sql - 如何按多列的组合进行计数和分组？

Question

对于如下表：

foo_table

id | str_col | bool_col

 1   "1234"     0
 2   "3215"     0
 3   "8132"     1
 4   NULL       1
 5   ""         1
 6   ""         0

我知道如何查询：

count(*) | bool_col
   3          0
   3          1

和

count(*) | isnull(str_col) or str_col = ""
   3          0
   3          1

但我怎么能得到类似的东西：

count(*) | bool_col | isnull(str_col) or str_col = ""
   2          0           0
   1          0           1
   1          1           0
   2          1           1

与此同时，我只是单独做：

select count(*) from foo_table where bool_col and (isnull(str_col) or str_col = "");
select count(*) from foo_table where not bool_col and (isnull(str_col) or str_col = "");
select count(*) from foo_table where bool_col and not (isnull(str_col) or str_col = "");
select count(*) from foo_table where not bool_col and not (isnull(str_col) or str_col = "");

Answer 1

尝试

SELECT COUNT(*), 
       bool_col, 
       CASE WHEN str_col IS NULL OR str_col = '' THEN 1 ELSE 0 END str_col
  FROM foo_table
 GROUP BY bool_col, 
       CASE WHEN str_col IS NULL OR str_col = '' THEN 1 ELSE 0 END

输出（MySql）：

| COUNT(*) | BOOL_COL | STR_COL |
---------------------------------
|        2 |        0 |       0 |
|        1 |        0 |       1 |
|        1 |        1 |       0 |
|        2 |        1 |       1 |

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Answer 2

SELECT COUNT(CASE 
 WHEN bool_col AND (isnull(str_col) or str_col = "")  THEN 1 
END) as c1,
COUNT(CASE 
 WHEN not bool_col and (isnull(str_col) or str_col = "") THEN 1
END) as c2,
COUNT(CASE 
 WHEN bool_col and not (isnull(str_col) or str_col = "") THEN 1
END) as c3,
COUNT(CASE 
 WHEN not bool_col and not (isnull(str_col) or str_col = "") THEN 1
END) as c4
FROM table1

Answer 3

在 oracle 中有一个内置函数用于调用cube

select  bool_col , 
        case when str_col is null or str_col = '' then 1 else 0 end str_col ,
        count(*)
from    table1
group by cube (bool_col , case when str_col is null or str_col = '' then 1 else 0 end)

cube会给你所有的组合。还有rollup一个是私人案例cube。