Question

我有这个问题：

SELECT points.subject_id, SUM(points_in_round) 
AS points_in_round 
FROM points, subjects 
WHERE subjects.subject_id = points.subject_id 
AND subjects.subject_kind_id = 'businessman' 
AND points.subject_id IN
(SELECT subject_id FROM residents 
WHERE economic_id = 'USA' AND round_number = 1)
GROUP BY points.subject_id 
ORDER BY points_in_round DESC

我想用EXIST替换IN clausule。我重写了这样的查询：

SELECT points.subject_id, SUM(points_in_round) 
AS points_in_round 
FROM points, subjects 
WHERE subjects.subject_id = points.subject_id 
AND subjects.subject_kind_id = 'businessman' 
AND EXISTS
(SELECT * FROM residents 
WHERE economic_id = 'USA' AND round_number = 1
AND points.subject_id = residents.subject_id)
GROUP BY points.subject_id 
ORDER BY points_in_round DESC

我得到相同的结果，但执行计划表示已编辑查询的查询成本为3100，而原始查询的查询成本为290.

我哪里错了？

谢谢大家

Answer 1

我们不知道您的表上存在哪些索引。请尝试以下查询，并根据需要添加适当的索引：

SELECT points.subject_id, SUM(points_in_round) AS points_in_round 
FROM points
LEFT JOIN subjects 
  ON subjects.subject_id = points.subject_id 
LEFT JOIN residents
  ON residents.subject_id = points.subject_id
WHERE subjects.subject_kind_id = 'businessman' 
  AND economic_id = 'USA' 
  AND round_number = 1
GROUP BY points.subject_id 
ORDER BY points_in_round DESC

MySQL用EXISTS替换IN，编辑查询要贵得多

1 个答案: