SQL将一天分段分为8点到18点统计没半个小时一分组分成22组

1、原始数据如下：

2、利用分组求和计算出一天内小时与分合计

select  ( case when length( extract(hour  from  cast (oper_date as timestamp)) )<2 then '0'|| extract(hour  from  cast (oper_date as timestamp)) else to_char( extract(hour  from  cast (oper_date as timestamp)) )end

 

   || case when  length( extract(minute  from  cast (oper_date as timestamp)) )<2 then  '0'|| extract(minute  from  cast (oper_date as timestamp)) else to_char( extract(minute  from  cast (oper_date as timestamp))) end
  
                )    aa,count(*) COUNT_P
                from STD_LOG 
                where OPER_DATE >= to_date('2018-08-01','yyyy-MM-dd')
                    and OPER_DATE < to_date('2018-08-02','yyyy-MM-dd')
                group by case when length( extract(hour  from  cast (oper_date as timestamp)))<2 then '0'|| extract(hour  from  cast (oper_date as timestamp)) else to_char(  extract(hour  from  cast (oper_date as timestamp))) end

 

   || case when  length( extract(minute  from  cast (oper_date as timestamp)))<2 then  '0'|| extract(minute  from  cast (oper_date as timestamp)) else to_char(  extract(minute  from  cast (oper_date as timestamp))) end
   order by aa

结果如下：主要将小时与分分别补0；主要使用函数cast与timestamp 将to_date转换为timestamp类型 然后用extract分别取出小时与分钟 然后合并 

3、最后 将小时小于8点的 默认为8点大于18点的默认为18点，将分钟小于30分钟的默认为00大于30分钟的默认为30，然后SUM

           
select （case when cast( substr(aa,0,2) as decimal)<8 then '08' when cast( substr(aa,0,2)as decimal)>18 then '18' else substr(aa,0,2) end  || ':'||

 

case when cast (  substr(aa,3) as decimal)<30 then '00' else '30' end ） hhmm  , sum (COUNT_P)
 
 from  (

 

select  ( case when length( extract(hour  from  cast (oper_date as timestamp)) )<2 then '0'|| extract(hour  from  cast (oper_date as timestamp)) else to_char( extract(hour  from  cast (oper_date as timestamp)) )end

 

   || case when  length( extract(minute  from  cast (oper_date as timestamp)) )<2 then  '0'|| extract(minute  from  cast (oper_date as timestamp)) else to_char( extract(minute  from  cast (oper_date as timestamp))) end
  
                )    aa,count(*) COUNT_P
                from STD_LOG 
                where OPER_DATE >= to_date('2018-08-01','yyyy-MM-dd')
                    and OPER_DATE < to_date('2018-08-02','yyyy-MM-dd')
                group by case when length( extract(hour  from  cast (oper_date as timestamp)))<2 then '0'|| extract(hour  from  cast (oper_date as timestamp)) else to_char(  extract(hour  from  cast (oper_date as timestamp))) end

 

   || case when  length( extract(minute  from  cast (oper_date as timestamp)))<2 then  '0'|| extract(minute  from  cast (oper_date as timestamp)) else to_char(  extract(minute  from  cast (oper_date as timestamp))) end
   order by aa
              
                ) a
  group by case when cast( substr(aa,0,2) as decimal)<8 then '08' when cast( substr(aa,0,2)as decimal)>18 then '18' else substr(aa,0,2) end  ||  ':'||

 

case when cast (  substr(aa,3) as decimal)<30 then '00' else '30' end
               
                order by hhmm
4、最终结果如下：