Time elapsed since variable

Question

I need time elapsed since $var

$var= [19:11:27 19:34:09]

$var was created from

date "+[%y:%m:%d %T]"

searching here for `[bash] time elapsed` shows 240+ Q/A. Did you look at any of them? Good luck. — shellter, Nov 28 '19 at 05:24
To get elapsed seconds: `elapsed=$(($(date +'%s')-$(date -d '19-11-27 19:34:09' +'%s')))`. To change seconds to day-hour-minute see https://stackoverflow.com/a/27820847/5411198 — builder-7000, Nov 28 '19 at 05:24
@sergio how do i change `date -d '19-11-27 19:34:09' to `date -d' — Hugo Slop, Nov 28 '19 at 05:35
https://stackoverflow.com/questions/59083053/date-utils-date-difference-piping — Hugo Slop, Nov 28 '19 at 06:39
https://stackoverflow.com/questions/59083053/date-utils-date-difference-piping/59083202#59083202 I answered my own question and this is the same issue here. — Hugo Slop, Nov 28 '19 at 07:04

score 1 · Answer 1 · answered Nov 28 '19 at 05:43

1

If mktime function of awk is available, how about:

awk -v var='[19:11:28 14:34:09]' '
BEGIN {
    split(var, a, "[^0-9]")
    elapsed = systime() - mktime(a[2]+2000" "a[3]" "a[4]" "a[5]" "a[6]" "a[7])
    d = int(elapsed / 86400)
    h = int((elapsed - d * 86400) / 3600)
    m = int((elapsed - d * 86400 - h * 3600) / 60) 
    printf("%d days %d hours %d mins passed.\n", d, h, m)
}'

answered Nov 28 '19 at 05:43

tshiono

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How do i use this... i put it in a script..and errors... i see this looks like unix... idk how to use it. Also seems unnecessary... d= h= and m=. – Hugo Slop Nov 28 '19 at 05:49
Please copy the code above and save as a file e.g. `file.awk`, then execute `bash file.awk` on the command line. If you still have an error, please let me know including the error message. BR. – tshiono Nov 28 '19 at 06:46

dash-o · Answer 2 · 2019-11-28T20:57:05.803

0

Few alternative,

If you can change the setting of var:

You are better using the SECONDS bash variable. You can then use bash math to calculate difference, and format the output.

 # Instead of var=...
START=$SECONDS
... do something that take long time
... to executed
FINISH=$SECONDS
ELAPSED=$((FINISH-START))

If you can NOT change the setting of var:

You will have to convert var to a format that date will accept, and use the methods suggested in the comments. Minimal effort to convert to ISO YYYY-MM-DD HH:MM:SS

VX=20${var/:/-}   # Replace first : with -, add 2000 to the yy part
VX=${var/:/-}     # Replace Second : with -
START=$(date -d "$VX" +'%s')
FINISH=(date +'%s')
ELAPSED=$((FINISH-START))

edited Nov 28 '19 at 20:57

answered Nov 28 '19 at 05:28

dash-o

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@HugoSlop The most difficult thing about your question is how to parse the date format you have (`[%y:%m:%d %T]`) into something more useful. If it's possible to *start* with a simpler representation -- like the number of seconds since the shell started (`$SECONDS`), the task gets much easier. Seconds since the unix epoch (`date +%s`) would also work well. – Gordon Davisson Nov 28 '19 at 06:15
1

@HugoSlop I've fixed the formatting, and added logic for case that the value of 'var' can not be modified. – dash-o Nov 28 '19 at 13:33

Time elapsed since variable

2 Answers2