手动将PostgreSQL二进制格式的时间戳转换为整数 [英] Manually convert PostgreSQL binary format of timestamp to integer

问题描述

我读到 PostgreSQL 在内部将时间戳存储为 8 字节/64 位整数.

I've read that PostgreSQL internally stores timestamps as an 8-byte/64-bit integer.

为了检查这一点，我创建了一个表，插入了一些时间戳并查询了 heap_page_items.
输出例如对于时间戳 2019-01-08 09:00:00 是 00 e4 c2 56 ed 21 02 00.

To check this I've created a table, inserted some timestamps and queried the heap_page_items.
The output e.g. for timestamp 2019-01-08 09:00:00 was 00 e4 c2 56 ed 21 02 00.

鉴于此十六进制序列，我如何手动计算自 PostgreSQL 纪元 (2000-01-01) 以来的微秒数?使用时间戳转换器，我得到了值 600253200000 作为结果.我必须如何解包"字节才能接收此值?

Given this hex sequence, how do I manually compute the number of microseconds since PostgreSQL epoch (2000-01-01)? Using a timestamp converter I got value 600253200000 as result. How do I have to "unpack" the bytes to receive this value?

推荐答案

您的分析几乎是完美的，而且您必须在小端机器上.

Your analysis is almost perfect, and you must be on a little-endian machine.

您不必解压该值，因为在您的 64 位架构上，8 字节整数可以放入 PostgreSQL Datum.

You don't have to unpack the value, because on your 64-bit architecture 8-byte integers can fit into a PostgreSQL Datum.

SELECT extract(epoch FROM TIMESTAMP '2019-01-08 09:00:00')
     - extract(epoch FROM TIMESTAMP '2000-01-01 00:00:00');

 ?column?  
-----------
 600253200
(1 row)

600253200 秒是 600253200000000 微秒.

600253200 seconds are 600253200000000 microseconds.

你在表格中看到的整数是0x000221ed56c2e400，对应十进制600253200000000:

The integer you see in the table is 0x000221ed56c2e400, which corresponds to decimal 600253200000000:

bash> printf "%ld\n" 0x000221ed56c2e400
600253200000000

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