Question

我的数据如下：

HOUR    COUNT
0       3
1       3
2       6
4       2
5       1
7       4
8       3
9       6
10      11  
... 
23      3

我需要数据看起来像这样：

HOUR    COUNT
0       3
1       3
2       6
3       0
4       2
5       1
6       0
7       4
8       3
9       6
10      11
... 
23      3

如果在某个小时（即小时2或6）没有发生任何事件的明确记录，我希望SQL插入小时数为0的小时。

以下是我的代码：

-- Columns selected
SELECT DISTINCT col1 AS 'HOUR', COUNT(col1) AS 'COUNT'

-- Database Used: DB
FROM DB

-- Filters
WHERE date_col BETWEEN '1/1/13' AND '1/2/13'
GROUP BY col1
ORDER BY col1

--*********************************************************************************
-- End Report

Answer 1

如果您没有小时表，则可以使用以下查询。根据您的平台，您可以使用功能代替CASE。

SELECT a.hour, 
       CASE 
         WHEN b.count IS NULL THEN 0 
         ELSE b.count 
       END AS 'count' 
FROM   (SELECT 1 AS 'hour' 
        UNION ALL 
        SELECT 2 
        UNION ALL 
        SELECT 3 
        UNION ALL 
        SELECT 4 
        UNION ALL 
        SELECT 5 
        UNION ALL 
        SELECT 6 
        UNION ALL 
        SELECT 7 
        UNION ALL 
        SELECT 8 
        UNION ALL 
        SELECT 9 
        UNION ALL 
        SELECT 10 
        UNION ALL 
        SELECT 11 
        UNION ALL 
        SELECT 12 
        UNION ALL 
        SELECT 13 
        UNION ALL 
        SELECT 14 
        UNION ALL 
        SELECT 15 
        UNION ALL 
        SELECT 16 
        UNION ALL 
        SELECT 17 
        UNION ALL 
        SELECT 18 
        UNION ALL 
        SELECT 19 
        UNION ALL 
        SELECT 20 
        UNION ALL 
        SELECT 21 
        UNION ALL 
        SELECT 22 
        UNION ALL 
        SELECT 23) a 
       LEFT JOIN tbl b 
              ON b.hour = a.hour

<强>结果

| HOUR | COUNT |
----------------
|    1 |     3 |
|    2 |     6 |
|    3 |     0 |
|    4 |     2 |
|    5 |     1 |
|    6 |     0 |
|    7 |     4 |
|    8 |     3 |
|    9 |     6 |
|   10 |    11 |
|   11 |    10 |
|   12 |    11 |
|   13 |     5 |
|   14 |     7 |
|   15 |    10 |
|   16 |     7 |
|   17 |     6 |
|   18 |     8 |
|   19 |     2 |
|   20 |     7 |
|   21 |     5 |
|   22 |     6 |
|   23 |     3 |

Answer 2

您需要一个案例陈述：

case count(col1)  as 'Count'  
    when null  
    THEN 0  
    ELSE 'Count'

Answer 3

更简单的查询以获得24小时 - 这仅用于构建小时表（如果需要）。使用Sysdate是可选的：

SELECT to_char(trunc(SYSDATE), 'hh24')-1 + LEVEL "24_hrs"
  FROM dual
CONNECT BY LEVEL <= 24;

Answer 4

您可以使用common table expression创建所有小时数：

WITH [Hours]([Hour])
AS
(
  SELECT 0 AS [Hour]
  UNION ALL
  SELECT [Hours].[Hour] + 1 AS [Hour]
  FROM [Hours]
  WHERE [Hours].[Hour] < 23
)

SELECT * FROM [Hours]

然后您可以使用这些时间加入您的查询：

WITH [Hours]([Hour])
AS
(
  SELECT 0 AS [Hour] --Don't forget 0, else you miss one hour on your day!
  UNION ALL
  SELECT [Hours].[Hour] + 1 AS [Hour]
  FROM [Hours]
  WHERE [Hours].[Hour] < 23
)

SELECT 
 [Hours].[Hour],
 COUNT([table].[col1]) AS [Count]
FROM [Hours]
LEFT JOIN [table] ON [Hours].[Hour] = [table].[col1]
WHERE [table].[date_col] BETWEEN '1/1/13' AND '1/2/13'
ORDER BY [Hour] ASC

SQL中不同记录的计数不存在不同记录的情况

4 个答案: