postgresql 计算时间差的秒数、天数实例

处理时间时用到了，记录一下。

时间差天数

select '2017-12-10'::date - '2017-12-01'::date;

时间差秒数

select extract(epoch FROM (now() - (now()-interval '1 day') )); 
select trunc(extract(epoch FROM (now() - (now()-interval '1 day') ))::numeric); 
select trunc(extract(epoch FROM (now() - (now()-interval '1 day') ))::numeric,1); 
select round(extract(epoch FROM (now() - (now()-interval '1 day') ))::numeric); 
select round(extract(epoch FROM (now() - (now()-interval '1 day') ))::numeric,1);

补充：postgresql计算2个日期之间工作日天数的方法

select date_part( 'day', minus_weekend(begin_date,end_date)) from table1 where name in ('a', 'b', 'c')

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postgresql 计算时间差的秒数、天数实例

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