Oracle 合并连续时间段

给出几个时间点,合并Oracle连续的时间点。
例如:
09:00-10:00,10:00-12:00,12:00-16:00 合并输出为09:00-16:00
09:00-10:00,10:00-12:00,12:00-16:00 ,17:00-18:00 输出为09:00-16:00 ,17:00-18:00
09:00-10:00,10:00-12:00,12:00-16:00 ,17:00-18:00,18:00-20:00 输出为 09:00-16:00 ,17:00-20:00

如下是使用lag,lead函数实现:
lag(column,N,defValue):取上一条记录
参数:column 取某列的值
N 取几条,默认取1条
defValue:如果没有取到值,返回该默认值。【比如取第一条的上一条,是取不到值的,如果给了该参数,则返回
该值,否则为null】
lead(column,N,defValue):取下一条记录


SQL语句如下:
select start_time,lead(priv,1,end_time) over(order by start_time) end_time from (
select start_time,end_time,(start_time-lag(end_time,1,1)over(order by start_time)) diff,
lag(end_time,1,1)over(order by end_time) priv
from (
select replace(substr(a,1,instr(a,'-',1,1)-1),':','') start_time,
replace(substr(a,instr(a,'-',-1,1)+1),':','') end_time
from (
select '09:00-10:00' a from dual union
select '10:00-11:00' a from dual union
select '11:00-12:00' a from dual union
select '13:00-14:00' a from dual union
select '15:00-16:00' a from dual union
select '17:00-18:00' a from dual union
select '18:00-20:00' a from dual union
select '20:00-21:00' a from dual union
select '21:00-22:00' a from dual union
select '23:00-24:00' a from dual union
select '25:00-26:00' a from dual ---这句是额外添加上,当然值只要不在24小时内就行,为了算法需要
)
)
) where diff<>0 and start_time<>'2500'; --2500是25:00的转换值,这里我们需要去掉'25:00-26:00'这条记录

执行结果如下:
Oracle 合并连续时间段

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