问题描述

我需要一个在 oracle 中执行此序列的 SINGLE 查询.

I need a SINGLE query that does this sequence in oracle.

select count(*) from table1
where request_time < timestamp'2012-05-19 12:00:00' and (end_time > timestamp'2012-05-19 12:00:00' or end_time=null);

select count(*) from table1
where request_time < timestamp'2012-05-19 13:00:00' and (end_time > timestamp'2012-05-19 13:00:00' or end_time=null);

select count(*) from table1
where request_time < timestamp'2012-05-19 14:00:00' and (end_time > timestamp'2012-05-19 14:00:00' or end_time=null);

select count(*) table1
where request_time < timestamp'2012-05-19 15:00:00' and (end_time > timestamp'2012-05-19 15:00:00' or end_time=null);

select count(*) from table1
where request_time < timestamp'2012-05-19 16:00:00' and (end_time > timestamp'2012-05-19 16:00:00' or end_time=null);

如您所见，小时正在一点一点增加.这是输出

As you see the hour is increasing one by one. here is the output

COUNT(*)               
1085                   

<小时>

COUNT(*)               
1233                   

<小时>

COUNT(*)               
1407                   

<小时>

COUNT(*)               
1322                   

<小时>

COUNT(*)               
1237

<小时>

我写了一个查询，但它没有给我正确的答案！

I have written a query but it does not give me the right answer!

select col1, count(*) from
(select TO_CHAR(request_time, 'YYYY-MM-DD HH24') as col1 from table1
 where request_time <= timestamp'2012-05-19 12:00:00' and (end_time >= timestamp'2012-05-19 12:00:00' or end_time=null))
group by col1 order by col1;

这个查询给了我一个结果集，它的 count(*) 的总和等于上面写的第一个查询！结果如下:

this query gives me a result set that sum of it's count(*) is equal to the first query written above! here is the result:

COL1          COUNT(*)               
------------- ---------------------- 
2012-05-19 07      22                     
2012-05-19 08      141                    
2012-05-19 09      322                    
2012-05-19 10      318                    
2012-05-19 11      282  

推荐答案

注意 trunc 表达式与日期值的用法.如果不在 sql*plus 中运行查询，则可以省略 alter session.

Note the usage of trunc expression with date values. You can omit the alter session if you are not running the query in sql*plus.

SQL> alter session set nls_date_format='yyyy-mm-dd hh24:mi:ss';

Session altered.

SQL> SELECT 
       trunc(created,'HH'), 
       count(*) 
     FROM 
       test_table 
     WHERE 
       created > trunc(SYSDATE -2) 
     group by trunc(created,'HH');


TRUNC(CREATED,'HH')   COUNT(*)
------------------- ----------
2012-05-21 09:00:00        748
2012-05-21 16:00:00         24
2012-05-21 17:00:00         12
2012-05-21 22:00:00        737
2012-05-21 23:00:00        182
2012-05-22 20:00:00         16
2012-05-22 21:00:00        293
2012-05-22 22:00:00        610

8 ROWS selected.

这篇关于计算oracle中两个日期之间每小时的记录数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！