获得列x的MAX，同时在列y上获得DISTINCT

Question

我想加入这些表（task_id with post_id），它将返回您在下面看到的所有内容。这很好，但我想接下来要做的是在保留具有最高版本的项目的同时删除task_id的重复项。

我有一个2个表，以便：

项目

ID title task_id project_id  version 
1 "Test"  123   456   1 
2 "Test 2"  124   456   1 
3 "Test 3"  125   456   1 
4 "X 3.1"  125   456   1.1 
5 "X 3.2"  125   456   1.2 

任务

ID post_id meta_key meta_value 
1  123 _completed  0 
4  124 _completed  0 
5  125 _completed  0 

，我有这样的SQL语句至今：

SELECT * 
FROM items t0 
INNER JOIN tasks AS t1 
ON t0.task_id = t1.post_id 
WHERE t1.meta_key = '_completed' 
AND project_id = 456 

返回：

ID title task_id project_id version ID post_id meta_key meta_value 
1 "Test"  123   456   1  1  123 _completed  0 
2 "Test 2"  124   456   1  1  124 _completed  0 
3 "Test 3"  125   456   1  1  125 _completed  0 
4 "X 3.1"  125   456   1.1  1  125 _completed  0 
5 "X 3.2"  125   456   1.2  1  125 _completed  0 

如何删除重复的TASK_ID的，但保持的最高版本，因此该表将是：

ID title task_id project_id version ID post_id meta_key meta_value 
1 "Test"  123   456   1  1  123 _completed  0 
2 "Test 2"  124   456   1  1  124 _completed  0 
5 "X 3.2"  125   456   1.2  1  125 _completed  0 

基本上我最近下令他们BY版本和使用GROUP BY TASK_ID但显然SQL做不让你这样做。

任何人都可以想出一个很好的方法来实现这一目标吗？

Answer 1

根据相关文章，我认为这应该工作。

SELECT * 
FROM items t0 
INNER JOIN tasks AS t1 ON t0.task_id = t1.post_id 
INNER JOIN 
    (SELECT task_id, MAX(version) AS MaxVersion 
    FROM items 
    GROUP BY task_id) groupedt0 
ON t0.task_id = groupedt0.task_id 
AND t0.version = groupedt0.MaxVersion 
AND project_id = 456 
AND t1.meta_key = '_completed' 
ORDER BY ID ASC 

参见：How can I SELECT rows with MAX(Column value), DISTINCT by another column in SQL?