Concatenate date variable to get epoch value in linux

Asked Jun 02 '21 at 20:00

Active Jun 02 '21 at 20:00

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I have got the date from a log file in the format YYYY-mm-dd HH:MM:SS, in a script i am trying to get its epoch value and then print it into another log file:

time_extract=$(tac /home/test.log | grep -m1 "finished" | awk '{print $1 " " $2}')
epoch_last_time=$(date -d $time_extract +%s)
echo $epoch_last_time >> epoch_times.log

But i get nothing.

I executed the command outside the script and i got its epoch value. The date i get from log is 2021-06-02 13:38:00 and the epoch result is 1622659080.

How can i get the value by executing it in a shell?

asked Jun 02 '21 at 20:00

Anell Zheng

1

Always quote variables unless you have a good reason not to. In this case you need to. – Barmar Jun 02 '21 at 20:03
1

```epoch_last_time=$(date -d "$time_extract" +%s)``` – Barmar Jun 02 '21 at 20:07
@Barmar Thanks for the explanation, – Anell Zheng Jun 02 '21 at 20:22
You said it works when you run outside of a script; what does the first line of your shell script look like? What OS distribution are you using? – EricD Jun 02 '21 at 21:24
BTW, `...anything... | grep foo | awk '{ print ... }'` can basically always be replaced with `...anything... | awk '/foo/ { print ... }'`; no reason for grep when you have awk at hand. – Charles Duffy Jun 02 '21 at 21:53
@EricD I am using linux, why it does not worked inside shell script was because the time extracted from log and used to get the epoch was not enclosed by quotes – Anell Zheng Jun 03 '21 at 14:49
@CharlesDuffy thanks for the comment, i will implement it. – Anell Zheng Jun 03 '21 at 14:50

Concatenate date variable to get epoch value in linux

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