blocks|key|1538851|text|可能最简单的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1538852|SELECT+EXTRACT(epoch+FROM+my_interval)/3600|code-block|syntax|javascript|1538853|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Probably the easiest way is:

<pre><code>SELECT EXTRACT(epoch FROM my_interval)/3600
</code></pre>

blocks|key|1538877|text|如果您想要整数，即天数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1538878|SELECT+(EXTRACT(epoch+FROM+(SELECT+(NOW()+-+'2014-08-02+08:10:56')))/86400)::int|code-block|syntax|javascript|1538879|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you want integer i.e. number of days:

<pre><code>SELECT (EXTRACT(epoch FROM (SELECT (NOW() - '2014-08-02 08:10:56')))/86400)::int
</code></pre>

blocks|key|1538916|text|+++++++++select+date+'now()'+-+date+'1955-12-15';|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1538917|下面是计算总天数的简单查询。|unstyled|1538918|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code> select date 'now()' - date '1955-12-15';
</code></pre>

Here is the simple query which calculates total no of days.

blocks|key|1538954|text|要获得天数，最简单的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1538955|SELECT+EXTRACT(DAY+FROM+NOW()+-+'2014-08-02+08:10:56');|code-block|syntax|javascript|1538956|据我所知，它将返回与以下相同的结果：|1538957|SELECT+(EXTRACT(epoch+FROM+(SELECT+(NOW()+-+'2014-08-02+08:10:56')))/86400)::int;|1538958|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

To get the number of days the easiest way would be:

<pre><code>SELECT EXTRACT(DAY FROM NOW() - '2014-08-02 08:10:56');
</code></pre>

As far as I know it would return the same as:

<pre><code>SELECT (EXTRACT(epoch FROM (SELECT (NOW() - '2014-08-02 08:10:56')))/86400)::int;
</code></pre>

blocks|key|1323773|text|select+floor((date_part('epoch',+order_time+-+'2016-09-05+00:00:00')+/+3600)),+count(*)
from+od_a_week
group+by+floor((date_part('epoch',+order_time+-+'2016-09-05+00:00:00')+/+3600));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1323774|::int转换遵循四舍五入原则。如果你想要一个不同的结果，比如向下舍入，你可以使用相应的数学函数，比如floor。|unstyled|offset|length|style|CODE|1323775|entityMap^0|0|0|5|1F|5|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]|$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|S|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>select floor((date_part('epoch', order_time - '2016-09-05 00:00:00') / 3600)), count(*)
from od_a_week
group by floor((date_part('epoch', order_time - '2016-09-05 00:00:00') / 3600));
</code></pre>

The <code>::int</code> conversion follows the principle of rounding.
If you want a different result such as rounding down, you can use the corresponding math function such as <code>floor</code>.

blocks|key|1323795|text|我正在使用PostgreSQL+11，我创建了一个函数来获取两个不同时间戳之间的小时数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1323796|create+function+analysis.calcHours(datetime1+timestamp,+datetime2+timestamp)
++++returns+integer
++++language+plpgsql+as+$$
++++declare
++++++++diff+interval;
++++begin
++++++++diff+=+datetime2+-+datetime1;
++++++++return+(abs(extract(days+from+diff))*24+%2B+abs(extract(hours+from+diff)))::integer;
++++end;+$$;|code-block|syntax|javascript|1323797|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I'm working with PostgreSQL 11, and I created a function to get the hours betweeen 2 differents timestamps
<pre><code>create function analysis.calcHours(datetime1 timestamp, datetime2 timestamp)
 returns integer
 language plpgsql as $$
 declare
 diff interval;
 begin
 diff = datetime2 - datetime1;
 return (abs(extract(days from diff))*24 + abs(extract(hours from diff)))::integer;
 end; $$;
</code></pre>

Say I have an interval like

<pre><code>4 days 10:00:00
</code></pre>

in postgres. How do I convert that to a number of hours (106 in this case?) Is there a function or should I bite the bullet and do something like

<pre><code>extract(days, my_interval) * 24 + extract(hours, my_interval)
</code></pre>

How do I convert an interval into a number of hours with postgres?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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假设我有一个间隔，比如4 days 10:00:00在postgres。如何将其转换为小时数(在本例中为106？)有没有一个函数，或者我应该咬紧牙关做一些类似这样的事情extract(days, my_interval) * 24 + extract(hours, my_interval)

问如何使用postgres将时间间隔转换为小时数？
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问如何使用postgres将时间间隔转换为小时数？EN