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I am doing some timezone calculations in bash. I'm getting some unexpected values when converting the timezone offset hour output to an integer to do some additional calculations.

Partial script:

offset=$(date +%z) echo "$offset" hours=$(( offset )) echo "$hours"

Output

-0400 -256

Desired Output (I accidentally omitted the need to divide by 100 for the final output)

-0400 -4

I think that the arithmetic is getting evaluated as octal. How can I evaluate the output from date +%z as decimal?

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3 Answers 3

Reset to default
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offset=$(date +%-z) would give an output of -400 in your case.

- after the % symbol removes zero padding.

[1] Relevant answer

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  • -400 hours? ;-) Commented Mar 28, 2018 at 17:58
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Using sed and bc:

date +%z | sed -E 's/^([+-])(..)(..)/scale=2;0\1(\2 + \3\/60)/' | bc

This will give you 2.00 back in the timezone I'm in (+0200).

With strange/unusual timezones:

$ echo '+0245' | sed -E 's/^([+-])(..)(..)/scale=2;0\1(\2 + \3\/60)/' | bc 2.75 $ echo '-0245' | sed -E 's/^([+-])(..)(..)/scale=2;0\1(\2 + \3\/60)/' | bc -2.75

The sed expression will turn the timezone into a "bc script". For the timezone +HHMM, the script will be

scale=2;0+(HH + MM/60)

For -HHMM it will be

scale=2;0-(HH + MM/60)

The zero is in there because my bc does not understand unary +.

If you only ever going to deal with full hour timezones, then you may use

date +%z | sed -E 's/^([+-])(..)../0\1\2/' | bc

which will deal you integers.

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  • bc... that's a new one for me. Commented Mar 28, 2018 at 18:13
  • @datUser Just a standard calculator. Takes expressions on standard input and produces result on standard output. Commented Mar 28, 2018 at 18:17
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If the timezone is a whole number of hours (GNU date):

$ date +%-:::z -4

Otherwise (assume -0427):

$ date +%-:::z | awk -F: '{x=$1;printf("%s%.2f\n",x>=0?"+":"-",(x>=0?x:-x)+$2/60)}' -4.45

Or, for older, more limited date implementations, us plain:

$ date +%z | awk -F '' '{printf("%s%.2f\n",$1,$2$3+($4$5)/60)}' -4.45

This is still not POSIX because a null FS is undefined for POSIX awk.

For POSIX we need to split characters with sed:

$ date +%z |\ sed -E 's/(.)(..)(..)/\1 \2 \3/' |\ awk '{ printf("%s%.2f\n",$1,$2+$3/60) }'
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  • Note that %:::z is specific to GNU date (from gnulib's fprintftime), it's not even in the GNU libc's strftime() (contrary to %-z which is also not standard but supported by GNU strftime()). Commented Mar 29, 2018 at 12:58
  • @StéphaneChazelas Yes, it is a GNU extension. Are you implying that a GNU date answer is invalid? Commented Mar 29, 2018 at 19:21
  • Not at all, it was a "Note" as additional information. I was the one giving you that upvote. Commented Mar 29, 2018 at 19:44
  • @StéphaneChazelas Simpler, older and POSIX compliant solutions added. Commented Mar 29, 2018 at 20:35

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