如何在不知道日期格式的情况下将日期字符串转换为时间戳 [英] How to convert date strings to timestamp without knowing the date format
问题描述
我正在尝试编写一个查询,以将值插入没有时区数据的时间戳
类型字段。该值来自CSV文件。
I am trying to write a query to insert a value into a timestamp with no timezone data
type field. The value is coming from CSV file.
我正在使用的版本是 PostgreSQL 8.1.21 。
CSV文件上传是由客户端完成的,并且具有日期列。日期有时以 '28 -Sep-13'
出现,有时以 '28 / 09/2013'
格式出现。
The CSV file upload is done by the client and it has a date column. The date sometimes comes as '28-Sep-13'
and sometimes as '28/09/2013'
formats.
我尝试使用以下内容将字符串转换为时间戳:
str_date :: timestamp
。
I tried to use the following to cast the string into timestamp:
str_date::timestamp
.
如果 str_date
类似于 '28 -Sep-13'
,但是如果传入日期的格式为 '28 / 09/2013'
,则不会发生此错误:
This works fine if str_date
is something like '28-Sep-13'
but it won't work if the incoming date has the format '28/09/2013'
, when this error occurs:
ERROR: date/time field value out of range: "28/09/2013"
HINT: Perhaps you need a different "datestyle" setting
基本上,客户保留更改上传的CSV文件中的日期格式。
是否可以根据其实际格式将日期字符串转换为时间戳?
Basically the client keeps changing the date format in the uploaded CSV file.
Is there a way to convert the date strings into timestamp depending on its actual format?
推荐答案
您需要将日期样式设置为 ISO,DMY。默认情况下将其设置为 ISO,MDY,这会导致您的示例失败:
You need to set your datestyle to "ISO, DMY". It is set to "ISO, MDY" by default, and would cause your example to fail:
> show datestyle;
DateStyle
-----------
ISO, MDY
(1 row)
> select '28-Sep-13'::date;
date
------------
2013-09-28
(1 row)
> select '28/09/2013'::date;
ERROR: date/time field value out of range: "28/09/2013"
LINE 1: select '28/09/2013'::date;
^
HINT: Perhaps you need a different "datestyle" setting.
> set datestyle = 'ISO, DMY';
SET
> select '28-Sep-13'::date;
date
------------
2013-09-28
(1 row)
> select '28/09/2013'::date;
date
------------
2013-09-28
(1 row)
(示例在PostgreSQL 9.1中完成,但DateStyle设置和相关行为是古老的,因此应该可以正常工作)
(examples done in PostgreSQL 9.1, but the DateStyle setting and associated behavior are ancient, so should work fine)
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