如何只计算两个日期之间的日期在postgres sql查询中。
假设我给了两个日期，如果相差一个月，那么它应该显示为30天。即使是几个月，也需要将我尝试过使用age（date，now（）:: timestamp的天数
转换为天数没有时区），但给出的是月-日期-年格式。例如：10周一24天13：57：40.5265。但是我需要遵循以下方式。
例如，如果相差两个月，那么我需要60天的输出。
Suppose I have given two dates, if difference is one month then it should be display as 30 days.Even months also need to convert into days I have tried with age( date,now()::timestamp without time zone) but it is giving months-dates-years format. ex: 10 mons 24 days 13:57:40.5265.But I need following way. For example if two months difference is there then I need output as 60 days.
不要使用 age（）函数用于日期/时间算术。它仅返回符号结果（足够好了用于人类表示，但对于日期/时间计算几乎没有意义；与标准差相比）。
Don't use the age() function for date/time arithmetic. It only returns "symbolic" results (which are good enough for human representation, but almost meaningless for date/time calculations; compared to the standard difference).
标准差异运算符（-）返回两者的基于天的结果日期，时间戳和带时区的时间戳（前者返回天为 int ，后两个返回天为基础的 interval s）：
The standard difference operator (-) returns day-based results for both date, timestamp and timestamp with time zone (the former returns days as int, the latter two return day-based intervals):
从基于天的间隔中，您可以使用 extract（）函数提取日期：
From the day-based intervals you can extract days with the extract() function:
select current_date - '2017-01-01',
extract(day from now()::timestamp - '2017-01-01 00:00:00'),
extract(day from now() - '2017-01-01 00:00:00Z');