解析带时区的时间格式在Presto SQL中

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英文:

Parsing time format with zone in presto sql

问题

如何将以下时间格式解析为Presto SQL中的时间戳,更具体地说是AWS Athena?

Mon Feb 27 2023 06:21:22 GMT+0530 (India Standard Time)

Mon Feb 27 2023 00:37:44 GMT-0500 (Eastern Standard Time),等等

我尝试了这个:

select date_parse(
    'Mon Feb 27 2023 06:21:22 GMT+0530 (India Standard Time)',
    '%a %b %d %Y %H:%i:%s GMT%z (%Z)')

但出现了以下错误:

INVALID_FUNCTION_ARGUMENT: Invalid format: "Mon Feb 27 2023 06:21:22 GMT+0530 (India Standard Time)" is malformed at "+0530 (India Standard Time)"

我不确定Presto SQL中是否存在%z%Z指示符。

请注意,时区可能是任何时区,而不仅仅是GMT+0530,所以我不能忽略时区。

英文:

How to parse the following time format to timestamp in presto sql, more specifically AWS Athena?

Mon Feb 27 2023 06:21:22 GMT+0530 (India Standard Time),

Mon Feb 27 2023 00:37:44 GMT-0500 (Eastern Standard Time), etc

I tried this,

select date_parse(
    'Mon Feb 27 2023 06:21:22 GMT+0530 (India Standard Time)',
    '%a %b %d %Y %H:%i:%s GMT%z (%Z)')

but got the following error

> INVALID_FUNCTION_ARGUMENT: Invalid format: "Mon Feb 27 2023 06:21:22 GMT+0530 (India Standard Time)" is malformed at "+0530 (India Standard Time)"

I'm not sure if the %z or %Z specifier is present in Presto sql.

Note that the timezone may be anything, not just GMT+0530, so i can't ignore the timezone.

答案1

得分: 1

首先,就我所见,印度标准时间东部标准时间似乎不在Athena支持的时区列表中,建议从字符串中移除此部分,然后使用支持JodaTime的parse_datetime进行转换:

-- 示例数据
with dataset(dt) as (
    values ('Mon Feb 27 2023 06:21:22 GMT+0530 (India Standard Time)'),
           ('Mon Feb 27 2023 00:37:44 GMT-0500 (Eastern Standard Time)')
)

-- 查询
SELECT
    PARSE_DATETIME(regexp_replace(dt, '\s*\(.*\)'), 'E MMM dd yyyy HH:mm:ss ''GMT''Z')
FROM dataset;

输出:

             _col0
--------------------------------
 2023-02-27 06:21:22.000 +05:30
 2023-02-27 00:37:44.000 -05:00
英文:

First of all as far as I can see India Standard Time and Eastern Standard Time are not in the list of timezones supported by Athena, I suggest to remove this part from the string and then switch to parse_datetime which supports JodaTime’s DateTimeFormat:

-- sample data
with dataset(dt) as (
    values ('Mon Feb 27 2023 06:21:22 GMT+0530 (India Standard Time)'),
           ('Mon Feb 27 2023 00:37:44 GMT-0500 (Eastern Standard Time)')
)

-- query
SELECT
    PARSE_DATETIME(regexp_replace(dt, '\s*\(.*\)'), 'E MMM dd yyyy HH:mm:ss ''GMT''Z')
FROM dataset;

Output:

             _col0
--------------------------------
 2023-02-27 06:21:22.000 +05:30
 2023-02-27 00:37:44.000 -05:00

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  • 本文由 发表于 2023年7月20日 14:02:48
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