排序GROUP BY agregats

Question

想象一下这个表t1，

+----------+-------+--------+ 
| group_id | name | age | 
+----------+-------+--------+ 
| 1  | A1 | 1 | 
| 1  | A2 | 2 | 
| 1  | A3 | 3 | 
| 2  | B1 | 4 | 
+----------+-------+--------+ 

使用MySQL中下面的查询，

SELECT group_id, name, COUNT(*) FROM t1 GROUP BY group_id 

我们得到，

+----------+-------+--------+----------+ 
| group_id | name | age | COUNT(*) | 
+----------+-------+--------+----------+ 
| 1  | A1 | 2 | 3  | 
| 2  | B1 | 4 | 1  | 
+----------+-------+--------+----------+ 

正如你可以在这里看到，这是可能的值name = A1和age = 2不是来自同一记录。

---

我的问题是，我怎么可以控制单个结果形成name和age列显示，所以内容是从一个记录？有没有办法以某种方式对它们进行排序？例如按相反顺序按年龄排序会给

+----------+-------+--------+----------+ 
| group_id | name | age | COUNT(*) | 
+----------+-------+--------+----------+ 
| 1  | A3 | 3 | 3  | 
| 2  | B1 | 4 | 1  | 
+----------+-------+--------+----------+ 

谢谢。

Answer 1

SELECT group_id, name, COUNT(*) 
FROM t1 
WHERE name IN ('xxx', 'yyy', ..., 'zzz') 
GROUP BY group_id 
SORT BY COUNT(*) 

Answer 2

我不知道完全控制，但你可以这样做

SELECT student_name, MIN(test_score), MAX(test_score) 
     FROM student 
     GROUP BY student_name; 

Answer 3

我不知道你为什么说你的查询工作。你也应该按名称组...

SELECT group_id, name, COUNT(*) FROM t1 GROUP BY group_id, name 

如果你想只是其中之一，请尝试：

SELECT group_id, MIN(name), COUNT(*) FROM t1 GROUP BY group_id