Differentiate between 2 and 02 in bash script

I have a bash script that takes the date, month and year as separate arguments. Using that, a url is constructed which then uses wget to fetch content and store it in an html file (say t.html). Now, the user may enter a 2 digit month/date (as in 02 instead of just 2 and vice-versa). How do I distinguish between the above two formats and correct for this from within the script?

The url works as follows:

Date: a 2 digit input is needed. So a 7 must be supplied as 07 for the url to be constructed properly. Here I am looking for a check that would append a zero to the date in case it is less than 10 and does not already have a zero in front. So, 7 should become 07 for the date field before the url is constructed.

Month: a 2 digit input is needed, but here the url automatically appends the 0 in case month < 10. So, if the user enters 2, then the url forms 02, but if the user enters 02, the url forms 002. Here, the 0 may need to be appended or removed.

P.S: I know this method followed by the url s*c&s, but I just need to work with it.

开发者_C百科Thanks,

Sriram

---

It's a little unclear what you're asking for. If you're only looking to strip one leading zero, then this will do:

month=${month#0}

This will turn 02 into 2, and 12 remains 12.

If you need to remove more than one zero (above 002 will turn into 02), you'll need to do something different, like using regular expressions

while [ -z "${month##0*}" ]; do month=${month#0}; done

or use regular expressions, like with sed

month=$(sed -e 's/^0*//'<<<"$month")

HTH

EDIT
As per your edit; as has already been suggested, use printf

month=$(printf %02d "$month")

this will turn 2 into 02, 12 remains as is, as does 123. If you want to force a two digit number or you don't have printf (which is a shell built-in in bash, and usually available otherwise too, so chances are pretty low), sed can help again

month=$(sed -e 's/.*\(..\)$/\1/'<<<"00$month")

which will prepend two zeros (002) and keep the last two characters (02), turning the empty string into 00 as well. This will turn a into 0a too though. Come to think of it, you don't need sed for this

month="00$month"
month="${month:0-2}"

(The 0- is required to disambiguate from default value expansion)

---

Just use printf.

For example $(printf "%02d" $day) to add the leading zero in case.
And $(printf "%1d" $month) to strip the zero in case.

---

To strip all leading zeros, regardless of the number and regardless of the number of final digits (it fixes "010" for example):

shopt -s extglob
month=009
month=${month##+(0)}
echo "$month"

Demo:

$ for a in 2 02 002 0002 20 020 00200; do echo ${a##+(0)}; done
2
2
2
2
20
20
200

To fix any date:

shopt -s extglob
fixdigits () {
    if (( ${#1} > 2 || ${#1} < 1 ))
    then
        echo "Invalid number of digits"
        return 1
    fi
    printf '%.2d\n' 0${1##+(0)}
}

Demo:

$ for a in 2 02 002 0002 20 020 00200; do fixdigits $a; done
02
02
Invalid number of digits
Invalid number of digits
02
Invalid number of digits
Invalid number of digits

Alternatively:

shopt -s extglob
fixdigits () {
    digits=${1##+(0)}
    if (( ${#digits} > 2 || ${#digits} < 1 ))
    then
        echo "Invalid number of digits"
        return 1
    fi
    printf '%.2s\n' 0$digits
}

Demo:

$ for a in 2 02 002 0002 20 020 00200; do fixdigits $a; done
02
02
02
02
02
02
Invalid number of digits