Sql日期到小时数（总计）

我有一个疑问：

   SELECT
     NUMTODSINTERVAL(
     SUM( TO_DATE( MT.TI_CONTR, 'HH24:MI' ) - TO_DATE( '00:00', 'HH24:MI' ) ),
     'DAY'
     ) AS total
   FROM MYTABLE MT;

执行此查询会得到以下结果：

+22 19:02:00.000000
+94 19:26:00.000000
+46 03:50:00.000000
+76 08:30:00.000000
+44 02:42:00.000000

当然，这是以天为单位的分组，一旦达到24小时

TI_CONTR列是一个varchar，以以下格式存储小时和分钟：hh:mm（例如“05:22”）

如何获得总小时数的结果（ex 252:20）

---

感谢

Oracle不允许您对间隔数据类型执行SUM（），因此最好使用老式的SUBSTR（）和数学来解决这一问题

with dat as (SELECT '19:02' t1_contr from dual 
             union all
             SELECT '19:26' t1_contr from dual 
             union all
             SELECT '03:50' t1_contr from dual 
             union all
             SELECT '08:30' t1_contr from dual 
             union all
             SELECT '02:42' t1_contr from dual 
             )
select to_char(sum(substr(t1_contr,1,2)) --sum the hours, then add 
             + trunc(sum(substr(t1_contr,4,2))/60)) --the hours portion of the summed minutes
             ||':'|| -- put in your separator
             to_char( mod(sum(substr(t1_contr,4,2)),60)) --and append the summed minutes after removing the hours
from dat   

---

Oracle不允许您对区间数据类型执行SUM（），因此最好使用老式的SUBSTR（）和数学来解决这个问题

with dat as (SELECT '19:02' t1_contr from dual 
             union all
             SELECT '19:26' t1_contr from dual 
             union all
             SELECT '03:50' t1_contr from dual 
             union all
             SELECT '08:30' t1_contr from dual 
             union all
             SELECT '02:42' t1_contr from dual 
             )
select to_char(sum(substr(t1_contr,1,2)) --sum the hours, then add 
             + trunc(sum(substr(t1_contr,4,2))/60)) --the hours portion of the summed minutes
             ||':'|| -- put in your separator
             to_char( mod(sum(substr(t1_contr,4,2)),60)) --and append the summed minutes after removing the hours
from dat