Question

我目前有一个类似于以下内容的查询

SELECT 
    p.*, total
FROM 
    scripts p
LEFT JOIN   
    (SELECT script_id, COUNT(*) as total FROM creator_log WHERE created_date > timestampadd(hour, -6, now()) GROUP BY script_id) p2
ON 
    p.script_id = p2.script_id
WHERE 
    (p2.total < 1 OR p2.total IS NULL) 
AND 
    p.script_status = '1'
ORDER BY 
    p.script_id DESC
LIMIT 
    1

然后我如何根据返回的count() script_id products从另一个名为p.script_id的表中找到count，然后确保SELECT p.*, total, active_total FROM scripts p LEFT JOIN (SELECT script_id, COUNT(*) as total FROM creator_log WHERE created_date > timestampadd(hour, -6, now()) GROUP BY script_id) p2 ON p.script_id = p2.script_id RIGHT JOIN (SELECT script_id, COUNT(*) as active_total FROM products WHERE last_used < timestampadd(hour, -6, now()) AND status = 1) p3 ON p.script_id = p3.script_id WHERE (p2.total < 1 OR p2.total IS NULL) AND (p3.active_total < 5 or p3.active_total IS NULL) AND p.script_status = '1' ORDER BY p.script_id DESC LIMIT 1 }小于5？

我认为我应该做某种正确的加入？

编辑我尝试了以下无效

{{1}}

Answer 1

不需要像这样进行内连接或完全连接（不是右或左）：

SELECT p.*, total
FROM scripts p
LEFT JOIN   
    (SELECT script_id, COUNT(*) as total 
     FROM creator_log 
     WHERE created_date > timestampadd(hour, -6, now()) 
     GROUP BY script_id) p2
   ON p.script_id = p2.script_id AND p2.total < 1
JOIN 
   (SELECT script_id, COUNT(*) as active_total 
    FROM products
    WHERE last_used < timestampadd(hour, -6, now()) AND status = 1) p3
  ON p.script_id = p3.script_id AND p3.active_total < 5
WHERE p.script_status = '1'
ORDER BY p.script_id DESC
LIMIT 1

注意，您可以向ON子句添加多个要求，从而使您的where子句更简单，更容易理解。

这是如何工作的：因为p3上的连接是两个脚本中的完整连接元素，并且p3必须存在。这使得p3中的元素如您所描述的那样“必需”。

mySQL中的LEFT和RIGHT JOIN

1 个答案: