Question

我的查询是：

SELECT business.name, AVG(rating.value) as total  
FROM business   
INNER JOIN rating   
   ON business.id = rating.business_id   
INNER JOIN directory   
   ON business.id = directory.item_id  
WHERE directory.directory_id = 13    
   AND directory.type = 'business'   
   AND rating.rating_id = '1'   
GROUP BY business.id   
ORDER BY total DESC

我的问题是，我想包括任何商家，即使他们没有与之相关的任何评级？

Answer 1

您需要使用LEFT JOIN代替INNER JOIN：

SELECT business.name, AVG(rating.value) as total  
FROM business   
INNER JOIN directory   
   ON business.id = directory.item_id  
LEFT JOIN rating   
   ON business.id = rating.business_id AND rating.rating_id = '1'
WHERE directory.directory_id = 13    
   AND directory.type = 'business'   
GROUP BY business.id   
ORDER BY total DESC

如果没有评级，则总计将为NULL。您可以在MySQL（如果需要）中处理此问题：

IFNULL(AVG(rating.value), 'No ratings') as total

另请注意，您需要将rating.rating_id = '1'移至JOIN条件。

Answer 2

加入LEFT JOIN - 表时，请执行INNER JOIN而不是rating。

Answer 3

你只需要改变......

INNER JOIN rating

到

LEFT JOIN rating

INNER JOIN返回两个表中都匹配的行。

Answer 4

在加入商家和评分时，将INNER JOIN更改为LEFT JOIN

Answer 5

您想要LEFT OUTER JOIN。

SELECT business.name, AVG(rating.value) as total  
FROM business   
LEFT OUTER JOIN rating   
   ON business.id = rating.business_id   
INNER JOIN directory   
   ON business.id = directory.item_id  
WHERE directory.directory_id = 13    
   AND directory.type = 'business'   
   AND rating.rating_id = '1'   
GROUP BY business.id   
ORDER BY total DESC

mysql查询帮助

5 个答案: