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I have a script which renders timestamps in the format YYYY'DDD'TTT where Y = year, D = day out of 365 and T = time in 1000th's of the day:

#!/bin/bash
clear
s=$(($(date +"%H*3600+%M*60+%S")))
t=$(($s * 5 / 432))
d=$(date +%j)
y=$(date +%Y)
printf "$y\`$d\`$t" "$y" "$d" "$t"

but I need to modify it so that D and T will always produce a three-character value (ie, '001' instead of '1' for Jan 1, '001' instead of '1' for the first minute of the day) - and I have no idea how.

Any help is hugely appreciated

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2 Answers

up vote 4 down vote

Perhaps you could take a look at the printf format string. The first parameter of printf should be a format string which includes placeholders for each of the arguments that follow.

You can include %d to represent an argument in signed decimal format, and you can prefix d by 0n for n characters of zero padding.

printf "%d/%d/%d" 2011 2 3

Will output 2011/2/3

printf "%04d/%03d/%05d" 2011 2 45

Should output 2011/002/00045

Hope this helps

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Thanks! That does help, but my real problem is that I can't just insert additional zeros, since after the first 99 values (days, ect), the value gains a third character on it's own. My script needs to pad values < 100 with one zero, and values < 10 with two zeros, if that makes sense... – Jack Anders Sep 29 '11 at 15:19
1  
That is what the printf "%03d" is doing. The '3' is 'minimum' width and the '0' means to pad with a zero character ('0') to fill the minimum width. See the examples that digitalsean gave. Notice that the "2011" did not have any zeros added, because it met the minimum width. – Arcege Sep 29 '11 at 16:15
up vote 2 down vote

Let date do the work!

date +%3j

Then, for the milliday part, a simple trick is to compute 1000 plus the number of millidays and strip away the leading 1. So for your script:

s=$(($(date +"%H*3600+%M*60+%S")))
t=$(($s * 5 / 432 + 1000))
date "+%Y'%3j'${t#1}"
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For using date alone, +1, but as shown, 2011-01-01 00:02 produces 2011'001'1 without the zero-padding .... Here is a tweaked variant: t=00$(($(($(date +"%H*3600+%M*60+%S"))) * 5 / 432)); date "+%Y'%3j'${t:(-3)}" – Peter.O Sep 30 '11 at 3:35
@fred Oops, good spot. I've edited to a more portable way of padding the number. – Gilles Sep 30 '11 at 7:20

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