Question

我是PHP和MySQL的新手 - ＆gt;我还不好。今天我遇到了这个问题。我有一个查询加入2个表'投票'和'故事' 这是：

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，以便仅从“故事”表中选择信息，其中字段显示= 1

一个简单的查询将如下所示：

SELECT * FROM stories WHERE showing = 1

但是我不知道如何在我加入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贴在那里。请务必按正确顺序放置。

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

添加它没有问题，因为它是连接的基表：

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。

无法弄清楚mysql查询

5 个答案: