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How many days until the end of the year

Terminal - How many days until the end of the year
echo "There are $(($(date +%j -d"Dec 31, $(date +%Y)")-$(date +%j))) left in year $(date +%Y)."
2010-02-06 00:15:40
User: unixhome
Functions: date echo
3
How many days until the end of the year

Alternatives

There are 8 alternatives - vote for the best!

Terminal - Alternatives

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What others think

Cute, pointless and too long. You don't need the year in the calculation. "Dec 31" will be assumed to be this year.

echo "There are $(($(date +%j -d"Dec 31")-$(date +%j))) left in year $(date +%Y)."
Comment by flatcap 238 weeks and 3 days ago

It doesn't work for me unless I use - (hypen). This way it doesn't pad the field and works correctly.

echo "There are $(($(date +%j -d"Dec 31")-$(date +%-j))) left in year $(date +%Y)."
Comment by sliceoflinux 238 weeks and 2 days ago

bash: 365-039: value too great for base (error token is "039")

Comment by RanyAlbeg 238 weeks and 1 day ago

Any shell experts? The bash shell chokes on any number ending on 8 or 9. The korn shell does not behave this way. Anyway, you can use the following as a workaround:

echo "There are $(($(date +%j -d"Dec 31") - $(echo $(date +%-j) | sed -e "s/^0*//"))) left in year $(date +%Y)."

or

echo "There are $(expr $(date +%j -d"Dec 31") - $(date +%-j)) left in year $(date +%Y)."

Comment by unixhome 237 weeks and 6 days ago

Ok, found the following in the Bash FAQ:

http://mywiki.wooledge.org/ArithmeticExpression

There is one common pitfall with arithmetic expressions in Bash: numbers with leading zeroes are treated as octal. This causes great confusion among people who are extracting zero-padded numbers from various sources (although dates are by far the most common) and then doing math on them without sanitizing them first. (It's especially bad if you write a program like this in March, test it, roll it out... and then it doesn't blow up until August 1.)

If you have leading-zero problems with Bash's built-in arithmetic, there are two possible solutions. The first is, obviously, to remove the leading zeroes from the numbers before doing math with them. This is not trivial in Bash, unfortunately, because Bash has no ability to perform substitutions on a variable using regular expressions (it can only do it with "glob" patterns).

Comment by unixhome 237 weeks and 6 days ago

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