问题与SQL添加

Question

请，

我想获得一些用户赢得点的总数值很少有区别的计数，但一切都从同一张表。

SQL：

SELECT o.author_id, SUM(o.vote_value) AS vote_value, n.best AS best_sum, 
       (SUM(o.vote_value) + (10 * n.best)) AS total 
       FROM comments o 
       LEFT JOIN (
        SELECT COUNT(best) AS best, author_id 
        FROM comments 
        WHERE best = 1 
        GROUP BY author_id 
       ) n ON o.author_id = n.author_id 
       GROUP BY o.author_id 
       ORDER BY total DESC 
       LIMIT 0, 4 

问题是，如果没有最好的（默认为0，并且最好是1）我得到了什么 “总” 和 “best_sum”。我需要0，如果用户没有“最好的”，这样的计算可以正确执行

对不起，我的英文不好，我认为你会明白 感谢

Answer 1

我认为你的问题是你筛选出最好不是1的行，所以它们不包含在组中。试试这个：

SELECT o.author_id, SUM(o.vote_value) AS vote_value, n.best AS best_sum, 
      (SUM(o.vote_value) + (10 * n.best)) AS total 
      FROM comments o 
      JOIN (
       SELECT SUM(best) AS best, author_id 
       FROM comments 
       GROUP BY author_id 
      ) n ON o.author_id = n.author_id 
      GROUP BY o.author_id 
      ORDER BY total DESC 
      LIMIT 0, 4 

此外，也许我失去了一些东西，但不会这项工作，保存子查询？

SELECT o.author_id, SUM(o.vote_value) AS vote_value, SUM(o.best) AS best_sum, 
      (SUM(o.vote_value) + (10 * SUM(o.best))) AS total 
      FROM comments o 
      GROUP BY o.author_id 
      ORDER BY total DESC 
      LIMIT 0, 4 

Answer 2

我假设你想有一个0值，如果最好= NULL尝试COALESCE

count(COALESCE(best,0) AS best) 

Answer 3

您应该使用COALESCE（n.best，0）

COALESCE返回第二个参数时，第一个是NULL

SELECT o.author_id, SUM(o.vote_value) AS vote_value, 
      COALESCE(n.best, 0) AS best_sum, 
      (SUM(o.vote_value) + (10 * COALESCE(n.best, 0))) AS total 
      FROM comments o 
      LEFT JOIN (
       SELECT COUNT(best) AS best, author_id 
       FROM comments 
       WHERE best = 1 
       GROUP BY author_id 
      ) n ON o.author_id = n.author_id 
      GROUP BY o.author_id 
      ORDER BY total DESC 
      LIMIT 0, 4