Read and convert date from file

Question

I'm using a CentOS and I want to write a shell script. So

I have a file with a date:

> 
> cat VM1_EOMAP_TIME.log
07 Sep 2022 16:30
>

And I want to get the minutes since current time and the date in the file

My idea is: ("EPOCH from current time" - "EPOCH from the date of that file") / 60

But I can't get the "EPOCH from current time"

> cat VM1_EOMAP_TIME.log
07 Sep 2022 16:30
 
> date --date='07 Sep 2022 16:30' +%s
1662568200
 
> date --date=$(cat VM1_EOMAP_TIME.log) +%s  
date: extra operand ‘2022’
Try 'date --help' for more information.

> date --date=`cat VM1_EOMAP_TIME.log` +%s
date: extra operand ‘2022’
Try 'date --help' for more information.

> TTT="07 Sep 2022 16:30"
> echo $TTT
07 Sep 2022 16:30

> date --date=$TTT +%s
date: extra operand ‘2022’
Try 'date --help' for more information.

Why I get "date: extra operand ‘2022’" error?

You need to quote your variables. `--date=$TTT` expands to `--date=07 Sep ...` so you have extra arguments. Try `--date="$TTT"`. — doneal24, Sep 07 '22 at 16:55
See [Why does my shell script choke on whitespace or other special characters?](https://unix.stackexchange.com/questions/131766/why-does-my-shell-script-choke-on-whitespace-or-other-special-characters) — steeldriver, Sep 07 '22 at 16:58

score 1 · Answer 1 · answered Sep 07 '22 at 17:05

1

If your bash is recent, the current time in epoch seconds is builtin: $EPOCHSECONDS

Otherwise, bash's printf can get it with the %(fmt)T directive and magic value "-1":

printf -v current_epoch '%(%s)T' -1

Also, bash has a builtin convenience for $(cat file) => $(<file):

log_epoch=$( date -d "$(<VM1_EOMAP_TIME.log)" "+%s" )

answered Sep 07 '22 at 17:05

glenn jackman

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And GNU `date` has `-f` to read from a file... – Kusalananda Sep 08 '22 at 10:03

score 1 · Answer 2 · answered Sep 07 '22 at 19:53

1

Your problem is quoting. And that you should use date -u to compensate for timezone.

$ echo "07 Sep 2022 16:30" > infile
$ var="$(<infile)"
$ cat infile; echo "$var"
07 Sep 2022 16:30
07 Sep 2022 16:30

$ date -ud "$(<infile)" ;   date -ud "$var" +%s
1662568200
1662568200

$ varlog=$(date -ud "$var" +%s)    #capture value for later

There are other simpler solutions in bash:

$ echo "$EPOCHSECONDS"; printf '%(%s)T' '-1'
1662582600
1662582600

$ var1="$EPOCHSECONDS"; printf -v var2 '%(%s)T' '-1'
$ echo "$var1 $var2"
1662568200 1662568200

So, the final math could be:

$ echo $((var2 - varlog))
10747

$ echo $(((var2-varlog)/60))
179

QuartzCristal

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3
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Maybe the times are wall clock time, so the timezone is implied as that of the caller. Using `-u` might break that – roaima Sep 07 '22 at 20:27
Well, the OP calculated the epoch to be `1662568200`. If you reverse that `date -ud @1662568200 +'%F %T %Z%z'` you get back `2022-09-07 16:30:00 UTC+0000`. Then, his computer is using `UTC` time (to match the time inside the file). But if his local time is New_York (for example) he would get: `TZ="America/New_York" date -d @1662568200 +'%F %T %Z%z'` --> `2022-09-07 12:30:00 EDT-0400`. @roaima – QuartzCristal Sep 07 '22 at 21:23
So, it is better to use UTC times, or add a timezone to the value given to avoid guessing. @roaima – QuartzCristal Sep 07 '22 at 21:29
If the OP is converting based on UTC then UTC it should be. Good spot! – roaima Sep 07 '22 at 22:30
1

See also `date -f`, if using GNU `date`. – Kusalananda Sep 08 '22 at 10:02

score 1 · Answer 3 · answered Sep 08 '22 at 10:27

1

$ let diff=($(date +%s -d 'today')-$(date +%s -f VM1_EOMAP_TIME.log))/60
$ echo "$diff"

answered Sep 08 '22 at 10:27

sseLtaH

2,706
1
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[Shellcheck](https://shellcheck.net) warns [Instead of `let expr`, prefer `(( expr ))`](https://www.shellcheck.net/wiki/SC2219). – glenn jackman Sep 08 '22 at 13:44

Read and convert date from file

3 Answers3