在 UNIX 时间戳 Shell/Bash 中使用时区转换日期 [英] Converting date with timezone in UNIX timestamp Shell/Bash

问题描述

我需要将格式为yyyy/mm/dd hh:mm:ss TZ"的字符串日期转换为 UNIX 时间(TZ = 时区).

I need to convert a date from string in the format "yyyy/mm/dd hh:mm:ss TZ" to UNIX time (TZ = Timezone).

到目前为止我所做的是使用

What I have done so far is to convert a date in the format "yyyy/mm/dd hh:mm:ss" without a timezone to timestamp by using

dateYMD="2019/2/28 12:23:11.46"
newt=$(date -d "${dateYMD}" +"%s")
echo ${newt}

我有以下结果.

1551349391

我的努力是找到如何将时区和日期/时间转换为时间戳(unix 时间).例如，我需要 4 个与 dateYMD 具有相同日期/时间但位于 4 个不同时区的变量，以便它们的时间戳不同.

My struggle is to find how both timezone and date/time can be converted to timestamp (unix time) . For example I need 4 variables with the same date/time as dateYMD but in 4 different timezones so that their timestamps would be different.

这是我尝试过的最新版本

Here is the latest I have tried

dateYMD="2017/09/09 08:58:09"
timez=$(TZ=Australia/Sydney date -d @$(date +%s -d "${dateYMD}"))
unixTimez=$( date --date "${timez}" +"%s" )
echo ${unixTimez}

这向我展示了以下错误

   date: invalid date ‘чт фев 28 21:23:11 AEDT 2019’

推荐答案

您不需要两次调用 date.只需将 TZ 设置为该变量所需的时区即可调用一次.

You don't need to call date twice. Just call it once with TZ set to the timezone you want for that variable.

timesydney=$(TZ=Australia/Sydney date -d "$dateYMD" +%s)
timenyc=$(TZ=US/Eastern date -d "$dateYMD" +%s)

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