Bash 将日期转换为时间戳
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Bash convert date to timestamp
提问by Lenny
I have a date in format 'YYYYMMDDHHMMSS'and I need to convert it to Unix timestamp.
我有一个格式的日期，'YYYYMMDDHHMMSS'我需要将其转换为 Unix 时间戳。
I tried to date -d '20140826225834'but I get 'invalid date'error. I asume that I would have to convert what I have ( 20140826225834) to accepted date and then convert it to timestamp?
Edit: I have sed this date from 2014-08-21_23.03.07- maybe it would be easier to convert this date type
我尝试过，date -d '20140826225834'但'invalid date'出现错误。我假设我必须将我拥有的 ( 20140826225834) 转换为接受日期，然后将其转换为时间戳？编辑：我已经将这个日期从2014-08-21_23.03.07- 也许转换这个日期类型会更容易
回答by fedorqui 'SO stop harming'
You should probably change the format of the date you get, so that datecan handle it. I change it to a YYYY/MM/DD HH:MM:SSformat with sed.
您可能应该更改您获得的日期的格式，以便date可以处理它。我将其更改为YYYY/MM/DD HH:MM:SS带有sed.
$ date -d"$(sed -r 's#(.{4})(.{2})(.{2})(.{2})(.{2})#// ::#' <<< "20140826225834")" "+%s"
1409086714
By pieces:
按件数：
$ sed -r 's#(.{4})(.{2})(.{2})(.{2})(.{2})#// ::#' <<< "20140826225834"
2014/08/26 22:58:34
$ date -d"2014/08/26 22:58:34"
Tue Aug 26 22:58:34 CEST 2014
$ date -d"2014/08/26 22:58:34" "+%s"
1409086714
回答by hek2mgl
You could use PHP, since PHP's strtotime()can parse your input format:
您可以使用 PHP，因为 PHPstrtotime()可以解析您的输入格式：
#!/bin/bash
input="20140826225834"
output=$(php -r 'echo strtotime("'"$input"'");')
echo "$output" # 1409086714
