我们有两张table：
1.地点
1.评论
表详细信息示例

CREATE TABLE places (   
id INT,     
name varchar(255),  
address varchar(255),   
type varchar(255),  
average_rating INT,     
price_point varchar(255),   
total reviews INT,
PRIMARY KEY id); 

INSERT INTO places VALUES  
('1', 'Hairs to You',   '45-45', 'Queens Boulevard',    'Beauty', '4.9',    '$$$',  '36'), 
('2',  'Doggonit!',      '100',   'Atlantic Ave',      'Pet Store', '3.1',  '$$',   '52'),
('3', 'Abra Kebabra',    '193',    'Sauthoff Way',  'Restaurant',   '3.3',  '$',    '315');

CREATE TABLE reviews (  
id INT,     
user_name varchar(255),     
place_id varchar(255),  
review_date DATE,   
rating INT,     
note varchar(255),
PRIMARY KEY id,
FOREIGN KEY (place_id) REFERENCES Places(id)
);

INSERT INTO reviews VALUES 
('149', '@pinkdeb',     '8',    '2019-07-25',   '4',    'Nice little place to grab a drink'),
('117', '@ahohl',       '16',   '2019-07-29',   '3',    'The produce is always bad but otherwise okay'),
('119', '@sammyantha',  '8',    '2019-07-30',   '4',    'LOVE how kitschy this place is! Bring your visiting friends');

• • 我希望找到拥有最多低于平均评分的点评的点评者。**

你认为这是正确的密码吗？

SELECT username, name, COUNT(*)
FROM reviews
CROSS JOIN places
  ON places.id = reviews.place_id
  WHERE rating < average_rating
  GROUP BY username
  ORDER BY COUNT(*) DESC
  LIMIT 1;

我想得到帮助，以防它是错误的。