Quicj Bash Scripting question

Kedelfor · 05-06-2005, 07:05 AM

I am trying to create a script that makes a directory for the day prior to today.

here is the code so far.

#!/bin/bash

#sets up variables
month=`date +'%m'`
day=$((`date +'%d'`- 1))
year=`date +'%y'`

date="$month.$day.$year"

mkdir /home/$date

It works fine, but the problem is that the date needs a zero in front of it, so that all directories have the same amount of characters. ie todays date would be 05.05.05.

I can get a zero in front of the 6 in 2 ways, but once the day is above 11 it gives me 3 characters. Basically i want the variable to have a length of 2 characters.

Right now i can get 05.5.05 as is, but i need it to be 05.05.05. I also need the 05.10.05 to stay like that and not have more than 2 characters in each field. I know it may not make sense, but I am unable to find how to make a variable have to return 2 characters only.

Thanks

trevelluk · 05-06-2005, 07:20 AM

Try something like this, after first setting day, month and year:

Code:

if [ $month -le 9 ]; then
    month="0$month"
fi
# And do the same for the others

Incidentally, have you considered what will happen if you run on the 1st of a month?

Kedelfor · 05-06-2005, 07:22 AM

No i didn't, but thank you for pointing that out. Any idea how I could handle that?

bathory · 05-06-2005, 07:40 AM

Just use this

Code:

day=`date -d yesterday +%d`

for the day

Kedelfor · 05-06-2005, 07:42 AM

I found a good way to do it, but thatnks for your help.

date=`date --date yesterday +'%m.%d.%y'`

This seems to be supported by the gnu version of date. I believe it should be on most if not all linux distros. If I am wrong someone please let me know. Thank you again for your prompt reply.

trevelluk · 05-06-2005, 07:52 AM

OK, something like this (hmm, this is more complicated than I thought actually):

Code:

# Set up inital values using date
.....

# Check for the first of January
if [ $day -eq 1 ] && [ $month -eq 1 ]; then
    day=31
    month=12
    let "year=$year-1"
elif [ $day -eq 1 ]
    let "month=$month-1"
    if [ $month -eq 4 ] || [ $month -eq 6 ] || ...continue for all the months with 30 days... || [ $month -eq 11 ]; then
        $day=30
    elif [ $month=2 ]; then
        # Februrary - WARNING this is wrong in leap years
        $day=28
    else
        $day=31
    fi
fi

# Check for length of variables, etc
...

I hope this is some help - I haven't worked out the leap year problem yet though.

trevelluk · 05-06-2005, 07:53 AM

Doh, after all that typing too! Maybe I should learn to read man pages properly