Postgres sql时区

Question

我尝试返回特定时区的值，并且我在回应时遇到了一些奇怪的行为：

SELECT created_at AT TIME ZONE 'US/Pacific' - created_at,
       NOW() AT TIME ZONE 'US/Pacific' - NOW(),
       NOW() now  FROM "my_table" 
ORDER by id DESC LIMIT 1

返回

10:00:00 | -10:00:00 | 2017-08-02 17:36:30.660477+03

为什么同一函数会从表中返回相反的值（-10 / + 10）以及动态时间？

谢谢！

Answer 1

类型timestamp without timezone被解释为“本地时区” 因此，您的created_at AT TIME ZONE 'US/Pacific'值将被视为给定时区的时间戳。

NOW()返回timestamp with timezone，因此您的NOW() AT TIME ZONE 'US/Pacific'值必须转换为其他时区。

参见文档：

• 8.5.3 Time Zones
• 9.9.3 AT TIME ZONE

修改

试试这个：

SELECT created_at AT TIME ZONE 'US/Pacific' AS crated_at_us_pacific, created_at,
       NOW() AT TIME ZONE 'US/Pacific' AS now_us_pacific, NOW() AS now
   FROM "my_table"
   ORDER by id DESC
   LIMIT 1

Answer 2

感谢Usagi Miyamoto的一些澄清，它让我走上正轨。

如果没有时区AT TIME ZONE的字段时间戳定义了特定的时区。所以首先声明localtime然后将其转换为特定的时区：所以修复的SQL是：

SELECT created_at AT TIME ZONE 'localtime' AT TIME ZONE 'US/Pacific' - created_at,
       NOW() AT TIME ZONE 'US/Pacific' - NOW(),
       NOW() now  FROM "my_table" 
ORDER by id DESC LIMIT 1

修改

更动态

SELECT created_at::timestamptz AT TIME ZONE 'US/Pacific' - created_at,
       NOW() AT TIME ZONE 'US/Pacific' - NOW(),
       NOW() now  FROM "my_table" 
ORDER by id DESC LIMIT 1

首先timestamptz将时间转换为当地时区，然后转换为您需要的时区AT TIME ZONE 'US/Pacific'