如何計數並按多列組合進行分組？

Question

對於表，例如：

foo_table 

id | str_col | bool_col 

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

我知道如何查詢兩個：

count(*) | bool_col 
    3   0 
    3   1 

和

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

，但我怎麼能得到的東西，如：

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

在此期間，我只是個別做：

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

嘗試

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 

輸出（MySQL的）：

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

SQLFiddle MySQL的

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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中存在的，在功能構建稱爲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會給你所有的組合。還有rollup這是一個私人案件cube。