postgreSql 日期减法

select date_part('minute', now() - create_date)  from employee;

 

update employee set issue_efficiency=date_part('days', issue_complete_date - review_pass_date)*3600*24+date_part('hour', issue_complete_date - review_pass_date)*3600+date_part('minute', issue_complete_date - review_pass_date)*60+date_part('second', issue_complete_date - review_pass_date) where id='12345678';
  

update employee set issue_efficiency=EXTRACT(EPOCH FROM (issue_complete_date - review_pass_date)) where issue_complete_date is not null and review_pass_date is not null ;
--获取两个时间相差的秒数