无法弄清mysql查询

Question

我是PHP和MySQL的新手->我还不好。 今天我遇到了问题。 我有一个将2个表“ votes”和“ stories”连接起来的查询，它是：

SELECT stories.*, SUM(votes.vote_value) as 'total_votes' 
FROM stories JOIN votes ON stories.id = votes.item_name WHERE stories.st_date >= DATE_SUB(NOW(), INTERVAL 32 DAY) 
GROUP BY stories.id 
ORDER BY total_votes ASC LIMIT 10

我需要对其进行修改，因此它仅从“故事”表中选择信息，其中显示字段= 1

一个简单的查询如下所示：

SELECT * FROM stories WHERE showing = 1

但是我不知道如何在连接两个数据库的第一个查询中实现它。

Answer 1

SELECT stories.*, SUM(votes.vote_value) as 'total_votes' 
FROM stories JOIN votes ON stories.id = votes.item_name $date 
WHERE showing=1
GROUP BY stories.id 
ORDER BY total_votes ASC LIMIT 10

只需将where粘贴。 但是，请确保以正确的顺序放置它。

Answer 2

SELECT stories.*, SUM(votes.vote_value) as 'total_votes' 
FROM stories, votes 
WHERE stories.id = votes.item_name AND stories.showing = 1 AND stories.st_date >= DATE_SUB(NOW(), INTERVAL 32 DAY)
GROUP BY stories.id 
ORDER BY total_votes ASC LIMIT 10

Answer 3

只需添加它就没有问题，因为它是联接的基表：

SELECT stories.*, SUM(votes.vote_value) as 'total_votes' 
FROM stories JOIN votes ON stories.id = votes.item_name $date 
WHERE stories.showing = 1
GROUP BY stories.id 
ORDER BY total_votes ASC LIMIT 10

Answer 4

将您的查询更改为

SELECT stories.*, SUM(votes.vote_value) as 'total_votes' FROM stories 
JOIN votes ON stories.id = votes.item_name 
WHERE stories.showing = 1 
GROUP BY stories.id 
ORDER BY total_votes ASC LIMIT 10

Answer 5

使用HAVING显示= 1，它应该可以解决您的问题。