Question

表：STATUS_TABLE

id | Status  | 
=================
1      true
2      false
3      false
4      true

如何使用Oracle在单独的列中获得True和False的计数？我必须只显示状态为真的记录。所以我把条件放在哪里。但我必须显示两者的数量。像

MARKED_RECORD  UNMARKED_RECORD
      2                2

Answer 1

像

这样的东西

SELECT  SUM(CASE WHEN Status = 'true' THEN 1 ELSE 0 END) MARKED_RECORD,
        SUM(CASE WHEN Status = 'false' THEN 1 ELSE 0 END) UNMARKED_RECORD
FROM    STATUS_TABLE

Answer 2

WITH src1 AS
(
        SELECT 1 id, 'TRUE'  status FROM dual UNION ALL
        SELECT 2 id, 'TRUE'  status FROM dual UNION ALL
        SELECT 3 id, 'FLASE' status FROM dual UNION ALL
        SELECT 4 id, 'TRUE'  status FROM dual UNION ALL
        SELECT 5 id, 'FLASE' status FROM dual UNION ALL
        SELECT 6 id, 'TRUE'  status FROM dual UNION ALL
        SELECT 7 id, 'FLASE' status FROM dual
)
, src AS
(
        SELECT  id
        ,       status
        ,       SUM(DECODE(status, 'TRUE',   1, NULL)) OVER (ORDER BY id) AS is_true
        ,       SUM(DECODE(status, 'FLASE',  1, NULL)) OVER (ORDER BY id) AS is_false
        FROM    src1
)
SELECT  s1.id
,       s1.status
,       s2.m_is_true
,       s2.m_is_false
FROM    src s1
CROSS   JOIN
(
        SELECT  MAX(is_true)  AS m_is_true
        ,       MAX(is_false) AS m_is_false
        FROM    src
) s2
WHERE   s1.status = 'TRUE'
;

结果：

ID       STATUS   M_IS_TRUE     M_IS_FALSE
1        TRUE     4             3        
2        TRUE     4             3        
4        TRUE     4             3        
6        TRUE     4             3

Answer 3

喜欢这个

select decode(status,'true','Marked','false','Unmarked')status, 
       count(status) Count 
from sample_table 
code layoutgroup by status

获取单独列中的记录计数

3 个答案: