SQLite-选择列的不同组合的列组

我有一个SQLite db，如下所示：

id | num1 | num2 | value 
---|------|------|------
1  |  2   |  1   | "foo"
2  |  1   |  1   | "foo"
3  |  2   |  4   | "foo"
4  |  2   |  3   | "foo"
5  |  1   |  1   | "foo"
6  |  1   |  1   | "foo"
7  |  1   |  3   | "foo"
8  |  1   |  2   | "foo"

我需要选择以下选项：

所有列 num1和num2的不同组合数 num1/num2组合的秩由num1和num2排序 生成的表应该如下所示：

id | num1 | num2 | value | num | rank
---|------|------|-------|-----|-----
1  |  2   |  1   | "foo" |  6  |   4  
2  |  1   |  1   | "foo" |  6  |   1  
3  |  2   |  4   | "foo" |  6  |   6  
4  |  2   |  3   | "foo" |  6  |   5  
5  |  1   |  1   | "foo" |  6  |   1  
6  |  1   |  1   | "foo" |  6  |   1  
7  |  1   |  3   | "foo" |  6  |   3  
8  |  1   |  2   | "foo" |  6  |   2  

我尝试了很多不同的子查询，但没有得到任何有用的结果

编辑：我通过以下查询获得了组合数so列num：

SELECT
*,
(
SELECT count (*)
FROM ( SELECT DISTINCT num1, num2
FROM table)
)
AS num
FROM table;

---

排名可以通过计算等于或小于当前行的不同行来计算。 如果num1较小，或者如果num1相等，num2较小，则行较小

...,
(SELECT COUNT(*)
 FROM (SELECT DISTINCT num1, num2
       FROM MyTable) AS T2
 WHERE T2.num1 < MyTable.num1
    OR (T2.num1  = MyTable.num1 AND
        T2.num2 <= MyTable.num2)
) AS num,
...