Question

我需要选择一个列表，显示雇用汽车的每个客户的客户ID，职位，名字和姓氏，按客户姓名的字母顺序排序，以及每个客户已经预订的数量。< / p>

我已完成第一部分，但不确定在何处计算预订数量。

以下表格是

create table customer
(customer_id  char(4) primary key not null,
customer_sname varchar (30) not null,
customer_fname varchar (30) not null,
customer_title varchar (6) not null,
customer_address1 varchar (35) not null,
customer_address2 varchar (35) null,
customer_postcode varchar (25) null,
customer_phone varchar (30) null,
customer_email varchar (40) null,
customer_di varchar (40) not null)
ENGINE=InnoDB;

create table car_booking
(booking_id INTEGER AUTO_INCREMENT primary key not null,
car_id char (4) not null,
customer_id char (4)  not null,
hire_sdate date not null,
hire_edate date not null)
engine=innodb

我已经完成了这个

select customer_id, customer_title, Customer_fname, customer_sname
from customer
where customer_id in
(select customer_id from car_booking )
order by customer_sname asc

由于

Answer 1

这将需要使用聚合函数（COUNT），GROUP BY子句和LEFT JOIN到CAR_BOOKING表：

   SELECT c.customer_id, c.customer_title, c.customer_fname, c.customer_sname,
          COALESCE(COUNT(*), 0) AS num_bookings
     FROM CUSTOMER c
LEFT JOIN CAR_BOOKING cb ON cb.customer_id = c.customer_id
 GROUP BY c.customer_id, c.customer_title, c.customer_fname, c.customer_sname
 ORDER BY c.customer_sname

因为有些列没有包含在像COUNT这样的聚合函数中，所以需要在GROUP BY子句中定义这些列。

我使用LEFT OUTER JOIN到CAR_BOOKINGS表来返回没有预订的客户 - 这些记录将在num_booking列中显示为零。您可以在查询中省略LEFT关键字，仅返回客户＆amp;计入预订。 COALESCE是一个将空值转换为所需值的标准函数 - 在这种情况下，计数为空...

Answer 2

在SQL Server中，我使用：

select c.customer_id,
       c.customer_title,           
       c.customer_fname,
       c.customer_sname,
       count (*)
  from cutomer c,
       car_booking cb
 where cb.customer_id = c.customer_id
 group by c.customer_id,
          c.customer_title,
          c.customer_fname,
          c.customer_sname

不熟悉MySQL，所以它可能会有所不同，但这是一般的想法。

Answer 3

select customer.customer_id, customer.customer_title, customer.customer_fname, customer.customer_sname, count(*) as Bookings
from customer JOIN car_booking ON customer.customer_id = car_booking.customer_id
GROUP BY customer.customer_id, customer.customer_title, customer.Customer_fname, customer.customer_sname
order by customer_sname asc

我需要帮助使用count来查询

3 个答案: