我的 mysql 查询中的错误与限制条件

Question

我有 3 张桌子 客户、时间和销售额 我想找出所有客户的年收入条件是没有孩子的客户和收入必须大于我们设定的限制 我的表结构 客户

CREATE TABLE `customers` (
  `customer_id` int(11) DEFAULT NULL,
  `account_num` double DEFAULT NULL,
  `lname` varchar(50) DEFAULT NULL,
  `fname` varchar(50) DEFAULT NULL,
  `mi` varchar(50) DEFAULT NULL,
  `address1` varchar(50) DEFAULT NULL,
  `address2` varchar(50) DEFAULT NULL,
  `address3` varchar(50) DEFAULT NULL,
  `address4` varchar(50) DEFAULT NULL,
  `postal_code` varchar(50) DEFAULT NULL,
  `region_id` int(11) DEFAULT NULL,
  `phone1` varchar(50) DEFAULT NULL,
  `phone2` varchar(50) DEFAULT NULL,
  `birthdate` datetime DEFAULT NULL,
  `marital_status` varchar(50) DEFAULT NULL,
  `yearly_income` varchar(50) DEFAULT NULL,
  `gender` varchar(50) DEFAULT NULL,
  `total_children` smallint(6) DEFAULT NULL,
  `num_children_at_home` smallint(6) DEFAULT NULL,
  `education` varchar(50) DEFAULT NULL,
  `member_card` varchar(50) DEFAULT NULL,
  `occupation` varchar(50) DEFAULT NULL,
  `houseowner` varchar(50) DEFAULT NULL,
  `num_cars_owned` smallint(6) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

销售额

CREATE TABLE `sales` (
  `product_id` int(11) DEFAULT NULL,
  `time_id` int(11) DEFAULT NULL,
  `customer_id` int(11) DEFAULT NULL,
  `store_id` int(11) DEFAULT NULL,
  `store_sales` float DEFAULT NULL,
  `store_cost` float DEFAULT NULL,
  `unit_sales` double DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

次

CREATE TABLE `times` (
  `time_id` int(11) DEFAULT NULL,
  `the_date` datetime DEFAULT NULL,
  `the_day` varchar(50) DEFAULT NULL,
  `the_month` varchar(50) DEFAULT NULL,
  `the_year` smallint(6) DEFAULT NULL,
  `day_of_month` smallint(6) DEFAULT NULL,
  `month_of_year` smallint(6) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

我的问题是：-查找没有孩子且 yearly_income 大于用户在运行查询时给出的限制的客户列表。

我的查询是

SET @limit=50;
SELECT customers.`fname`, customers.`lname` ,ROUND(SUM(sales.store_sales)) as income,times.the_year
 FROM `sales` 
 LEFT JOIN times
 ON sales.time_id=times.time_id
 LEFT JOIN customers 
ON customers.customer_id=sales.customer_id 
WHERE income>@limit AND `total_children`=0
GROUP BY sales.customer_id,times.the_year 

我收到此错误：#1054 - 'where 子句'中的未知列 'income'

Answer 1

您别名为income 的数量是一个集合，因此在WHERE 子句中引用它是没有意义的。将此 WHERE 逻辑移动到 HAVING 子句：

SET @limit=50;
SELECT
    c.fname,
    c.lname,
    ROUND(SUM(s.store_sales)) AS income,
    t.the_year
FROM sales s 
LEFT JOIN times t
    ON s.time_id = t.time_id
LEFT JOIN customers c
    ON c.customer_id = s.customer_id 
WHERE
    total_children = 0
GROUP BY
    c.customer_id,
    t.the_year 
HAVING
    ROUND(SUM(s.store_sales)) > @limit;

请注意，从技术上讲，我们可以在 HAVING 子句中使用别名：

HAVING income > @limit;

但这不能移植到大多数其他数据库。另外，我在查询中引入了别名，这样更容易阅读。

Answer 2

对于原始表中不存在的聚合列，您应该使用having 子句而不是where

SET @limit=50;
SELECT customers.`fname`, customers.`lname` ,ROUND(SUM(sales.store_sales)) as income,times.the_year
 FROM `sales` 
 LEFT JOIN times
 ON sales.time_id=times.time_id
 LEFT JOIN customers 
ON customers.customer_id=sales.customer_id 
WHERE `total_children`=0
GROUP BY customers.customer_id,times.the_year 
having income>@limit 

Answer 3

income是别名，所以不能在where条件下使用。

SET @limit=50;
SELECT * FROM
(
SELECT customers.`fname`, customers.`lname` ,ROUND(SUM(sales.store_sales)) 
       as income,times.the_year
FROM `sales` LEFT JOIN times ON sales.time_id=times.time_id
             LEFT JOIN customers ON customers.customer_id=sales.customer_id 
WHERE  `total_children`=0
GROUP BY sales.customer_id,times.the_year 
)t WHERE income>@limit 

Answer 4

查找没有孩子和年收入的客户列表 大于用户在运行查询时给出的限制。

您确定不想要这样的简单查询吗？

SELECT *
FROM customers AS c 
WHERE yearly_income > @limit  -- but why is yearly_income a VarChar(50)?
AND total_children=0