way to join more than three tables?

我正在开发这个酒店预订系统。

我有多个 table 客户、预订、账单、房间、房型、代理。

我正在收据。

所有这些 table 都已连接,但我只收到一行信息,而不是三行收据信息。

我只是一个初学者,如果我的查询不够复杂或经验不足,请多多包涵。

顺便说一句,我正在使用 sqlfiddle。

这是我的查询:

SELECT r.bill_ID
     , c.name
     , b.payment_type
     , b.card_number
     , a.agency_name
     , mt.description
     , SUM(x.fee) Amenities
     , mt.room_rate Room
  FROM reservation r
  JOIN bill b
    ON b.bill_ID = r.bill_ID
  JOIN agency a
    ON a.agency_ID = r.agency_ID
  JOIN room m 
    ON m.room_ID = r.room_ID
  JOIN amenities x 
    ON b.amenity_ID = x.amenity_ID
  JOIN customer c
    ON c.cust_ID = r.cust_ID
  JOIN roomType mt
    ON mt.type_ID = m.type_ID;

这是我的 table:

CREATE TABLE roomType(
  type_ID varchar(15) PRIMARY KEY,
  description varchar(45),
  room_rate int);
                      
CREATE TABLE amenities(
  amenity_ID varchar(15) PRIMARY KEY,
  amenity_name varchar(45),
  fee decimal(4,1));
  
CREATE TABLE agency(
  agency_ID varchar(15) PRIMARY KEY,
  agency_name varchar(45));
  
CREATE TABLE customer(
  cust_ID varchar(15) PRIMARY KEY,
  name varchar(45),
  address varchar(45),
  contact_no varchar(45),
  email varchar(45),
  agency_ID varchar(15),
  FOREIGN KEY (agency_ID) REFERENCES agency(agency_ID));
  
CREATE TABLE bill(
  bill_ID varchar(15) PRIMARY KEY,
  cust_ID varchar(15),
  amenity_ID varchar(15),
  agency_ID varchar(15),
  payment_type varchar(5),
  card_number int,
  amount int,
  FOREIGN KEY (agency_ID) REFERENCES agency(agency_ID),
  FOREIGN KEY (amenity_ID) REFERENCES amenities(amenity_ID),
  FOREIGN KEY (cust_ID) REFERENCES customer(cust_ID));
  
CREATE TABLE room(
  room_ID varchar(15) PRIMARY KEY,
  type_ID varchar(15),
  room_num int,
  max_guest int,
  availability varchar(15),
  FOREIGN KEY (type_ID) REFERENCES roomType(type_ID));
  
CREATE TABLE reservation(
  rsrvtn_ID varchar(15) PRIMARY KEY,
  bill_ID varchar(15),
  cust_ID varchar(15),
  room_ID varchar(15),
  agency_ID varchar(15),
  check_in date,
  check_out date,
  num_guest int,
  book_status varchar(10),
  FOREIGN KEY (agency_ID) REFERENCES agency(agency_ID),
  FOREIGN KEY (bill_ID) REFERENCES bill(bill_ID),
  FOREIGN KEY (cust_ID) REFERENCES customer(cust_ID),
  FOREIGN KEY (room_ID) REFERENCES room(room_ID));
  
CREATE TABLE administrator(
  admin_ID varchar(15) PRIMARY KEY,
  admin_lvl varchar(15),
  rsrvtn_ID varchar(15),
  bill_ID varchar(15),
  room_ID varchar(15),
  cust_ID varchar(15),
  FOREIGN KEY (cust_ID) REFERENCES customer(cust_ID),
  FOREIGN KEY (rsrvtn_ID) REFERENCES reservation(rsrvtn_ID),
  FOREIGN KEY (bill_ID) REFERENCES bill(bill_ID),
  FOREIGN KEY (room_ID) REFERENCES room(room_ID));

您的查询有一个 SUM() 使其成为聚合查询。没有 GROUP BY returns 正好一行的聚合查询。或者错误。

您的查询应该返回一个错误。为什么? SELECT 列与 GROUP BY 列不一致。所有未聚合的列都应该在 GROUP BY:

GROUP BY r.bill_ID, c.name, b.payment_type, b.card_number, 
         a.agency_name, mt.description, mt.room_rate

这是常识。大多数数据库都需要它。它甚至是 MySQL 支持的当前版本。

不幸的是,曾几何时,MySQL 支持您问题中的语法。从 MySQL 8.0 开始,这不再是默认行为。它由系统设置 only_full_group_by 控制,在 documentation.

中进行了解释